1) The document provides lessons on solving quadratic equations using various methods like factoring, completing the square, and the quadratic formula. 2) It includes examples of solving quadratic equations by factoring polynomials, using the zero product property to set factors equal to zero, and finding the solutions. 3) The quadratic formula is derived by completing the square on the general quadratic equation ax2 + bx + c = 0, resulting in the formula x = (-b ± √(b2 - 4ac))/2a to solve for real solutions.

1. 𝐏𝐓𝐒 𝟑
Bridge to Calculus Workshop
Summer 2020
Lesson 10
Solving Quadratic
Equations
“The only way to learn
mathematics is to do mathematics.”
- Paul Halmos -

2. Lehman College, Department of Mathematics
Basic Quadratic Equations (1 of 4)
A quadratic equation is any equation of the form:
where 𝑎, 𝑏, and 𝑐 are real numbers, with 𝑎 ≠ 0. A
quadratic equation equates a degree 2 polynomial to 0.
Example 1. Solve the quadratic equation:
Solution. We first factor the polynomial:
Zero Product Rule: Let 𝑎 and 𝑏 be two real numbers, if
Then, either 𝑎 is zero or 𝑏 is zero or both are zero.
𝑎𝑥2
+ 𝑏𝑥 + 𝑐 = 0
𝑥2
− 3𝑥 + 2 = 0
𝑥2
− 3𝑥 + 2 = 0
𝑥 − 1 𝑥 − 2 = 0
𝑎 ⋅ 𝑏 = 0

3. Lehman College, Department of Mathematics
Basic Quadratic Equations (2 of 4)
Solution (cont’d). We first factor the polynomial:
From the Zero Product Rule, we conclude that:
Finally, this means that:
Example 2. Solve the quadratic equation:
Solution. We first factor the polynomial:
We conclude that:
𝑥2
− 3𝑥 + 2 = 0
𝑥 − 1 𝑥 − 2 = 0
𝑥 − 1 = 0 𝑥 − 2 = 0or
𝑥 = 1 or 𝑥 = 2.
𝑥2
− 9 = 0
𝑥2
− 9 = 0
𝑥 − 3 𝑥 + 3 = 0
𝑥 = 3 or 𝑥 = −3.
Difference of two squares
Write the original equation

4. Lehman College, Department of Mathematics
Basic Quadratic Equations (3 of 4)
Example 3. Solve the quadratic equation:
Solution. We first factor the polynomial:
We conclude that:
Finally, we obtain:
Example 4. Solve the quadratic equation:
Solution. Note that the leading coefficient is not 1, but
the greatest common integer factor is 1.
2𝑥2
− 2𝑥 − 12 = 0
2 𝑥 − 3 𝑥 + 2 = 0
𝑥 + 2 = 0.𝑥 − 3 = 0 or
𝑥 = 3 or 𝑥 = −2.
2𝑥2 − 2𝑥 − 12 = 0
2 𝑥2 − 𝑥 − 6 = 0 Common integer factor
6𝑥2
+ 𝑥 − 2 = 0
Write the original equation

5. Lehman College, Department of Mathematics
Factor 1 Factor 2 Sum
Basic Quadratic Equations (4 of 4)
Solution. We will factor the quadratic using factoring
number and grouping. Factoring number:
Split the linear term using the factors obtained:
It follows that:
3𝑥 + 2 2𝑥 − 1 = 0
3𝑥 + 2 = 0 2𝑥 − 1 = 0or 𝑥 = −
2
3
,
1
2
1 −12 −11
2
6𝑥2 + 𝑥 − 2 = 0
−6
3 −4 −1
−4
6𝑥2 − 3𝑥 + 4𝑥 − 2 = 0
3𝑥 2𝑥 − 1 + 2 2𝑥 − 1 = 0
so
−12
−3, 4

6. Lehman College, Department of Mathematics
Square of a Binomial Rule (1 of 1)
Example 5. Recall the square of a binomial rule:
Solution. By the distributive property:
Example 6. Factor the following quadratic:
Solution. The leading coefficient is 1 and the constant
term is the square of half the linear coefficient:
𝑎 + 𝑏 𝑎 + 𝑏 = 𝑎 𝑎 + 𝑏 + 𝑏 𝑎 + 𝑏
= 𝑎2 + 𝑎𝑏 + 𝑏𝑎 + 𝑏2
= 𝑎2
+ 2𝑎𝑏 + 𝑏2
𝑎 + 𝑏 2
= 𝑎2
+ 2𝑎𝑏 + 𝑏2
𝑥2
+ 6𝑥 + 9
𝑥2
+ 6𝑥 + 9 = 𝑥 + 3 2

7. Lehman College, Department of Mathematics
Completing the Square Method (1 of 6)
Example 6. Solve the following quadratic equation:
Solution. The quadratic is a difference of two squares:
It follows that 𝑥 = 2, −2 are the solutions. Rewrite as:
We see that taking square roots yields only one solution
To get the correct pair of solutions, we write:
𝑥2
− 4 = 0
𝑥2 − 4 = 0
𝑥2
− 22
= 0
𝑥 − 2 𝑥 + 2 = 0
𝑥2
− 4 = 0
𝑥2
= 4
Write the original equation
Add 4 to both sides
𝑥2 = 4 Take square roots
𝑥2 = ± 4

