SlideShare a Scribd company logo
1 of 51
Solving 2nd Degree Equations ax2 + bx + c = 0
Solving 2nd Degree Equations ax2 + bx + c = 0
To solve ax2 + bx + c = 0:
1st: Use the square-root method if the x-term is missing.
Solving 2nd Degree Equations ax2 + bx + c = 0
To solve ax2 + bx + c = 0:
1st: Use the square-root method if the x-term is missing.
2nd: Try factoring it into two binomials.
Solving 2nd Degree Equations ax2 + bx + c = 0
To solve ax2 + bx + c = 0:
1st: Use the square-root method if the x-term is missing.
2nd: Try factoring it into two binomials.
3rd: Use the quadratic formula (QF).
Solving 2nd Degree Equations ax2 + bx + c = 0
The Square-Root Method
To solve ax2 + bx + c = 0:
1st: Use the square-root method if the x-term is missing.
2nd: Try factoring it into two binomials.
3rd: Use the quadratic formula (QF).
Solving 2nd Degree Equations ax2 + bx + c = 0
The Square-Root Method
Use the square-root method if there is no x-term.
To solve ax2 + bx + c = 0:
1st: Use the square-root method if the x-term is missing.
2nd: Try factoring it into two binomials.
3rd: Use the quadratic formula (QF).
Solving 2nd Degree Equations ax2 + bx + c = 0
The Square-Root Method
Use the square-root method if there is no x-term.
I. Solve for the x2 and get the form x2 = d.
To solve ax2 + bx + c = 0:
1st: Use the square-root method if the x-term is missing.
2nd: Try factoring it into two binomials.
3rd: Use the quadratic formula (QF).
Solving 2nd Degree Equations ax2 + bx + c = 0
The Square-Root Method
Use the square-root method if there is no x-term.
I. Solve for the x2 and get the form x2 = d.
II. Then x = Β±οƒ–d are the roots.
To solve ax2 + bx + c = 0:
1st: Use the square-root method if the x-term is missing.
2nd: Try factoring it into two binomials.
3rd: Use the quadratic formula (QF).
Solving 2nd Degree Equations ax2 + bx + c = 0
The Square-Root Method
Use the square-root method if there is no x-term.
I. Solve for the x2 and get the form x2 = d.
II. Then x = Β±οƒ–d are the roots.
Example A. Solve 3x2 – 7 = 0
To solve ax2 + bx + c = 0:
1st: Use the square-root method if the x-term is missing.
2nd: Try factoring it into two binomials.
3rd: Use the quadratic formula (QF).
Solving 2nd Degree Equations ax2 + bx + c = 0
The Square-Root Method
Use the square-root method if there is no x-term.
I. Solve for the x2 and get the form x2 = d.
II. Then x = Β±οƒ–d are the roots.
Example A. Solve 3x2 – 7 = 0
3x2 – 7 = 0 solve for x2
To solve ax2 + bx + c = 0:
1st: Use the square-root method if the x-term is missing.
2nd: Try factoring it into two binomials.
3rd: Use the quadratic formula (QF).
Solving 2nd Degree Equations ax2 + bx + c = 0
The Square-Root Method
Use the square-root method if there is no x-term.
I. Solve for the x2 and get the form x2 = d.
II. Then x = Β±οƒ–d are the roots.
Example A. Solve 3x2 – 7 = 0
3x2 – 7 = 0 solve for x2
3x2 = 7
x2 = 7
3
To solve ax2 + bx + c = 0:
1st: Use the square-root method if the x-term is missing.
2nd: Try factoring it into two binomials.
3rd: Use the quadratic formula (QF).
Solving 2nd Degree Equations ax2 + bx + c = 0
The Square-Root Method
Use the square-root method if there is no x-term.
I. Solve for the x2 and get the form x2 = d.
II. Then x = Β±οƒ–d are the roots.
Example A. Solve 3x2 – 7 = 0
3x2 – 7 = 0 solve for x2
3x2 = 7
x2 = take square root
x = Β±οƒ–7/3
7
3
To solve ax2 + bx + c = 0:
1st: Use the square-root method if the x-term is missing.
2nd: Try factoring it into two binomials.
3rd: Use the quadratic formula (QF).
Solving 2nd Degree Equations ax2 + bx + c = 0
The Square-Root Method
Use the square-root method if there is no x-term.
I. Solve for the x2 and get the form x2 = d
II. Then x = Β±οƒ–d are the roots.
Example A. Solve 3x2 – 7 = 0
3x2 – 7 = 0 solve for x2
3x2 = 7
x2 = take square root
x = Β±οƒ–7/3 ο‚» Β±1.53
7
3
To solve ax2 + bx + c = 0:
1st: Use the square-root method if the x-term is missing.
2nd: Try factoring it into two binomials.
3rd: Use the quadratic formula (QF).
Solving 2nd Degree Equations ax2 + bx + c = 0
The Square-Root Method
Use the square-root method if there is no x-term.
I. Solve for the x2 and get the form x2 = d
II. Then x = Β±οƒ–d are the roots.
Example A. Solve 3x2 – 7 = 0
3x2 – 7 = 0 solve for x2
3x2 = 7
x2 = take square root
x = Β±οƒ–7/3 ο‚» Β±1.53
7
3
exact answers
To solve ax2 + bx + c = 0:
1st: Use the square-root method if the x-term is missing.
2nd: Try factoring it into two binomials.
3rd: Use the quadratic formula (QF).
Solving 2nd Degree Equations ax2 + bx + c = 0
The Square-Root Method
Use the square-root method if there is no x-term.
I. Solve for the x2 and get the form x2 = d
II. Then x = Β±οƒ–d are the roots.
Example A. Solve 3x2 – 7 = 0
3x2 – 7 = 0 solve for x2
3x2 = 7
x2 = take square root
x = Β±οƒ–7/3 ο‚» Β±1.53
7
3
exact answers
approx. answers
To solve ax2 + bx + c = 0:
1st: Use the square-root method if the x-term is missing.
2nd: Try factoring it into two binomials.
3rd: Use the quadratic formula (QF).
Factoring Method
Factor the equation into the form (#x + #)(#x + #) = 0.
Solving 2nd Degree Equations ax2 + bx + c = 0
Factoring Method
Factor the equation into the form (#x + #)(#x + #) = 0.
Hint: Find b2 – 4ac, if it is not a perfect square then the
equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…)
Solving 2nd Degree Equations ax2 + bx + c = 0
Example B. Which of the following equations is factorable?
Solve it by factoring if it is.
a. 3x2 – 2x + 8 = 0
Factoring Method
Factor the equation into the form (#x + #)(#x + #) = 0.
Hint: Find b2 – 4ac, if it is not a perfect square then the
equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…)
Solving 2nd Degree Equations ax2 + bx + c = 0
b. 3x2 – 2x – 8 = 0
Example B. Which of the following equations is factorable?
Solve it by factoring if it is.
a. 3x2 – 2x + 8 = 0
a = 3, b = –2, c =8
Factoring Method
Factor the equation into the form (#x + #)(#x + #) = 0.
Hint: Find b2 – 4ac, if it is not a perfect square then the
equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…)
Solving 2nd Degree Equations ax2 + bx + c = 0
b. 3x2 – 2x – 8 = 0
Example B. Which of the following equations is factorable?
Solve it by factoring if it is.
a. 3x2 – 2x + 8 = 0
a = 3, b = –2, c =8
so b2 – 4ac
= (–2)2 – 4(3)(8)
Factoring Method
Factor the equation into the form (#x + #)(#x + #) = 0.
Hint: Find b2 – 4ac, if it is not a perfect square then the
equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…)
Solving 2nd Degree Equations ax2 + bx + c = 0
b. 3x2 – 2x – 8 = 0
Example B. Which of the following equations is factorable?
