This will help you on how to solve quadratic equations by factoring.
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You will learn how to get the value of a, b and c given a quadratic equations.
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This will help you on how to solve quadratic equations by factoring.
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https://tinyurl.com/ybo27k2u
You will learn how to get the value of a, b and c given a quadratic equations.
For more instructional resources, CLICK me here! πππ
https://tinyurl.com/y9muob6q
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https://tinyurl.com/ycjp8r7u
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This powerpoint presentation discusses or talks about the topic or lesson Functions. It also discusses and explains the rules, steps and examples of Quadratic Functions.
You will learn how to solve quadratic equations by extracting square roots.
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This powerpoint presentation discusses or talks about the topic or lesson Direct Variations. It also discusses and explains the rules, concepts, steps and examples of Direct Variations.
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You will learn how to solve quadratic equations by extracting square roots.
For more instructional resources, CLICK me here! πππ
https://tinyurl.com/y9muob6q
LIKE and FOLLOW me here! πππ
https://tinyurl.com/ycjp8r7u
https://tinyurl.com/ybo27k2u
This will help you in factoring sum and difference of two cubes.
For more instructional resources, CLICK me here! οοο
https://tinyurl.com/y9muob6q
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https://tinyurl.com/ycjp8r7u
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This powerpoint presentation discusses or talks about the topic or lesson Direct Variations. It also discusses and explains the rules, concepts, steps and examples of Direct Variations.
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2. β’Any value that satisfies an equation is called a
solution.
β’The set of solution that satisfy an equation is
called solution set.
β’The solution is also called root.
In solving quadratic equations, it means
finding its solution(s) or root(s) that will satisfy
the given equation.
14. Steps in Solving Quadratic Equation
by Factoring.
1. Write the quadratic equation in standard form.
2. Find the factors of the quadratic expression.
3. Apply the Zero Product Property.
4. Solve each resulting equation.
5. Check the values of the variable obtained by substituting each in the
original equation.
22. Steps in Solving Quadratic Equation by
Completing the Square
1. If the value of a = 1, proceed to step 2. Otherwise, divide both
sides of the equation by a.
2. Group all the terms containing a variable on one side of the
equation and the constant term on the other side. That is
ax2 + bx = c.
3. Complete the square of the resulting binomial by adding the
square of the half of b on both sides of the equation.
π
2
2
4. Rewrite the perfect square trinomial as the square of binomial.
π₯ +
π
2
2
.
5. Use extracting square the square root to solve for x.
23. Example 1:
π₯2 + 2π₯ β 8 = 0
Since a = 1, letβs proceed with step 2.
Group all the terms containing a variable on one side of the equation
and the constant term on the other side. That is ax2 + bx = c.
π₯2 + 2π₯ β 8 = 0
π₯2 + 2π₯ β 8 + 8 = 0 + 8
π₯2 + 2π₯ = 8
Complete the square of the resulting binomial by adding the
square of the half of b on both sides of the equation.
π
2
2
π
2
2
=
2
2
2
= 12 = π
π₯2 + 2π₯ + 1 = 8 + 1
π₯2 + 2π₯ + 1 = 9
Rewrite the perfect square trinomial as the square of binomial.
π₯ +
π
2
2
.
(π₯ + 1)2 = 9
Use extracting square the square root to solve for x.
π₯ + 1 2 = 9
π₯ + 1 = Β±3
x + 1 = 3 x + 1 = β3
x +1 β 1 = 3 β 1 x + 1 β 1 = β3 β 1
x = 2 x = β 4
Checking:
x = 2 x = β 4
π₯2 + 2π₯ β 8 = 0 π₯2 + 2π₯ β 8 = 0
(2)2 + 2(2) β 8 = 0 (-4) + 2(-4) β 8 = 0
4 + 4 β 8 = 0 16 β 8 β 8 = 0
8 β 8 = 0 8 β 8 = 0
0 = 0 0 = 0
24. Example 2:
2π₯2 β 12π₯ = 54
Since a β 1, we divide both sides of the equation by a. a = 2
2π₯2
2
β
12π₯
2
=
54
2
β π₯2
β 6π₯ = 27
Group all the terms containing a variable on one side of the equation
and the constant term on the other side. That is ax2 + bx = c.
π₯2 β 6π₯ = 27
Complete the square of the resulting binomial by adding the
square of the half of b on both sides of the equation.
π
2
2
π
2
2
=
6
2
2
= 32 = π
π₯2 β 6π₯ + 9 = 27 + 9
π₯2 β 6π₯ + 9 = 36
Rewrite the perfect square trinomial as the square of binomial.
π₯ +
π
2
2
.
(π₯ β 3)2 = 36
Use extracting square the square root to solve for x.
π₯ β 3 2 = 36
π₯ β 3 = Β±6
x β 3 = 6 x β 3 = β6
x β 3 + 3 = 6 + 3 x β 3 = β6 + 3
x = 9 x = β 3
Checking:
x = 9 x = β 3
2π₯2 β 12π₯ = 54 2π₯2 β 12π₯ = 54
2(9)2 β 12(9) = 54 2(-3)2 β 12(-3) = 54
2(81) β 108 = 54 2(9) +36 = 54
162 β 108 = 54 18 + 36 = 54
54 = 54 54 = 54
25. Example 3:
4π₯2 + 16π₯ β 9 = 0
Since a β 1, we divide both sides of the equation by a. a = 4
4π₯2
4
+
16π₯
4
β
9
4
=
0
4
β π₯2 + 4π₯ β
9
4
= 0
Group all the terms containing a variable on one side of the equation
and the constant term on the other side. That is ax2 + bx = c.
π₯2
+ 4π₯ β
9
4
= 0 β π₯2
+ 4π₯ β
9
4
+
9
4
= 0 +
9
4
ππ + ππ =
π
π
Complete the square of the resulting binomial by adding the
square of the half of b on both sides of the equation.
π
2
2
π
2
2
=
4
2
2
= 22 = 4
π₯2 + 4π₯ + 4 =
9
4
+ 4
π₯2
+ 4π₯ + 4 =
25
4
Rewrite the perfect square trinomial as the square of binomial.
π₯ +
π
2
2
.
(π₯ + 2)2 =
25
4
Use extracting square the square root to solve for x.
π₯ + 2 2 =
25
4
π₯ + 2 = Β±
5
2
π₯ + 2 =
5
2
π₯ + 2 = β
5
2
π₯ + 2 β 2 =
5
2
β 2 π₯ + 2 β 2 = β
5
2
β 2
π =
π
π
π = β
π
π
Checking:
x =
1
2
x = β
9
2
4π₯2 + 16π₯ β 9 = 0 4π₯2 + 16π₯ β 9 = 0
4
1
2
2
+ 16
1
2
β 9 = 0 4 β
9
2
2
+ 16 β
9
2
β 9 = 0
4
1
4
+ 8 β 9 = 0 4
81
4
β 72 β 9 = 0
1 + 8 β 9 = 0 81 β 72 β 9 = 0
0 = 0 0 = 0