Solving Equations with the Quadratic Formula By completing the square once for the general equation a x 2 + b x + c = 0, you can develop a formula that.

Solving Equations with the Quadratic Formula

By completing the square once for the general equation a x 2 + b x + c = 0, you can develop a formula that gives the solutions of any quadratic equation. The formula for the solutions is called the quadratic formula.

THE QUADRATIC FORMULA

Let a, b, and c be real numbers such that a 0. The solutions of the quadratic equation a x 2 + b x + c = 0 are:

Remember that before you apply the quadratic formula to a quadratic equation, you must write the equation in standard form, a x 2 + b x + c = 0.

x = – b ± b 2 – 4ac

2a

Solve 2x 2 + x = 5.

SOLUTION

Write original equation. 2x 2 + x = 5

2x 2 + x – 5 = 0 Write in standard form.

Quadratic formula

a = 2, b = 1, c = – 5

Simplify.

Solving a Quadratic Equation with Two Real Solutions

x = –b ± b 2 – 4ac

2a

x = –1 ± 1 2 – 4(2)(–5)

2(2)

x = –1 ± 414

Solve 2x 2 + x = 5.

SOLUTION

The solutions are:

1.35

CHECK Graph y = 2 x 2 + x – 5 and note that the x-intercepts are about 1.35 and about –1.85.

–1.85

Solving a Quadratic Equation with Two Real Solutions

and x =–1 – 41

4–1 + 41

4x =

Solving a Quadratic Equation with One Real Solution

Solve x 2 – x = 5x – 9.

SOLUTION

Write original equation. x 2 – x = 5x – 9

x 2 – 6x + 9 = 0 a = 1, b = –6, c = 9

Quadratic formula

Simplify.

x = 3

The solution is 3.

Simplify.

x = 6 ± 0

2

x = 6 ± (– 6) 2 – 4(1)(9)

2(1)

Solving a Quadratic Equation with One Real Solution

Solve x 2 – x = 5x – 9.

CHECK Graph y = x 2 – 6 x + 9 and note that the only x -intercept is 3. Alternatively, substitute 3 for x in the original equation.

6 = 6

3 2 – 3 = 5(3) – 9?

9 – 3 = 15 – 9?

Solving a Quadratic Equation with Two Imaginary Solutions

Solve –x 2 + 2x = 2.

SOLUTION

Write original equation. –x 2 + 2x = 2

–x 2 + 2x – 2 = 0 a = –1, b = 2, c = –2

Quadratic formula

Simplify.

x = 1 ± i

x = –2 ± 2 i

–2Write using the imaginary unit i.

The solutions are 1 + i and 1 – i.

Simplify.

x = –2 ± – 4

–2

x = –2 ± 2 2 – 4(–1)(–2)

2(–1)

Solving a Quadratic Equation with Two Imaginary Solutions

Solve –x 2 + 2x = 2.

SOLUTION

CHECK Graph y = –x 2 + 2 x – 2 and note that there are no x-intercepts. So, the original equation has no real solutions. To check the imaginary solutions 1 + i and 1 – i, substitute them into the original equation. The check for 1 + i is shown.

2 = 2

–(1 + i ) 2 + 2(1 + i ) = 2?

–2i + 2 + 2i = 2?

In the quadratic formula, the expression b 2 – 4 a c under the radical sign is called the discriminant of the associated equation a x 2 + b x + c = 0.

Consider the quadratic equation a x 2 + b x + c = 0.

NUMBER AND TYPE OF SOLUTIONS OF A QUADRATIC EQUATION

discriminant

You can use the discriminant of a quadratic equation to determine the equation’s number and type of solutions.

If b 2 – 4 a c > 0, then the equation has two real solutions.

Solving Equations with the Quadratic Formula

If b 2 – 4 a c = 0, then the equation has one real solution.

If b 2 – 4 a c < 0, then the equation has two imaginary solutions.

–b ± b 2 – 4ac2a

x =

Find the discriminant of the quadratic equation and give the number and type of solutions of the equation.