Solve it by factoring if it is.
a. 3x2 – 2x + 8 = 0
a = 3, b = –2, c =8
so b2 – 4ac
= (–2)2 – 4(3)(8)
= 4 – 96 = –92
Factoring Method
Factor the equation into the form (#x + #)(#x + #) = 0.
Hint: Find b2 – 4ac, if it is not a perfect square then the
equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…)
Solving 2nd Degree Equations ax2 + bx + c = 0
b. 3x2 – 2x – 8 = 0
Example B. Which of the following equations is factorable?
Solve it by factoring if it is.
a. 3x2 – 2x + 8 = 0
a = 3, b = –2, c =8
so b2 – 4ac
= (–2)2 – 4(3)(8)
= 4 – 96 = –92
Not a perfect square
so it's not factorable
Factoring Method
Factor the equation into the form (#x + #)(#x + #) = 0.
Hint: Find b2 – 4ac, if it is not a perfect square then the
equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…)
Solving 2nd Degree Equations ax2 + bx + c = 0
b. 3x2 – 2x – 8 = 0
Example B. Which of the following equations is factorable?
Solve it by factoring if it is.
a. 3x2 – 2x + 8 = 0
a = 3, b = –2, c =8
so b2 – 4ac
= (–2)2 – 4(3)(8)
= 4 – 96 = –92
Not a perfect square
so it's not factorable
Factoring Method
Factor the equation into the form (#x + #)(#x + #) = 0.
Hint: Find b2 – 4ac, if it is not a perfect square then the
equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…)
Solving 2nd Degree Equations ax2 + bx + c = 0
b. 3x2 – 2x – 8 = 0
a = 3, b = –2, c = –8
Example B. Which of the following equations is factorable?
Solve it by factoring if it is.
a. 3x2 – 2x + 8 = 0
a = 3, b = –2, c =8
so b2 – 4ac
= (–2)2 – 4(3)(8)
= 4 – 96 = –92
Not a perfect square
so it's not factorable
Factoring Method
Factor the equation into the form (#x + #)(#x + #) = 0.
Hint: Find b2 – 4ac, if it is not a perfect square then the
equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…)
Solving 2nd Degree Equations ax2 + bx + c = 0
b. 3x2 – 2x – 8 = 0
a = 3, b = –2, c = –8
b2 – 4ac = (–2)2 – 4(3)(-8)
Example B. Which of the following equations is factorable?
Solve it by factoring if it is.
a. 3x2 – 2x + 8 = 0
a = 3, b = –2, c =8
so b2 – 4ac
= (–2)2 – 4(3)(8)
= 4 – 96 = –92
Not a perfect square
so it's not factorable
Factoring Method
Factor the equation into the form (#x + #)(#x + #) = 0.
Hint: Find b2 – 4ac, if it is not a perfect square then the
equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…)
Solving 2nd Degree Equations ax2 + bx + c = 0
b. 3x2 – 2x – 8 = 0
a = 3, b = –2, c = –8
b2 – 4ac = (–2)2 – 4(3)(-8)
= 4 + 96 = 100
Example B. Which of the following equations is factorable?
Solve it by factoring if it is.
a. 3x2 – 2x + 8 = 0
a = 3, b = –2, c =8
so b2 – 4ac
= (–2)2 – 4(3)(8)
= 4 – 96 = –92
Not a perfect square
so it's not factorable
Factoring Method
Factor the equation into the form (#x + #)(#x + #) = 0.
Hint: Find b2 – 4ac, if it is not a perfect square then the
equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…)
Solving 2nd Degree Equations ax2 + bx + c = 0
b. 3x2 – 2x – 8 = 0
a = 3, b = –2, c = –8
b2 – 4ac = (–2)2 – 4(3)(-8)
= 4 + 96 = 100 = 102
Example B. Which of the following equations is factorable?
Solve it by factoring if it is.
a. 3x2 – 2x + 8 = 0
a = 3, b = –2, c =8
so b2 – 4ac
= (–2)2 – 4(3)(8)
= 4 – 96 = –92
Not a perfect square
so it's not factorable
Factoring Method
Factor the equation into the form (#x + #)(#x + #) = 0.
Hint: Find b2 – 4ac, if it is not a perfect square then the
equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…)
Solving 2nd Degree Equations ax2 + bx + c = 0
b. 3x2 – 2x – 8 = 0
a = 3, b = –2, c = –8
b2 – 4ac = (–2)2 – 4(3)(-8)
= 4 + 96 = 100 = 102
So its factorable.
Example B. Which of the following equations is factorable?
Solve it by factoring if it is.
a. 3x2 – 2x + 8 = 0
a = 3, b = –2, c =8
so b2 – 4ac
= (–2)2 – 4(3)(8)
= 4 – 96 = –92
Not a perfect square
so it's not factorable
Factoring Method
Factor the equation into the form (#x + #)(#x + #) = 0.
Hint: Find b2 – 4ac, if it is not a perfect square then the
equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…)
Solving 2nd Degree Equations ax2 + bx + c = 0
b. 3x2 – 2x – 8 = 0
a = 3, b = –2, c = –8
b2 – 4ac = (–2)2 – 4(3)(-8)
= 4 + 96 = 100 = 102
So its factorable.
3x2 – 2x – 8 = 0
(3x + 4)(x – 2) = 0
Example B. Which of the following equations is factorable?
Solve it by factoring if it is.
a. 3x2 – 2x + 8 = 0
a = 3, b = –2, c =8
so b2 – 4ac
= (–2)2 – 4(3)(8)
= 4 – 96 = –92
Not a perfect square
so it's not factorable
Factoring Method
Factor the equation into the form (#x + #)(#x + #) = 0.
Hint: Find b2 – 4ac, if it is not a perfect square then the
equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…)
Solving 2nd Degree Equations ax2 + bx + c = 0
b. 3x2 – 2x – 8 = 0
a = 3, b = –2, c = –8
b2 – 4ac = (–2)2 – 4(3)(-8)
= 4 + 96 = 100 = 102
So its factorable.
3x2 – 2x – 8 = 0
(3x + 4)(x – 2) = 0
x = -4/3, x = 2
Solving 2nd Degree Equations ax2 + bx + c = 0
Use the quadratic formula (QF)
–bΒ±οƒ– b2 – 4ac
2a
Solving 2nd Degree Equations ax2 + bx + c = 0
Use the quadratic formula (QF)
The roots for the equation ax2 + bx + c = 0 are
x =
–bΒ±οƒ– b2 – 4ac
2a
Solving 2nd Degree Equations ax2 + bx + c = 0
Use the quadratic formula (QF)
The roots for the equation ax2 + bx + c = 0 are
x =
b2 – 4ac is called the discriminant because its value
indicates what type of roots there are.
–bΒ±οƒ– b2 – 4ac
2a
Solving 2nd Degree Equations ax2 + bx + c = 0
Use the quadratic formula (QF)
The roots for the equation ax2 + bx + c = 0 are
x =
b2 – 4ac is called the discriminant because its value
indicates what type of roots there are. Specifically,
if b2 – 4ac is a perfect square, we have fractional roots,
–bΒ±οƒ– b2 – 4ac
2a
Solving 2nd Degree Equations ax2 + bx + c = 0
Use the quadratic formula (QF)
The roots for the equation ax2 + bx + c = 0 are
x =
b2 – 4ac is called the discriminant because its value
indicates what type of roots there are. Specifically,
if b2 – 4ac is a perfect square, we have fractional roots,
if b2 – 4ac < 0 there is no real roots.
–bΒ±οƒ– b2 – 4ac
2a
Solving 2nd Degree Equations ax2 + bx + c = 0
Example C. Solve 3x2 – 2x – 2 = 0
Use the quadratic formula (QF)
The roots for the equation ax2 + bx + c = 0 are
x =
b2 – 4ac is called the discriminant because its value
indicates what type of roots there are. Specifically,
if b2 – 4ac is a perfect square, we have fractional roots,
if b2 – 4ac < 0 there is no real roots.
–bΒ±οƒ– b2 – 4ac
2a
Solving 2nd Degree Equations ax2 + bx + c = 0
Example C. Solve 3x2 – 2x – 2 = 0
Check if it is factorable: a = 3, b = –2, c = –2
Use the quadratic formula (QF)
The roots for the equation ax2 + bx + c = 0 are
x =
