5-6
The Quadratic Formula
You can use the Quadratic Formula to solve any
quadratic equation that is written in standard form,
including equations with real solutions or complex
solutions.
Holt Algebra 2
5-6
The Quadratic Formula
Example 1: Quadratic Functions with Real Zeros
Find the zeros of f(x)= 2x2 – 16x + 27 using
the Quadratic Formula.
2x2 – 16x + 27 = 0
Set f(x) = 0.
Write the Quadratic
Formula.
Substitute 2 for a, –16
for b, and 27 for c.
Simplify.
Write in simplest form.
Holt Algebra 2
5-6
The Quadratic Formula
Check It Out! Example 1a
Find the zeros of f(x) = x2 + 3x – 7 using the
Quadratic Formula.
x2 + 3x – 7 = 0
Set f(x) = 0.
Write the Quadratic Formula.
Substitute 1 for a, 3 for b,
and –7 for c.
Simplify.
Write in simplest form.
Holt Algebra 2
5-6
The Quadratic Formula
Example 2: Quadratic Functions with Complex Zeros
Find the zeros of f(x) = 4x2 + 3x + 2 using the
Quadratic Formula.
f(x)= 4x2 + 3x + 2
Set f(x) = 0.
Write the Quadratic Formula.
Substitute 4 for a, 3 for b, and
2 for c.
Simplify.
Write in terms of i.
Holt Algebra 2
5-6
The Quadratic Formula
Check It Out! Example 2
Find the zeros of g(x) = 3x2 – x + 8 using the
Quadratic Formula.
Set f(x) = 0
Write the Quadratic Formula.
Substitute 3 for a, –1 for b,
and 8 for c.
Simplify.
Write in terms of i.
Holt Algebra 2
5-6
The Quadratic Formula
The discriminant is part of the Quadratic
Formula that you can use to determine the
number of real roots of a quadratic equation.
Holt Algebra 2
5-6
The Quadratic Formula
Holt Algebra 2
5-6
The Quadratic Formula
Example 3A: Analyzing Quadratic Equations by Using
the Discriminant
Find the type and number of solutions for the
equation.
x2 + 36 = 12x
x2 – 12x + 36 = 0
b2 – 4ac
(–12)2 – 4(1)(36)
144 – 144 = 0
b2 – 4ac = 0
The equation has one distinct real solution.
Holt Algebra 2
5-6
The Quadratic Formula
Example 3B: Analyzing Quadratic Equations by Using
the Discriminant
Find the type and number of solutions for the
equation.
x2 + 40 = 12x
x2 – 12x + 40 = 0
b2 – 4ac
(–12)2 – 4(1)(40)
144 – 160 = –16
b2 –4ac < 0
The equation has two distinct nonreal complex
solutions.
Holt Algebra 2
5-6
The Quadratic Formula
Example 3C: Analyzing Quadratic Equations by Using
the Discriminant
Find the type and number of solutions for the
equation.
x2 + 30 = 12x
x2 – 12x + 30 = 0
b2 – 4ac
(–12)2 – 4(1)(30)
144 – 120 = 24
b2 – 4ac > 0
The equation has two distinct real solutions.
Holt Algebra 2
5-6
The Quadratic Formula
HW pg. 361
#’s 18-23, 30-35
Holt Algebra 2