8. Lehman College, Department of Mathematics
Completing the Square Method (2 of 6)
Example 7. Solve the following quadratic equation:
Solution. We could easily find that:
However, we will learn how to Complete the Square:
It follows that:
𝑥2
+ 2𝑥 − 3 = 0
𝑥2 + 2𝑥 − 3 = 0 Write the original equation
Add and subtract the square
of half the linear coefficient
𝑥2
+ 2𝑥 + 1 − 1 − 3 = 0
Complete the square
(𝑥2+2𝑥 + 1) + (−1 − 3) = 0
𝑥 + 1 2
− 4 = 0
𝑥 + 1 2 = 4 Add 4 to both sides
𝑥 = −3 or 𝑥 = 1
(𝑥 + 1)2= ± 4 Take square roots
𝑥 + 1 = ±2
𝑥 = −1 + 2 = 1 or 𝑥 = −1 − 2 = −3

9. Lehman College, Department of Mathematics
Completing the Square Method (3 of 6)
Example 6. Solve the following quadratic equation:
Solution. The integer factors 2 are 1 and 2, whose sum
is 3, not 4. So, we complete the square.
It follows that:
𝑥2
+ 6𝑥 + 2 = 0
𝑥2 + 6𝑥 + 2 = 0 Write the original equation
Add and subtract the square
of half the linear coefficient
𝑥2
+ 6𝑥 + 9 − 9 + 2 = 0
Complete the square
(𝑥2+ 6𝑥 + 9) + (2 − 9) = 0
𝑥 + 3 2 − 7 = 0
𝑥 + 3 2
= 7 Add 7 to both sides
(𝑥 + 3)2= ± 7 Take square roots
𝑥 + 3 = ± 7
𝑥 = −3 + 7 or 𝑥 = −3 − 7

10. Lehman College, Department of Mathematics
Completing the Square Method (4 of 6)
Example 7. Solve the following quadratic equation:
Solution. We factor the 2, then complete the square.
It follows that:
2𝑥2
− 4𝑥 − 4 = 0
2𝑥2
− 4𝑥 − 4 = 0 Write the original equation
Add and subtract the square
of half the linear coefficient
𝑥2
− 2𝑥 + 1 − 1 − 2 = 0
Complete the square
(𝑥2−2𝑥 + 1) + (−1 − 2) = 0
𝑥 − 1 2 − 3 = 0
𝑥 − 1 2
= 3 Add 2 to both sides
(𝑥 − 1)2= ± 3 Take square roots
𝑥 − 1 = ± 3
𝑥 = 1 + 3 or 𝑥 = 1 − 3
𝑥2 − 2𝑥 − 2 = 0 Divide both sides by 2

11. Lehman College, Department of Mathematics
Completing the Square Method (5 of 6)
Example 8. Solve the following quadratic equation:
Solution. We factor the 2 from the first two terms.
2𝑥2
+ 4𝑥 − 1 = 0
2𝑥2
+ 4𝑥 − 1 = 0 Write the original equation
Add and subtract the
square of half the linear
coefficient
2 𝑥2
+ 2𝑥 + 1 − 1 − 1 = 0
Complete the square
2(𝑥2 + 2𝑥 + 1) + 2 −1 − 1 = 0
2 𝑥 + 1 2
− 3 = 0
2 𝑥 + 1 2
= 3 Add 3 to both sides
2 𝑥2 + 2𝑥 − 1 = 0 Factor the 2 from terms
𝑥 + 1 2
=
3
2
Divide by 2 on both sides

12. Lehman College, Department of Mathematics
Completing the Square Method (6 of 6)
Solution (cont’d). From the previous slide:
(𝑥 + 1)2 = ±
3
2
Take square roots
𝑥 + 1 2 =
3
2
𝑥 + 1 = ±
3
2
Subtract 1 from both sides𝑥 = −1 ±
3
2

13. Lehman College, Department of Mathematics
Quadratic Formula (1 of 8)
Example 8. Solve the quadratic equation, where 𝑎 ≠ 0:
Solution. We factor the 𝑎, then complete the square.
𝑎𝑥2
+ 𝑏𝑥 + 𝑐 = 0
𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 Write the original equation
Add and subtract
the square of half
the linear coefficient
𝑎 𝑥2 +
𝑏
𝑎
𝑥 +
𝑏
2𝑎
2
−
𝑏
2𝑎
2
+ 𝑐 = 0
Complete the square𝑎 𝑥 +
𝑏
2𝑎
2
− 𝑎
𝑏
2𝑎
2
+ 𝑐 = 0
𝑎 𝑥2 +
𝑏
𝑎
𝑥 + 𝑐 = 0 Factor the 𝑎 from first 2 terms
𝑎 𝑥 +
𝑏
2𝑎
2
= 𝑎
𝑏
2𝑎
2
− 𝑐