b2 – 4ac is called the discriminant because its value
indicates what type of roots there are. Specifically,
if b2 – 4ac is a perfect square, we have fractional roots,
if b2 – 4ac < 0 there is no real roots.
–bΒ±οƒ– b2 – 4ac
2a
Solving 2nd Degree Equations ax2 + bx + c = 0
Example C. Solve 3x2 – 2x – 2 = 0
Check if it is factorable: a = 3, b = –2, c = –2
So b2 – 4ac = (–2)2 – 4(3)(–2) = 28.
Use the quadratic formula (QF)
The roots for the equation ax2 + bx + c = 0 are
x =
b2 – 4ac is called the discriminant because its value
indicates what type of roots there are. Specifically,
if b2 – 4ac is a perfect square, we have fractional roots,
if b2 – 4ac < 0 there is no real roots.
–bΒ±οƒ– b2 – 4ac
2a
Solving 2nd Degree Equations ax2 + bx + c = 0
Example C. Solve 3x2 – 2x – 2 = 0
Check if it is factorable: a = 3, b = –2, c = –2
So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Not factorable!
Use the quadratic formula (QF)
The roots for the equation ax2 + bx + c = 0 are
x =
b2 – 4ac is called the discriminant because its value
indicates what type of roots there are. Specifically,
if b2 – 4ac is a perfect square, we have fractional roots,
if b2 – 4ac < 0 there is no real roots.
–bΒ±οƒ– b2 – 4ac
2a
Solving 2nd Degree Equations ax2 + bx + c = 0
Example C. Solve 3x2 – 2x – 2 = 0
Check if it is factorable: a = 3, b = –2, c = –2
So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Not factorable!
Use QF, we get
x = – (–2) Β± οƒ– 28
2(3)
Use the quadratic formula (QF)
The roots for the equation ax2 + bx + c = 0 are
x =
b2 – 4ac is called the discriminant because its value
indicates what type of roots there are. Specifically,
if b2 – 4ac is a perfect square, we have fractional roots,
if b2 – 4ac < 0 there is no real roots.
–bΒ±οƒ– b2 – 4ac
2a
Solving 2nd Degree Equations ax2 + bx + c = 0
Example C. Solve 3x2 – 2x – 2 = 0
Check if it is factorable: a = 3, b = –2, c = –2
So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Not factorable!
Use QF, we get
x = – (–2) Β± οƒ– 28
2(3)
=
2 Β± οƒ– 28
6
Use the quadratic formula (QF)
The roots for the equation ax2 + bx + c = 0 are
x =
b2 – 4ac is called the discriminant because its value
indicates what type of roots there are. Specifically,
if b2 – 4ac is a perfect square, we have fractional roots,
if b2 – 4ac < 0 there is no real roots.
–bΒ±οƒ– b2 – 4ac
2a
Solving 2nd Degree Equations ax2 + bx + c = 0
Example C. Solve 3x2 – 2x – 2 = 0
Check if it is factorable: a = 3, b = –2, c = –2
So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Not factorable!
Use QF, we get
x = – (–2) Β± οƒ– 28
2(3)
=
2 Β± οƒ– 28
6
=
2 Β± 2οƒ–7
6
Use the quadratic formula (QF)
The roots for the equation ax2 + bx + c = 0 are
x =
b2 – 4ac is called the discriminant because its value
indicates what type of roots there are. Specifically,
if b2 – 4ac is a perfect square, we have fractional roots,
if b2 – 4ac < 0 there is no real roots.
–bΒ±οƒ– b2 – 4ac
2a
Solving 2nd Degree Equations ax2 + bx + c = 0
Example C. Solve 3x2 – 2x – 2 = 0
Check if it is factorable: a = 3, b = –2, c = –2
So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Not factorable!
Use QF, we get
x = – (–2) Β± οƒ– 28
2(3)
=
2 Β± οƒ– 28
6
=
2 Β± 2οƒ–7
6
=
1 Β± οƒ–7
3
Use the quadratic formula (QF)
The roots for the equation ax2 + bx + c = 0 are
x =
b2 – 4ac is called the discriminant because its value
indicates what type of roots there are. Specifically,
if b2 – 4ac is a perfect square, we have fractional roots,
if b2 – 4ac < 0 there is no real roots.
–bΒ±οƒ– b2 – 4ac
2a
Solving 2nd Degree Equations ax2 + bx + c = 0
Example C. Solve 3x2 – 2x – 2 = 0
Check if it is factorable: a = 3, b = –2, c = –2
So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Not factorable!
Use QF, we get
x = – (–2) Β± οƒ– 28
2(3)
=
2 Β± οƒ– 28
6
=
2 Β± 2οƒ–7
6
=
1 Β± οƒ–7
3
ο‚» {1.22
–0.549
Use the quadratic formula (QF)
The roots for the equation ax2 + bx + c = 0 are
x =
b2 – 4ac is called the discriminant because its value
indicates what type of roots there are. Specifically,
if b2 – 4ac is a perfect square, we have fractional roots,
if b2 – 4ac < 0 there is no real roots.
Example D. Solve 3x2 – 2x + 2 = 0
Solving 2nd Degree Equations ax2 + bx + c = 0
Example D. Solve 3x2 – 2x + 2 = 0
To check if it is factorable; a = 3, b = -2, c = -2
Solving 2nd Degree Equations ax2 + bx + c = 0
Example D. Solve 3x2 – 2x + 2 = 0
To check if it is factorable; a = 3, b = -2, c = -2
So b2 – 4ac = (–2)2 – 4(3)(2) = –20,
Solving 2nd Degree Equations ax2 + bx + c = 0
Example D. Solve 3x2 – 2x + 2 = 0
To check if it is factorable; a = 3, b = -2, c = -2
So b2 – 4ac = (–2)2 – 4(3)(2) = –20,
–20 is not a real number, so there is no real solution.
Solving 2nd Degree Equations ax2 + bx + c = 0
Example D. Solve 3x2 – 2x + 2 = 0
To check if it is factorable; a = 3, b = -2, c = -2
So b2 – 4ac = (–2)2 – 4(3)(2) = –20,
–20 is not a real number, so there is no real solution.
Solving 2nd Degree Equations ax2 + bx + c = 0
Solving 2nd Degree Equations ax2 + bx + c = 0
1. 3x2 – 12 = 0
Exercise A. Solve. Use the square root methods.
Give both the exact answers and the numerical answers.
If the answer is not real, state so.
2. 3x2 – 15 = 0 3. 3x2 + 15 = 0
4. x2 – 3 = –x2 + 15 5. –6 = 4x2 – 15 6. (x – 2)2 = 2
7. 4 = (2x – 3)2
Exercise B. Solve by factoring and by the quadratic formula.
14. x2 – 3x = 10 15. x(x – 2) = 24 16. 2x2 = 3(x + 1) – 1
8. x2 – 3x = 4 9. x2 – 15 = 2x 10. x2 + 5x + 12 = 0
11. –x2 – 2x + 8 = 0 12. 9 – 3x – 2x2 = 0 13. 2x2 – x – 1 = 0
Exercise C. Solve by the quadratic formula.
17. x2 – x + 1 = 0 18. x2 – x – 1 = 0
19. x2 – 3x – 2 = 0 20. x2 – 2x + 3 = 0
If the answers are not real numbers, state so.
21. 2x2 – 3x – 1 = 0 22. 3x2 = 2x + 3
Solving 2nd Degree Equations ax2 + bx + c = 0
Exercise C. Solve by the quadratic formula.
If the answers are not real numbers, state so.
23. 2(x2 – 1) + x = 4 24. (x – 1)(x + 1) = 2x(x + 2)
25.
(x – 1)
(x + 2)
=
2x
(x + 2)
26.
(x + 1)
(x + 2)
=
(x + 2)
(2x + 1)
1
Π€ 1 – Π€
1 – Π€
Π€1
Π€ =
27. Cut a stick of unit length (1) into two parts in a manner
such that β€œthe ratio of the whole to the large part is the same
as the ratio of the large part to the small part”
In picture,
the golden ratio Π€ (phi) is the ratio 1 : Π€. Find the exact and
approximate value of Π€. (Google β€œgolden ratio” )
Solving 2nd Degree Equations ax2 + bx + c = 0
HW answers
1. x = Β±2 3. no real answers 5. x = Β±3/2
7. x = Β½, 5/2 9. x = –3, 5 11. x = –4, 2
17. no real answer 19. x =
13. x = –1/2, 1 15. x = 6, –4
3 ±√17
2 21. x =
3 ±√17
4
23. x = –2, 3
27. Π€ = –1 +√5
2
25. x = –1, –2
approx. 0.618

