Quadratic Formula Part 2

Solving Quadratic

Equations by the

Quadratic Formula

➢ Solve quadratic equations by using

the Quadratic Formula.

➢ Determine the number of solutions

of a quadratic equation by using the
discriminant.

Objectives

2-5 Quadratic Formula

The roots of a quadratic equation
are…
• the solutions for the variable.

• related to the zeros of the corresponding function.

• related to the x-intercepts of the graph of the corresponding

function.

No real number x-intercepts
No real number zeros

No real number roots
…solutions

One real x-intercept
One real zeros

Two equal real roots
One real solution

Two real x-intercepts
Two real zeros

Two distinct real roots
Two real solutions

The roots of the quadratic equation
ax2 + bx + c = 0 can be found by using the
quadratic formula:

The Quadratic Formula

Multiple Choice

What is this formula?

• This is the standard formula.

• This is the quadratic formula.

• This is the square root formula.

• This is the Pythagorean formula

Two Equal Real Roots

Solve x2 + 3x - 2 = 0.

x = – b+ b² – 4ac

2a

Quadratic formula

x = – 3+ 3² – 4(1)(–2)

2(1)

a = 1, b = 3, c = –2

Simplify. PEMDAS

x =– 3+

17

2

The solutions are x =– 3+

17

2

0.56or

x =– 3–

17

2

– 3.56.

CHECKGraph y = x² + 3x – 2 and note

that the x-intercepts are
approx. 0.56 and –3.56.

Two distinct real roots

Solve 2x² - 5x + 2 = 0.

a = 2, b = -5, c = 2

Solving Quadratic Equations Using the Quadratic Formula

Two distinct real roots

Solve x² - 5x + 7 = 0.

Solving Quadratic Equations that have No Real Roots

No real roots

Multiple Choice

Determine the values of a, b, and c for the quadratic equation:

4x² – 8x = 3

• a = 4, b = -8, c = 3

• a = 4, b =-8, c =-3

• a = 4, b = 8, c = 3

• a = 4, b = 8, c = -3

Solve x² - 6x + 9 = 0.

Solving Quadratic Equations with Two Equal Real Roots

Two Equal real roots

Example 1A: Using the Quadratic Formula

Solve using the Quadratic Formula.

6x² + 5x – 4 = 0

6x² + 5x + (–4) = 0. Identify a, b, and c.

Use the Quadratic Formula.

Simplify
.

Substitute 6 for a, 5 for b,

and –4 for c.

Simplify.

Write as two equations.

Solve each equation.

Example 1B: Using the Quadratic Formula

Solve using the Quadratic Formula.

x² = x + 20

1x² + (–1x) + (–20) = 0 Write in standard form. Identify

a, b, and c.

Use the quadratic formula.

Simplify.

Substitute 1 for a, –1 for b,

and –20 for c.

x = 5 or x = –4

Simplify.

Write as two equations.

Solve each equation.

Multiple Choice

Solve Using the Quadratic Formula
 x² + 4x - 40 = -8

• -10 & -4

• -4 & 10

• -8 & 4

• 8 & -4

WHY USE THE

QUADRATIC FORMULA?

The quadratic formula allows you to solve

ANY quadratic equation, even if you

cannot factor it.

An important piece of the quadratic formula

is what’s under the radical:

b² – 4ac

This piece is called the discriminant.

The quadratic formula

will give the roots of the quadratic equation.
From the quadratic formula, the radicand,
b2 - 4ac, will determine the Nature of the Roots.

By the nature of the roots, we mean:
• whether the equation has real roots or imaginary
• if there are real roots, whether they are different or
equal

The radicand b² - 4ac is called the discriminant
of the equation ax² + bx + c = 0 because it discriminates
among the three cases that can occur.

Determining TheNature of the Roots

WHAT THE DISCRIMINANT

TELLS YOU!

Value of the Discriminant

Nature of the Solutions

Negative

2 imaginary solutions

Zero

1 Real Solution

Positive – perfect square

2 Reals- Rational

Positive – non-perfect

square

2 Reals- Irrational

The discriminant describes the Nature of the Roots
of a Quadratic Equation

If b² - 4ac > 0, then there are two
different real roots.

If b² - 4ac < 0, then there are no
real roots.

If b² - 4ac = 0, then there are
two equal real roots.

Multiple Choice

What is the nature of the roots?

3x² - 7x + 2 = 0

• 2 real, rational roots

• 2 imaginary roots

• 1 real, rational root

• 2 real, irrational roots

Use the discriminant to determine the nature of the roots.

Nature of

Roots

Equation

ax² + bx + c = 0

Discriminant

b² – 4ac

a.2x² + 6x + 5 = 0

b.x² – 7 = 0

c.4x² – 12x + 9 = 0

6² – 4(2)(5) = –4

No real roots

0² – 4(1)(– 7) = 28

Two distinct
real roots

(–12)² –4(4)(9) = 0

Two equal real
roots

3x² – 2x + 2 = 0

2x² + 11x + 12 = 0

x² + 8x + 16 = 0

a = 3, b = –2, c = 2

a = 2, b = 11, c = 12

a = 1, b = 8, c = 16

b² – 4ac

b² – 4ac

b2 – 4ac

(–2)² – 4(3)(2)

11² – 4(2)(12)

8² – 4(1)(16)

4 – 24

121 – 96

64 – 64

–20

25
0

b² – 4ac is negative.

There are no real

solutions.

b² – 4ac is positive.

There are two real

solutions.

b² – 4ac is zero.

There is one real

solution.

Example 3: Using the Discriminant

Find the number of solutions of each equation
using the discriminant.

A.

B.

C.