14. Lehman College, Department of Mathematics
Quadratic Formula (2 of 8)
Solution (cont’d). From the previous slide:
Bring right-hand side to
common denominator.
Simplify RHS𝑎 𝑥 +
𝑏
2𝑎
2
= 𝑎
𝑏
2𝑎
2
− 𝑐 = 𝑎
𝑏2
4𝑎2
− 𝑐
=
𝑏2
4𝑎
−
𝑐
1
4𝑎
4𝑎
=
𝑏2
− 4𝑎𝑐
4𝑎
𝑥 +
𝑏
2𝑎
2
=
𝑏2
− 4𝑎𝑐
4𝑎2
𝑥 +
𝑏
2𝑎
2
= ±
𝑏2 − 4𝑎𝑐
4𝑎2
If 𝑏2 − 4𝑎𝑐 ≥ 0 take
square roots of both
sides.
𝑥 +
𝑏
2𝑎
= ±
𝑏2 − 4𝑎𝑐
2𝑎
Divide both sides by 𝑎.

15. Lehman College, Department of Mathematics
Quadratic Formula (3 of 8)
Solution (cont’d). From the previous slide:
This is the well-known quadratic formula. It yields real
roots when the quantity ∆ = 𝑏2
− 4𝑎𝑐 ≥ 0. The number
∆ is called the discriminant of the quadratic formula.
If ∆ = 0, the quadratic is a perfect square. If ∆ itself is a
perfect square, the quadratic is factorable in integers.
Subtract
𝑥 +
𝑏
2𝑎
= ±
𝑏2 − 4𝑎𝑐
2𝑎
𝑏
2𝑎
𝑥 = ±
𝑏2 − 4𝑎𝑐
2𝑎
−
𝑏
2𝑎
Bring right-hand side to
a common denominator.=
−𝑏 ± 𝑏2 − 4𝑎𝑐
2𝑎

16. Lehman College, Department of Mathematics
Quadratic Formula (4 of 8)
Quadratics for which the discriminant ∆ = 𝑏2
− 4𝑎𝑐 is
negative are called irreducible quadratics.
Example 9. Calculate the discriminant of the following:
Solution. We use the formula ∆ = 𝑏2
− 4𝑎𝑐:
In Example 9(c), since ∆ = 49 is a perfect square, then
the quadratic 6𝑥2
+ 𝑥 − 2 is factorable in integers:
𝑥2
+ 1(a) 𝑥2
+ 3𝑥 + 4(b) 6𝑥2
− 𝑥 − 2(c)
∆ =(a) 02 − 4 1 1 = −4 < 0 irreducible quadratic
∆ =(b) 32
− 4 1 4 = −7 < 0 irreducible quadratic
∆ =(c) (−1)2−4 6 −2 = 49 > 0 factorable quadratic
6𝑥2
− 𝑥 − 2 = (3𝑥 − 2)(2𝑥 + 1)

17. Lehman College, Department of Mathematics
Quadratic Formula (5 of 8)
Example 10. Solve the following quadratic equation:
Solution 1. Since half the linear coefficient is not an
integer, completing the square will involve fractions, so
we will use the quadratic formula:
2𝑥2
+ 3𝑥 − 1 = 0
∆ = 32 − 4 2 −1 =
𝑥 =
Find the discriminant
Nonnegative discriminant
17
−3 ± 17
2 2
−3 ± 17
2 2
−3 ± 17
2 2
−3 ± 17
2 2
−3 ± 17
2 2
= −
3
4
±
17
4

18. Lehman College, Department of Mathematics
Quadratic Formula (6 of 8)
Example 10. Solve the quadratic equation:
Solution 2. We factor the 2, then complete the square.
2𝑥2
+ 3𝑥 − 1 = 0
2𝑥2 + 3𝑥 − 1 = 0 Write the original equation
Add and subtract
the square of half
the linear coefficient
2 𝑥2 +
3
2
𝑥 +
3
4
2
−
3
4
2
− 1 = 0
Complete the square2 𝑥 +
3
4
2
− 2
3
4
2
− 1 = 0
2 𝑥2 +
3
2
𝑥 − 1 = 0 Factor 2 from first two terms
2 𝑥 +
3
4
2
= 2
3
4
2
+ 1

19. Lehman College, Department of Mathematics
Quadratic Formula (7 of 8)
Solution (cont’d). From the previous slide:
Bring right-hand side to
common denominator.
Simplify RHS2 𝑥 +
3
4
2
= 2
3
4
2
+ 1 = 2
9
16
+ 1
=
9
8
+
1
1
8
8
=
9 + 8
8
𝑥 +
3
4
2
=
17
16
𝑥 +
3
4
2
= ±
17
16
Take square roots of
both sides.
𝑥 +
3
4
= ±
17
4
Divide both sides by 2.

20. Lehman College, Department of Mathematics
Quadratic Formula (8 of 8)
Solution (cont’d). From the previous slide:
Subtract
𝑥 +
3
4
= ±
17
4
3
4
𝑥 = ±
17
4
−
3
4