More Related Content

What's hot

8 inequalities and sign charts x
8 inequalities and sign charts x8 inequalities and sign charts x
8 inequalities and sign charts xmath260
Β 
Mathematics 9 Quadratic Functions (Module 1)
Mathematics 9 Quadratic Functions (Module 1)Mathematics 9 Quadratic Functions (Module 1)
Mathematics 9 Quadratic Functions (Module 1)Juan Miguel Palero
Β 
2.4 grapgs of second degree functions
2.4 grapgs of second degree functions2.4 grapgs of second degree functions
2.4 grapgs of second degree functionsmath260
Β 
Module 3 quadratic functions
Module 3   quadratic functionsModule 3   quadratic functions
Module 3 quadratic functionsdionesioable
Β 
Add maths module form 4 & 5
Add maths module form 4 & 5Add maths module form 4 & 5
Add maths module form 4 & 5smktsj2
Β 
0101: Graphing Quadratic Functions
0101: Graphing Quadratic Functions0101: Graphing Quadratic Functions
0101: Graphing Quadratic Functionskijo13
Β 
Additional Mathematics form 4 (formula)
Additional Mathematics form 4 (formula)Additional Mathematics form 4 (formula)
Additional Mathematics form 4 (formula)Fatini Adnan
Β 
class-X Maths Book | Chapter-2 | Slides By MANAV |
class-X Maths Book | Chapter-2 | Slides By MANAV |class-X Maths Book | Chapter-2 | Slides By MANAV |
class-X Maths Book | Chapter-2 | Slides By MANAV |Slides With MANAV
Β 
19 more parabolas a&amp; hyperbolas (optional) x
19 more parabolas a&amp; hyperbolas (optional) x19 more parabolas a&amp; hyperbolas (optional) x
19 more parabolas a&amp; hyperbolas (optional) xmath260
Β 
Higher Maths 2.1.2 - Quadratic Functions
Higher Maths 2.1.2 - Quadratic FunctionsHigher Maths 2.1.2 - Quadratic Functions
Higher Maths 2.1.2 - Quadratic Functionstimschmitz
Β 
2.9 graphs of factorable polynomials
2.9 graphs of factorable polynomials2.9 graphs of factorable polynomials
2.9 graphs of factorable polynomialsmath260
Β 
10 rectangular coordinate system x
10 rectangular coordinate system x10 rectangular coordinate system x
10 rectangular coordinate system xmath260
Β 
Algebra Presentation on Topic Modulus Function and Polynomials
Algebra Presentation on Topic Modulus Function and PolynomialsAlgebra Presentation on Topic Modulus Function and Polynomials
Algebra Presentation on Topic Modulus Function and PolynomialsKanyaJyesta1
Β 
Linear ineqns. and statistics
Linear ineqns. and statisticsLinear ineqns. and statistics
Linear ineqns. and statisticsindu psthakur
Β 
1.3 sign charts and inequalities
1.3 sign charts and inequalities1.3 sign charts and inequalities
1.3 sign charts and inequalitiesmath123c
Β 
Pptpersamaankuadrat 150205080445-conversion-gate02
Pptpersamaankuadrat 150205080445-conversion-gate02Pptpersamaankuadrat 150205080445-conversion-gate02
Pptpersamaankuadrat 150205080445-conversion-gate02MasfuahFuah
Β 
Module 1 quadratic functions
Module 1   quadratic functionsModule 1   quadratic functions
Module 1 quadratic functionsdionesioable
Β 
Form 5 Additional Maths Note
Form 5 Additional Maths NoteForm 5 Additional Maths Note
Form 5 Additional Maths NoteChek Wei Tan
Β 
1.5 notation and algebra of functions
1.5 notation and algebra of functions1.5 notation and algebra of functions
1.5 notation and algebra of functionsmath123c
Β 
Solving linear & quadratic equations
Solving linear & quadratic equationsSolving linear & quadratic equations
Solving linear & quadratic equationsErlyn Geronimo
Β 

What's hot (20)

8 inequalities and sign charts x
8 inequalities and sign charts x8 inequalities and sign charts x
8 inequalities and sign charts x
Β 
Mathematics 9 Quadratic Functions (Module 1)
Mathematics 9 Quadratic Functions (Module 1)Mathematics 9 Quadratic Functions (Module 1)
Mathematics 9 Quadratic Functions (Module 1)
Β 
2.4 grapgs of second degree functions
2.4 grapgs of second degree functions2.4 grapgs of second degree functions
2.4 grapgs of second degree functions
Β 
Module 3 quadratic functions
Module 3   quadratic functionsModule 3   quadratic functions
Module 3 quadratic functions
Β 
Add maths module form 4 & 5
Add maths module form 4 & 5Add maths module form 4 & 5
Add maths module form 4 & 5
Β 
0101: Graphing Quadratic Functions
0101: Graphing Quadratic Functions0101: Graphing Quadratic Functions
0101: Graphing Quadratic Functions
Β 
Additional Mathematics form 4 (formula)
Additional Mathematics form 4 (formula)Additional Mathematics form 4 (formula)
Additional Mathematics form 4 (formula)
Β 
class-X Maths Book | Chapter-2 | Slides By MANAV |
class-X Maths Book | Chapter-2 | Slides By MANAV |class-X Maths Book | Chapter-2 | Slides By MANAV |
class-X Maths Book | Chapter-2 | Slides By MANAV |
Β 
19 more parabolas a&amp; hyperbolas (optional) x
19 more parabolas a&amp; hyperbolas (optional) x19 more parabolas a&amp; hyperbolas (optional) x
19 more parabolas a&amp; hyperbolas (optional) x
Β 
Higher Maths 2.1.2 - Quadratic Functions
Higher Maths 2.1.2 - Quadratic FunctionsHigher Maths 2.1.2 - Quadratic Functions
Higher Maths 2.1.2 - Quadratic Functions
Β 
2.9 graphs of factorable polynomials
2.9 graphs of factorable polynomials2.9 graphs of factorable polynomials
2.9 graphs of factorable polynomials
Β 
10 rectangular coordinate system x
10 rectangular coordinate system x10 rectangular coordinate system x
10 rectangular coordinate system x
Β 
Algebra Presentation on Topic Modulus Function and Polynomials
Algebra Presentation on Topic Modulus Function and PolynomialsAlgebra Presentation on Topic Modulus Function and Polynomials
Algebra Presentation on Topic Modulus Function and Polynomials
Β 
Linear ineqns. and statistics
Linear ineqns. and statisticsLinear ineqns. and statistics
Linear ineqns. and statistics
Β 
1.3 sign charts and inequalities
1.3 sign charts and inequalities1.3 sign charts and inequalities
1.3 sign charts and inequalities
Β 
Pptpersamaankuadrat 150205080445-conversion-gate02
Pptpersamaankuadrat 150205080445-conversion-gate02Pptpersamaankuadrat 150205080445-conversion-gate02
Pptpersamaankuadrat 150205080445-conversion-gate02
Β 
Module 1 quadratic functions
Module 1   quadratic functionsModule 1   quadratic functions
Module 1 quadratic functions
Β 
Form 5 Additional Maths Note
Form 5 Additional Maths NoteForm 5 Additional Maths Note
Form 5 Additional Maths Note
Β 
1.5 notation and algebra of functions
1.5 notation and algebra of functions1.5 notation and algebra of functions
1.5 notation and algebra of functions
Β 
Solving linear & quadratic equations
Solving linear & quadratic equationsSolving linear & quadratic equations
Solving linear & quadratic equations
Β 

Similar to 1.1 review solving 2nd degree equations

5 2nd degree equations and the quadratic formula
5 2nd degree equations and the quadratic formula5 2nd degree equations and the quadratic formula
5 2nd degree equations and the quadratic formulaTzenma
Β 
Quadratic Equation
Quadratic EquationQuadratic Equation
Quadratic EquationNayanKohare
Β 
C2 st lecture 2 handout
C2 st lecture 2 handoutC2 st lecture 2 handout
C2 st lecture 2 handoutfatima d
Β 
Solving Quadratic Equations
Solving Quadratic EquationsSolving Quadratic Equations
Solving Quadratic Equationsdowne1mf
Β 
Quadratic equation
Quadratic equation Quadratic equation
Quadratic equation Shivangi Tidke
Β 
March 26, 2015
March 26, 2015March 26, 2015
March 26, 2015khyps13
Β 
Quadratic equations
Quadratic equationsQuadratic equations
Quadratic equationsA M
Β 
Quadratic equations
Quadratic equationsQuadratic equations
Quadratic equationsMervin Dayrit
Β 
Sample fin
Sample finSample fin
Sample finmath123a
Β 
April 1, 2014
April 1, 2014April 1, 2014
April 1, 2014khyps13
Β 
March 27, 2015
March 27, 2015March 27, 2015
March 27, 2015khyps13
Β 
presentation_quadraticequations-111211090004-phpapp02_1524500815_313961.pptx
presentation_quadraticequations-111211090004-phpapp02_1524500815_313961.pptxpresentation_quadraticequations-111211090004-phpapp02_1524500815_313961.pptx
presentation_quadraticequations-111211090004-phpapp02_1524500815_313961.pptxDeepNavi2
Β 
sim-140907230908-phpapp01.pptx
sim-140907230908-phpapp01.pptxsim-140907230908-phpapp01.pptx
sim-140907230908-phpapp01.pptxJeffreyEnriquez10
Β 
quadraticequations-111211090004-phpapp02 (1).pdf
quadraticequations-111211090004-phpapp02 (1).pdfquadraticequations-111211090004-phpapp02 (1).pdf
quadraticequations-111211090004-phpapp02 (1).pdfNehaJain840096
Β 
MIT Math Syllabus 10-3 Lesson 7: Quadratic equations
MIT Math Syllabus 10-3 Lesson 7: Quadratic equationsMIT Math Syllabus 10-3 Lesson 7: Quadratic equations
MIT Math Syllabus 10-3 Lesson 7: Quadratic equationsLawrence De Vera
Β 
Lecture quadratic equations good one
Lecture quadratic equations good oneLecture quadratic equations good one
Lecture quadratic equations good oneHazel Joy Chong
Β 

Similar to 1.1 review solving 2nd degree equations (20)

