Lesson 10: Solving Quadratic Equations

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 -

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

Solution (cont’d). We first factor the polynomial:
From the Zero Product Rule, we conclude that:
Finally, this means that:
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

We conclude that:
Finally, we obtain:
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

Factor 1 Factor 2 Sum
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

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

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
Add 4 to both sides
𝑥2 = 4 Take square roots
𝑥2 = ± 4

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

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
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

Solution. We factor the 2, then complete the square.
It follows that:
2𝑥2
− 4𝑥 − 4 = 0
2𝑥2
− 4𝑥 − 4 = 0 Write the original equation
𝑥2
− 2𝑥 + 1 − 1 − 2 = 0
Complete the square
(𝑥2−2𝑥 + 1) + (−1 − 2) = 0
𝑥 − 1 2 − 3 = 0
𝑥 − 1 2
(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

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
2 𝑥2 + 2𝑥 − 1 = 0 Factor the 2 from terms
𝑥 + 1 2
=
3
2
Divide by 2 on both sides

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

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

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.

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
a common denominator.=
−𝑏 ± 𝑏2 − 4𝑎𝑐
2𝑎

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)

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

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

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.

Subtract
𝑥 +
3
4
= ±
17
4
3
4
𝑥 = ±
17
4
−
3
4

Lesson 10: Solving Quadratic Equations

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