5 2nd degree equations and the quadratic formula
5 2nd degree equations and the quadratic formula5 2nd degree equations and the quadratic formula
5 2nd degree equations and the quadratic formula
Β 
Quadratic Equation
Quadratic EquationQuadratic Equation
Quadratic Equation
Β 
C2 st lecture 2 handout
C2 st lecture 2 handoutC2 st lecture 2 handout
C2 st lecture 2 handout
Β 
Solving Quadratic Equations
Solving Quadratic EquationsSolving Quadratic Equations
Solving Quadratic Equations
Β 
Quadratic Equations
Quadratic EquationsQuadratic Equations
Quadratic Equations
Β 
Quadratic equation
Quadratic equation Quadratic equation
Quadratic equation
Β 
Chithra
ChithraChithra
Chithra
Β 
March 26, 2015
March 26, 2015March 26, 2015
March 26, 2015
Β 
Quadratic equations
Quadratic equationsQuadratic equations
Quadratic equations
Β 
QUADRATIC.pptx
QUADRATIC.pptxQUADRATIC.pptx
QUADRATIC.pptx
Β 
Quadratic equations
Quadratic equationsQuadratic equations
Quadratic equations
Β 
Sample fin
Sample finSample fin
Sample fin
Β 
April 1, 2014
April 1, 2014April 1, 2014
April 1, 2014
Β 
March 27, 2015
March 27, 2015March 27, 2015
March 27, 2015
Β 
presentation_quadraticequations-111211090004-phpapp02_1524500815_313961.pptx
presentation_quadraticequations-111211090004-phpapp02_1524500815_313961.pptxpresentation_quadraticequations-111211090004-phpapp02_1524500815_313961.pptx
presentation_quadraticequations-111211090004-phpapp02_1524500815_313961.pptx
Β 
sim-140907230908-phpapp01.pptx
sim-140907230908-phpapp01.pptxsim-140907230908-phpapp01.pptx
sim-140907230908-phpapp01.pptx
Β 
Factoring2
Factoring2Factoring2
Factoring2
Β 
quadraticequations-111211090004-phpapp02 (1).pdf
quadraticequations-111211090004-phpapp02 (1).pdfquadraticequations-111211090004-phpapp02 (1).pdf
quadraticequations-111211090004-phpapp02 (1).pdf
Β 
MIT Math Syllabus 10-3 Lesson 7: Quadratic equations
MIT Math Syllabus 10-3 Lesson 7: Quadratic equationsMIT Math Syllabus 10-3 Lesson 7: Quadratic equations
MIT Math Syllabus 10-3 Lesson 7: Quadratic equations
Β 
Lecture quadratic equations good one
Lecture quadratic equations good oneLecture quadratic equations good one
Lecture quadratic equations good one
Β 

More from math123c

0. exponents y
0. exponents y0. exponents y
0. exponents ymath123c
Β 
123c test 4 review b
123c test  4 review b123c test  4 review b
123c test 4 review bmath123c
Β 
6 binomial theorem
6 binomial theorem6 binomial theorem
6 binomial theoremmath123c
Β 
5.5 permutations and combinations
5.5 permutations and combinations5.5 permutations and combinations
5.5 permutations and combinationsmath123c
Β 
5.4 trees and factorials
5.4 trees and factorials5.4 trees and factorials
5.4 trees and factorialsmath123c
Β 
5.3 geometric sequences
5.3 geometric sequences5.3 geometric sequences
5.3 geometric sequencesmath123c
Β 
5.2 arithmetic sequences
5.2 arithmetic sequences5.2 arithmetic sequences
5.2 arithmetic sequencesmath123c
Β 
5.1 sequences
5.1 sequences5.1 sequences
5.1 sequencesmath123c
Β 
4.5 matrix notation
4.5 matrix notation4.5 matrix notation
4.5 matrix notationmath123c
Β 
4.4 system of linear equations 2
4.4 system of linear equations 24.4 system of linear equations 2
4.4 system of linear equations 2math123c
Β 
4.3 system of linear equations 1
4.3 system of linear equations 14.3 system of linear equations 1
4.3 system of linear equations 1math123c
Β 
4.2 stem parabolas revisited
4.2 stem parabolas revisited4.2 stem parabolas revisited
4.2 stem parabolas revisitedmath123c
Β 
4.1 stem hyperbolas
4.1 stem hyperbolas4.1 stem hyperbolas
4.1 stem hyperbolasmath123c
Β 
3.4 ellipses
3.4 ellipses3.4 ellipses
3.4 ellipsesmath123c
Β 
3.3 conic sections circles
3.3 conic sections circles3.3 conic sections circles
3.3 conic sections circlesmath123c
Β 
3.2 more on log and exponential equations
3.2 more on log and exponential equations3.2 more on log and exponential equations
3.2 more on log and exponential equationsmath123c
Β 
3.1 properties of logarithm
3.1 properties of logarithm3.1 properties of logarithm
3.1 properties of logarithmmath123c
Β 
2.5 calculation with log and exp
2.5 calculation with log and exp2.5 calculation with log and exp
2.5 calculation with log and expmath123c
Β 
2.4 introduction to logarithm
2.4 introduction to logarithm2.4 introduction to logarithm
2.4 introduction to logarithmmath123c
Β 
2.3 continuous compound interests
2.3 continuous compound interests2.3 continuous compound interests
2.3 continuous compound interestsmath123c
Β 

More from math123c (20)

0. exponents y
0. exponents y0. exponents y
0. exponents y
Β 
123c test 4 review b
123c test  4 review b123c test  4 review b
123c test 4 review b
Β 
6 binomial theorem
6 binomial theorem6 binomial theorem
6 binomial theorem
Β 
5.5 permutations and combinations
5.5 permutations and combinations5.5 permutations and combinations
5.5 permutations and combinations
Β 
5.4 trees and factorials
5.4 trees and factorials5.4 trees and factorials
5.4 trees and factorials
Β 
5.3 geometric sequences
5.3 geometric sequences5.3 geometric sequences
5.3 geometric sequences
Β 
5.2 arithmetic sequences
5.2 arithmetic sequences5.2 arithmetic sequences
5.2 arithmetic sequences
Β 
5.1 sequences
5.1 sequences5.1 sequences
5.1 sequences
Β 
4.5 matrix notation
4.5 matrix notation4.5 matrix notation
4.5 matrix notation
Β 
4.4 system of linear equations 2
4.4 system of linear equations 24.4 system of linear equations 2
4.4 system of linear equations 2
Β 
4.3 system of linear equations 1
4.3 system of linear equations 14.3 system of linear equations 1
4.3 system of linear equations 1
Β 
4.2 stem parabolas revisited
4.2 stem parabolas revisited4.2 stem parabolas revisited
4.2 stem parabolas revisited
Β 
4.1 stem hyperbolas
4.1 stem hyperbolas4.1 stem hyperbolas
4.1 stem hyperbolas
Β 
3.4 ellipses
3.4 ellipses3.4 ellipses
3.4 ellipses
Β 
3.3 conic sections circles
3.3 conic sections circles3.3 conic sections circles
3.3 conic sections circles
Β 
3.2 more on log and exponential equations
3.2 more on log and exponential equations3.2 more on log and exponential equations
3.2 more on log and exponential equations
Β 
3.1 properties of logarithm
3.1 properties of logarithm3.1 properties of logarithm
3.1 properties of logarithm
Β 
2.5 calculation with log and exp
2.5 calculation with log and exp2.5 calculation with log and exp
2.5 calculation with log and exp
Β 
2.4 introduction to logarithm
2.4 introduction to logarithm2.4 introduction to logarithm
2.4 introduction to logarithm
Β 
2.3 continuous compound interests
2.3 continuous compound interests2.3 continuous compound interests
2.3 continuous compound interests
Β 

Recently uploaded

#StandardsGoals for 2024: What’s new for BISAC - Tech Forum 2024
#StandardsGoals for 2024: What’s new for BISAC - Tech Forum 2024#StandardsGoals for 2024: What’s new for BISAC - Tech Forum 2024
#StandardsGoals for 2024: What’s new for BISAC - Tech Forum 2024BookNet Canada
Β 
My Hashitalk Indonesia April 2024 Presentation
My Hashitalk Indonesia April 2024 PresentationMy Hashitalk Indonesia April 2024 Presentation
My Hashitalk Indonesia April 2024 PresentationRidwan Fadjar
Β 
Factors to Consider When Choosing Accounts Payable Services Providers.pptx
Factors to Consider When Choosing Accounts Payable Services Providers.pptxFactors to Consider When Choosing Accounts Payable Services Providers.pptx
Factors to Consider When Choosing Accounts Payable Services Providers.pptxKatpro Technologies
Β 
Transcript: #StandardsGoals for 2024: What’s new for BISAC - Tech Forum 2024
Transcript: #StandardsGoals for 2024: What’s new for BISAC - Tech Forum 2024Transcript: #StandardsGoals for 2024: What’s new for BISAC - Tech Forum 2024
Transcript: #StandardsGoals for 2024: What’s new for BISAC - Tech Forum 2024BookNet Canada
Β 
Tech-Forward - Achieving Business Readiness For Copilot in Microsoft 365
Tech-Forward - Achieving Business Readiness For Copilot in Microsoft 365Tech-Forward - Achieving Business Readiness For Copilot in Microsoft 365
Tech-Forward - Achieving Business Readiness For Copilot in Microsoft 3652toLead Limited
Β 
Beyond Boundaries: Leveraging No-Code Solutions for Industry Innovation
Beyond Boundaries: Leveraging No-Code Solutions for Industry InnovationBeyond Boundaries: Leveraging No-Code Solutions for Industry Innovation
Beyond Boundaries: Leveraging No-Code Solutions for Industry InnovationSafe Software
Β 
Kotlin Multiplatform & Compose Multiplatform - Starter kit for pragmatics
Kotlin Multiplatform & Compose Multiplatform - Starter kit for pragmaticsKotlin Multiplatform & Compose Multiplatform - Starter kit for pragmatics
Kotlin Multiplatform & Compose Multiplatform - Starter kit for pragmaticscarlostorres15106
Β 
The Codex of Business Writing Software for Real-World Solutions 2.pptx
The Codex of Business Writing Software for Real-World Solutions 2.pptxThe Codex of Business Writing Software for Real-World Solutions 2.pptx
The Codex of Business Writing Software for Real-World Solutions 2.pptxMalak Abu Hammad
Β 
Breaking the Kubernetes Kill Chain: Host Path Mount
Breaking the Kubernetes Kill Chain: Host Path MountBreaking the Kubernetes Kill Chain: Host Path Mount
Breaking the Kubernetes Kill Chain: Host Path MountPuma Security, LLC
Β 
IAC 2024 - IA Fast Track to Search Focused AI Solutions
IAC 2024 - IA Fast Track to Search Focused AI SolutionsIAC 2024 - IA Fast Track to Search Focused AI Solutions
IAC 2024 - IA Fast Track to Search Focused AI SolutionsEnterprise Knowledge
Β 
Azure Monitor & Application Insight to monitor Infrastructure & Application
Azure Monitor & Application Insight to monitor Infrastructure & ApplicationAzure Monitor & Application Insight to monitor Infrastructure & Application
Azure Monitor & Application Insight to monitor Infrastructure & ApplicationAndikSusilo4
Β 
Integration and Automation in Practice: CI/CD in MuleΒ Integration and Automat...
Integration and Automation in Practice: CI/CD in MuleΒ Integration and Automat...Integration and Automation in Practice: CI/CD in MuleΒ Integration and Automat...
Integration and Automation in Practice: CI/CD in MuleΒ Integration and Automat...Patryk Bandurski
Β 
Injustice - Developers Among Us (SciFiDevCon 2024)
Injustice - Developers Among Us (SciFiDevCon 2024)Injustice - Developers Among Us (SciFiDevCon 2024)
Injustice - Developers Among Us (SciFiDevCon 2024)Allon Mureinik
Β 
Unblocking The Main Thread Solving ANRs and Frozen Frames
Unblocking The Main Thread Solving ANRs and Frozen FramesUnblocking The Main Thread Solving ANRs and Frozen Frames
Unblocking The Main Thread Solving ANRs and Frozen FramesSinan KOZAK
Β 
FULL ENJOY πŸ” 8264348440 πŸ” Call Girls in Diplomatic Enclave | Delhi
FULL ENJOY πŸ” 8264348440 πŸ” Call Girls in Diplomatic Enclave | DelhiFULL ENJOY πŸ” 8264348440 πŸ” Call Girls in Diplomatic Enclave | Delhi
FULL ENJOY πŸ” 8264348440 πŸ” Call Girls in Diplomatic Enclave | Delhisoniya singh
Β 
Neo4j - How KGs are shaping the future of Generative AI at AWS Summit London ...
Neo4j - How KGs are shaping the future of Generative AI at AWS Summit London ...Neo4j - How KGs are shaping the future of Generative AI at AWS Summit London ...
Neo4j - How KGs are shaping the future of Generative AI at AWS Summit London ...Neo4j
Β 
08448380779 Call Girls In Civil Lines Women Seeking Men
08448380779 Call Girls In Civil Lines Women Seeking Men08448380779 Call Girls In Civil Lines Women Seeking Men
08448380779 Call Girls In Civil Lines Women Seeking MenDelhi Call girls
Β 
The 7 Things I Know About Cyber Security After 25 Years | April 2024
The 7 Things I Know About Cyber Security After 25 Years | April 2024The 7 Things I Know About Cyber Security After 25 Years | April 2024
The 7 Things I Know About Cyber Security After 25 Years | April 2024Rafal Los
Β 
A Domino Admins Adventures (Engage 2024)
A Domino Admins Adventures (Engage 2024)A Domino Admins Adventures (Engage 2024)
A Domino Admins Adventures (Engage 2024)Gabriella Davis
Β 
From Event to Action: Accelerate Your Decision Making with Real-Time Automation
From Event to Action: Accelerate Your Decision Making with Real-Time AutomationFrom Event to Action: Accelerate Your Decision Making with Real-Time Automation
From Event to Action: Accelerate Your Decision Making with Real-Time AutomationSafe Software
Β 

Recently uploaded (20)

#StandardsGoals for 2024: What’s new for BISAC - Tech Forum 2024
#StandardsGoals for 2024: What’s new for BISAC - Tech Forum 2024#StandardsGoals for 2024: What’s new for BISAC - Tech Forum 2024
#StandardsGoals for 2024: What’s new for BISAC - Tech Forum 2024
Β 
My Hashitalk Indonesia April 2024 Presentation
My Hashitalk Indonesia April 2024 PresentationMy Hashitalk Indonesia April 2024 Presentation
My Hashitalk Indonesia April 2024 Presentation
Β 
Factors to Consider When Choosing Accounts Payable Services Providers.pptx
Factors to Consider When Choosing Accounts Payable Services Providers.pptxFactors to Consider When Choosing Accounts Payable Services Providers.pptx
Factors to Consider When Choosing Accounts Payable Services Providers.pptx
Β 
Transcript: #StandardsGoals for 2024: What’s new for BISAC - Tech Forum 2024
Transcript: #StandardsGoals for 2024: What’s new for BISAC - Tech Forum 2024Transcript: #StandardsGoals for 2024: What’s new for BISAC - Tech Forum 2024
Transcript: #StandardsGoals for 2024: What’s new for BISAC - Tech Forum 2024
Β 
Tech-Forward - Achieving Business Readiness For Copilot in Microsoft 365
Tech-Forward - Achieving Business Readiness For Copilot in Microsoft 365Tech-Forward - Achieving Business Readiness For Copilot in Microsoft 365
Tech-Forward - Achieving Business Readiness For Copilot in Microsoft 365
Β 
Beyond Boundaries: Leveraging No-Code Solutions for Industry Innovation
Beyond Boundaries: Leveraging No-Code Solutions for Industry InnovationBeyond Boundaries: Leveraging No-Code Solutions for Industry Innovation
Beyond Boundaries: Leveraging No-Code Solutions for Industry Innovation
Β 
Kotlin Multiplatform & Compose Multiplatform - Starter kit for pragmatics
Kotlin Multiplatform & Compose Multiplatform - Starter kit for pragmaticsKotlin Multiplatform & Compose Multiplatform - Starter kit for pragmatics
Kotlin Multiplatform & Compose Multiplatform - Starter kit for pragmatics
Β 
The Codex of Business Writing Software for Real-World Solutions 2.pptx
The Codex of Business Writing Software for Real-World Solutions 2.pptxThe Codex of Business Writing Software for Real-World Solutions 2.pptx
The Codex of Business Writing Software for Real-World Solutions 2.pptx
Β 
Breaking the Kubernetes Kill Chain: Host Path Mount
Breaking the Kubernetes Kill Chain: Host Path MountBreaking the Kubernetes Kill Chain: Host Path Mount
Breaking the Kubernetes Kill Chain: Host Path Mount
Β 
IAC 2024 - IA Fast Track to Search Focused AI Solutions
IAC 2024 - IA Fast Track to Search Focused AI SolutionsIAC 2024 - IA Fast Track to Search Focused AI Solutions
IAC 2024 - IA Fast Track to Search Focused AI Solutions
Β 
Azure Monitor & Application Insight to monitor Infrastructure & Application
Azure Monitor & Application Insight to monitor Infrastructure & ApplicationAzure Monitor & Application Insight to monitor Infrastructure & Application
Azure Monitor & Application Insight to monitor Infrastructure & Application
Β 
Integration and Automation in Practice: CI/CD in MuleΒ Integration and Automat...
Integration and Automation in Practice: CI/CD in MuleΒ Integration and Automat...Integration and Automation in Practice: CI/CD in MuleΒ Integration and Automat...
Integration and Automation in Practice: CI/CD in MuleΒ Integration and Automat...
Β 
Injustice - Developers Among Us (SciFiDevCon 2024)
Injustice - Developers Among Us (SciFiDevCon 2024)Injustice - Developers Among Us (SciFiDevCon 2024)
Injustice - Developers Among Us (SciFiDevCon 2024)
Β 
Unblocking The Main Thread Solving ANRs and Frozen Frames
Unblocking The Main Thread Solving ANRs and Frozen FramesUnblocking The Main Thread Solving ANRs and Frozen Frames
Unblocking The Main Thread Solving ANRs and Frozen Frames
Β 
FULL ENJOY πŸ” 8264348440 πŸ” Call Girls in Diplomatic Enclave | Delhi
FULL ENJOY πŸ” 8264348440 πŸ” Call Girls in Diplomatic Enclave | DelhiFULL ENJOY πŸ” 8264348440 πŸ” Call Girls in Diplomatic Enclave | Delhi
FULL ENJOY πŸ” 8264348440 πŸ” Call Girls in Diplomatic Enclave | Delhi
Β 
Neo4j - How KGs are shaping the future of Generative AI at AWS Summit London ...
Neo4j - How KGs are shaping the future of Generative AI at AWS Summit London ...Neo4j - How KGs are shaping the future of Generative AI at AWS Summit London ...
Neo4j - How KGs are shaping the future of Generative AI at AWS Summit London ...
Β 
08448380779 Call Girls In Civil Lines Women Seeking Men
08448380779 Call Girls In Civil Lines Women Seeking Men08448380779 Call Girls In Civil Lines Women Seeking Men
08448380779 Call Girls In Civil Lines Women Seeking Men
Β 
The 7 Things I Know About Cyber Security After 25 Years | April 2024
The 7 Things I Know About Cyber Security After 25 Years | April 2024The 7 Things I Know About Cyber Security After 25 Years | April 2024
The 7 Things I Know About Cyber Security After 25 Years | April 2024
Β 
A Domino Admins Adventures (Engage 2024)
A Domino Admins Adventures (Engage 2024)A Domino Admins Adventures (Engage 2024)
A Domino Admins Adventures (Engage 2024)
Β 
From Event to Action: Accelerate Your Decision Making with Real-Time Automation
From Event to Action: Accelerate Your Decision Making with Real-Time AutomationFrom Event to Action: Accelerate Your Decision Making with Real-Time Automation
From Event to Action: Accelerate Your Decision Making with Real-Time Automation
Β 

1.1 review solving 2nd degree equations

  • 1. Solving 2nd Degree Equations ax2 + bx + c = 0
  • 2. Solving 2nd Degree Equations ax2 + bx + c = 0 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing.
  • 3. Solving 2nd Degree Equations ax2 + bx + c = 0 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials.
  • 4. Solving 2nd Degree Equations ax2 + bx + c = 0 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
  • 5. Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
  • 6. Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
  • 7. Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d. To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
  • 8. Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d. II. Then x = Β±οƒ–d are the roots. To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
  • 9. Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d. II. Then x = Β±οƒ–d are the roots. Example A. Solve 3x2 – 7 = 0 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
  • 10. Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d. II. Then x = Β±οƒ–d are the roots. Example A. Solve 3x2 – 7 = 0 3x2 – 7 = 0 solve for x2 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
  • 11. Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d. II. Then x = Β±οƒ–d are the roots. Example A. Solve 3x2 – 7 = 0 3x2 – 7 = 0 solve for x2 3x2 = 7 x2 = 7 3 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
  • 12. Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d. II. Then x = Β±οƒ–d are the roots. Example A. Solve 3x2 – 7 = 0 3x2 – 7 = 0 solve for x2 3x2 = 7 x2 = take square root x = Β±οƒ–7/3 7 3 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
  • 13. Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d II. Then x = Β±οƒ–d are the roots. Example A. Solve 3x2 – 7 = 0 3x2 – 7 = 0 solve for x2 3x2 = 7 x2 = take square root x = Β±οƒ–7/3 ο‚» Β±1.53 7 3 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
  • 14. Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d II. Then x = Β±οƒ–d are the roots. Example A. Solve 3x2 – 7 = 0 3x2 – 7 = 0 solve for x2 3x2 = 7 x2 = take square root x = Β±οƒ–7/3 ο‚» Β±1.53 7 3 exact answers To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
  • 15. Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d II. Then x = Β±οƒ–d are the roots. Example A. Solve 3x2 – 7 = 0 3x2 – 7 = 0 solve for x2 3x2 = 7 x2 = take square root x = Β±οƒ–7/3 ο‚» Β±1.53 7 3 exact answers approx. answers To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
  • 16. Factoring Method Factor the equation into the form (#x + #)(#x + #) = 0. Solving 2nd Degree Equations ax2 + bx + c = 0
  • 17. Factoring Method Factor the equation into the form (#x + #)(#x + #) = 0. Hint: Find b2 – 4ac, if it is not a perfect square then the equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…) Solving 2nd Degree Equations ax2 + bx + c = 0
  • 18. Example B. Which of the following equations is factorable? Solve it by factoring if it is. a. 3x2 – 2x + 8 = 0 Factoring Method Factor the equation into the form (#x + #)(#x + #) = 0. Hint: Find b2 – 4ac, if it is not a perfect square then the equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…) Solving 2nd Degree Equations ax2 + bx + c = 0 b. 3x2 – 2x – 8 = 0
  • 19. Example B. Which of the following equations is factorable? Solve it by factoring if it is. a. 3x2 – 2x + 8 = 0 a = 3, b = –2, c =8 Factoring Method Factor the equation into the form (#x + #)(#x + #) = 0. Hint: Find b2 – 4ac, if it is not a perfect square then the equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…) Solving 2nd Degree Equations ax2 + bx + c = 0 b. 3x2 – 2x – 8 = 0
  • 20. Example B. Which of the following equations is factorable? Solve it by factoring if it is. a. 3x2 – 2x + 8 = 0 a = 3, b = –2, c =8 so b2 – 4ac = (–2)2 – 4(3)(8) Factoring Method Factor the equation into the form (#x + #)(#x + #) = 0. Hint: Find b2 – 4ac, if it is not a perfect square then the equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…) Solving 2nd Degree Equations ax2 + bx + c = 0 b. 3x2 – 2x – 8 = 0
  • 21. Example B. Which of the following equations is factorable? Solve it by factoring if it is. a. 3x2 – 2x + 8 = 0 a = 3, b = –2, c =8 so b2 – 4ac = (–2)2 – 4(3)(8) = 4 – 96 = –92 Factoring Method Factor the equation into the form (#x + #)(#x + #) = 0. Hint: Find b2 – 4ac, if it is not a perfect square then the equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…) Solving 2nd Degree Equations ax2 + bx + c = 0 b. 3x2 – 2x – 8 = 0
  • 22. Example B. Which of the following equations is factorable? Solve it by factoring if it is. a. 3x2 – 2x + 8 = 0 a = 3, b = –2, c =8 so b2 – 4ac = (–2)2 – 4(3)(8) = 4 – 96 = –92 Not a perfect square so it's not factorable Factoring Method Factor the equation into the form (#x + #)(#x + #) = 0. Hint: Find b2 – 4ac, if it is not a perfect square then the equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…) Solving 2nd Degree Equations ax2 + bx + c = 0 b. 3x2 – 2x – 8 = 0
  • 23. Example B. Which of the following equations is factorable? Solve it by factoring if it is. a. 3x2 – 2x + 8 = 0 a = 3, b = –2, c =8 so b2 – 4ac = (–2)2 – 4(3)(8) = 4 – 96 = –92 Not a perfect square so it's not factorable Factoring Method Factor the equation into the form (#x + #)(#x + #) = 0. Hint: Find b2 – 4ac, if it is not a perfect square then the equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…) Solving 2nd Degree Equations ax2 + bx + c = 0 b. 3x2 – 2x – 8 = 0 a = 3, b = –2, c = –8
  • 24. Example B. Which of the following equations is factorable? Solve it by factoring if it is. a. 3x2 – 2x + 8 = 0 a = 3, b = –2, c =8 so b2 – 4ac = (–2)2 – 4(3)(8) = 4 – 96 = –92 Not a perfect square so it's not factorable Factoring Method Factor the equation into the form (#x + #)(#x + #) = 0. Hint: Find b2 – 4ac, if it is not a perfect square then the equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…) Solving 2nd Degree Equations ax2 + bx + c = 0 b. 3x2 – 2x – 8 = 0 a = 3, b = –2, c = –8 b2 – 4ac = (–2)2 – 4(3)(-8)
  • 25. Example B. Which of the following equations is factorable? Solve it by factoring if it is. a. 3x2 – 2x + 8 = 0 a = 3, b = –2, c =8 so b2 – 4ac = (–2)2 – 4(3)(8) = 4 – 96 = –92 Not a perfect square so it's not factorable Factoring Method Factor the equation into the form (#x + #)(#x + #) = 0. Hint: Find b2 – 4ac, if it is not a perfect square then the equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…) Solving 2nd Degree Equations ax2 + bx + c = 0 b. 3x2 – 2x – 8 = 0 a = 3, b = –2, c = –8 b2 – 4ac = (–2)2 – 4(3)(-8) = 4 + 96 = 100
  • 26. Example B. Which of the following equations is factorable? Solve it by factoring if it is. a. 3x2 – 2x + 8 = 0 a = 3, b = –2, c =8 so b2 – 4ac = (–2)2 – 4(3)(8) = 4 – 96 = –92 Not a perfect square so it's not factorable Factoring Method Factor the equation into the form (#x + #)(#x + #) = 0. Hint: Find b2 – 4ac, if it is not a perfect square then the equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…) Solving 2nd Degree Equations ax2 + bx + c = 0 b. 3x2 – 2x – 8 = 0 a = 3, b = –2, c = –8 b2 – 4ac = (–2)2 – 4(3)(-8) = 4 + 96 = 100 = 102
  • 27. Example B. Which of the following equations is factorable? Solve it by factoring if it is. a. 3x2 – 2x + 8 = 0 a = 3, b = –2, c =8 so b2 – 4ac = (–2)2 – 4(3)(8) = 4 – 96 = –92 Not a perfect square so it's not factorable Factoring Method Factor the equation into the form (#x + #)(#x + #) = 0. Hint: Find b2 – 4ac, if it is not a perfect square then the equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…) Solving 2nd Degree Equations ax2 + bx + c = 0 b. 3x2 – 2x – 8 = 0 a = 3, b = –2, c = –8 b2 – 4ac = (–2)2 – 4(3)(-8) = 4 + 96 = 100 = 102 So its factorable.
  • 28. Example B. Which of the following equations is factorable? Solve it by factoring if it is. a. 3x2 – 2x + 8 = 0 a = 3, b = –2, c =8 so b2 – 4ac = (–2)2 – 4(3)(8) = 4 – 96 = –92 Not a perfect square so it's not factorable Factoring Method Factor the equation into the form (#x + #)(#x + #) = 0. Hint: Find b2 – 4ac, if it is not a perfect square then the equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…) Solving 2nd Degree Equations ax2 + bx + c = 0 b. 3x2 – 2x – 8 = 0 a = 3, b = –2, c = –8 b2 – 4ac = (–2)2 – 4(3)(-8) = 4 + 96 = 100 = 102 So its factorable. 3x2 – 2x – 8 = 0 (3x + 4)(x – 2) = 0
  • 29. Example B. Which of the following equations is factorable? Solve it by factoring if it is. a. 3x2 – 2x + 8 = 0 a = 3, b = –2, c =8 so b2 – 4ac = (–2)2 – 4(3)(8) = 4 – 96 = –92 Not a perfect square so it's not factorable Factoring Method Factor the equation into the form (#x + #)(#x + #) = 0. Hint: Find b2 – 4ac, if it is not a perfect square then the equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…) Solving 2nd Degree Equations ax2 + bx + c = 0 b. 3x2 – 2x – 8 = 0 a = 3, b = –2, c = –8 b2 – 4ac = (–2)2 – 4(3)(-8) = 4 + 96 = 100 = 102 So its factorable. 3x2 – 2x – 8 = 0 (3x + 4)(x – 2) = 0 x = -4/3, x = 2
  • 30. Solving 2nd Degree Equations ax2 + bx + c = 0 Use the quadratic formula (QF)
  • 31. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x =
  • 32. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are.
  • 33. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots,
  • 34. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
  • 35. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Example C. Solve 3x2 – 2x – 2 = 0 Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
  • 36. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Example C. Solve 3x2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
  • 37. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Example C. Solve 3x2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
  • 38. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Example C. Solve 3x2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Not factorable! Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
  • 39. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Example C. Solve 3x2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Not factorable! Use QF, we get x = – (–2) Β± οƒ– 28 2(3) Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
  • 40. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Example C. Solve 3x2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Not factorable! Use QF, we get x = – (–2) Β± οƒ– 28 2(3) = 2 Β± οƒ– 28 6 Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
  • 41. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Example C. Solve 3x2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Not factorable! Use QF, we get x = – (–2) Β± οƒ– 28 2(3) = 2 Β± οƒ– 28 6 = 2 Β± 2οƒ–7 6 Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
  • 42. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Example C. Solve 3x2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Not factorable! Use QF, we get x = – (–2) Β± οƒ– 28 2(3) = 2 Β± οƒ– 28 6 = 2 Β± 2οƒ–7 6 = 1 Β± οƒ–7 3 Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
  • 43. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Example C. Solve 3x2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Not factorable! Use QF, we get x = – (–2) Β± οƒ– 28 2(3) = 2 Β± οƒ– 28 6 = 2 Β± 2οƒ–7 6 = 1 Β± οƒ–7 3 ο‚» {1.22 –0.549 Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
  • 44. Example D. Solve 3x2 – 2x + 2 = 0 Solving 2nd Degree Equations ax2 + bx + c = 0
  • 45. Example D. Solve 3x2 – 2x + 2 = 0 To check if it is factorable; a = 3, b = -2, c = -2 Solving 2nd Degree Equations ax2 + bx + c = 0
  • 46. Example D. Solve 3x2 – 2x + 2 = 0 To check if it is factorable; a = 3, b = -2, c = -2 So b2 – 4ac = (–2)2 – 4(3)(2) = –20, Solving 2nd Degree Equations ax2 + bx + c = 0
  • 47. Example D. Solve 3x2 – 2x + 2 = 0 To check if it is factorable; a = 3, b = -2, c = -2 So b2 – 4ac = (–2)2 – 4(3)(2) = –20, –20 is not a real number, so there is no real solution. Solving 2nd Degree Equations ax2 + bx + c = 0
  • 48. Example D. Solve 3x2 – 2x + 2 = 0 To check if it is factorable; a = 3, b = -2, c = -2 So b2 – 4ac = (–2)2 – 4(3)(2) = –20, –20 is not a real number, so there is no real solution. Solving 2nd Degree Equations ax2 + bx + c = 0
  • 49. Solving 2nd Degree Equations ax2 + bx + c = 0 1. 3x2 – 12 = 0 Exercise A. Solve. Use the square root methods. Give both the exact answers and the numerical answers. If the answer is not real, state so. 2. 3x2 – 15 = 0 3. 3x2 + 15 = 0 4. x2 – 3 = –x2 + 15 5. –6 = 4x2 – 15 6. (x – 2)2 = 2 7. 4 = (2x – 3)2 Exercise B. Solve by factoring and by the quadratic formula. 14. x2 – 3x = 10 15. x(x – 2) = 24 16. 2x2 = 3(x + 1) – 1 8. x2 – 3x = 4 9. x2 – 15 = 2x 10. x2 + 5x + 12 = 0 11. –x2 – 2x + 8 = 0 12. 9 – 3x – 2x2 = 0 13. 2x2 – x – 1 = 0 Exercise C. Solve by the quadratic formula. 17. x2 – x + 1 = 0 18. x2 – x – 1 = 0 19. x2 – 3x – 2 = 0 20. x2 – 2x + 3 = 0 If the answers are not real numbers, state so. 21. 2x2 – 3x – 1 = 0 22. 3x2 = 2x + 3
  • 50. Solving 2nd Degree Equations ax2 + bx + c = 0 Exercise C. Solve by the quadratic formula. If the answers are not real numbers, state so. 23. 2(x2 – 1) + x = 4 24. (x – 1)(x + 1) = 2x(x + 2) 25. (x – 1) (x + 2) = 2x (x + 2) 26. (x + 1) (x + 2) = (x + 2) (2x + 1) 1 Π€ 1 – Π€ 1 – Π€ Π€1 Π€ = 27. Cut a stick of unit length (1) into two parts in a manner such that β€œthe ratio of the whole to the large part is the same as the ratio of the large part to the small part” In picture, the golden ratio Π€ (phi) is the ratio 1 : Π€. Find the exact and approximate value of Π€. (Google β€œgolden ratio” )
  • 51. Solving 2nd Degree Equations ax2 + bx + c = 0 HW answers 1. x = Β±2 3. no real answers 5. x = Β±3/2 7. x = Β½, 5/2 9. x = –3, 5 11. x = –4, 2 17. no real answer 19. x = 13. x = –1/2, 1 15. x = 6, –4 3 ±√17 2 21. x = 3 ±√17 4 23. x = –2, 3 27. Π€ = –1 +√5 2 25. x = –1, –2 approx. 0.618