SOLUTION

x 2 – 6 x + 10 = 0

x 2 – 6 x + 10 = 0 Two imaginary: 3 ± i

a x + b x + c = 0

EQUATION DISCRIMINANT SOLUTION(S)

b 2 – 4a c

(–6) 2 – 4(1)(10) = – 4

Using the Discriminant

x = –b ± b 2 – 4a c

2a


x 2 – 6 x + 10 = 0


a x + b x + c = 0


b 2 – 4a c

(–6) 2 – 4(1)(10) = – 4


x 2 – 6 x + 9 = 0 (–6) 2 – 4(1)(9) = 0 One real: 3

x 2 – 6 x + 9 = 0

SOLUTION

x = –b ± b 2 – 4a c

2a


x 2 – 6 x + 10 = 0


a x + b x + c = 0


b 2 – 4a c

(–6) 2 – 4(1)(10) = – 4


x 2 – 6 x + 9 = 0 (–6) 2 – 4(1)(9) = 0 One real: 3

x 2 – 6 x + 9 = 0 x 2 – 6 x + 8 = 0

x 2 – 6 x + 8 = 0 (–6) 2 – 4(1)(8) = 4 Two real: 2, 4

SOLUTION

x = –b ± b 2 – 4a c

2a

In the previous example you may have noticed that the number of real solutions of x 2 – 6 x + c = 0 can be changed just by changing the value of c.

y = x 2 – 6 x + 10

y = x 2 – 6 x + 9

y = x 2 – 6 x + 8

Graph is above x-axis (no x-intercept)

Graph touches x-axis (one x-intercept)

Graph crosses x-axis (two x-intercepts)


A graph can help you see why this occurs. By changing c, you can move the graph of y = x 2 – 6x + c up or down in the coordinate plane.

If the graph is moved too high, it won’t have an x-intercept and the equation x 2 – 6 x + c = 0 won’t have a real-number solution.

Using the Quadratic Formula in Real Life

Previously you studied the model h = –16t 2 + h 0 for the height of an object that is dropped. For an object that is launched or thrown, an extra term v 0 t must be added to the model to account for the object’s initial vertical velocity v 0.

Models

h = –16 t 2 + h 0

h = –16 t 2 + v 0 t + h 0

Object is dropped.

Object is launched or thrown.

Labels

h = height

t = time in motion

h 0 = initial height

v 0 = initial vertical velocity

(feet)

(seconds)

(feet)

(feet per second)

Using the Quadratic Formula in Real Life

The initial velocity of a launched object can be positive, negative, or zero.

v 0 > 0 v 0 < 0 v 0 = 0

If the object is launched upward, its initial vertical velocity is positive (v 0 > 0).

If the object is launched downward, its initial vertical velocity is negative (v 0 < 0).

If the object is launched parallel to the ground, its initial vertical velocity is zero (v 0 = 0).

Solving a Vertical Motion Problem

A baton twirler tosses a baton into the air. The baton leaves the twirler’s hand 6 feet above the ground and has an initial vertical velocity of 45 feet per second. The twirler catches the baton when it falls back to a height of 5 feet. For how long is the baton in the air?

SOLUTION

h = –16 t 2 + v 0 t + h 0 Write height model.

h = 5, v 0 = 45, h 0=6

Since the baton is thrown (not dropped), use the model h = –16t 2 + v 0 t + h 0 with v

0 = 45 and h 0 = 6. To determine how long the baton is in the air, find the value of

t for which h = 5.

5 = –16 t 2 + 45t + 6

0 = –16 t 2 + 45t + 1 a = –16, b = 45, c = 1

Solving a Vertical Motion Problem

A baton twirler tosses a baton into the air. The baton leaves the twirler’s hand 6 feet above the ground and has an initial vertical velocity of 45 feet per second. The twirler catches the baton when it falls back to a height of 5 feet. For how long is the baton in the air?

t – 0.022 or t 2.8

Quadratic formula

a = –16, b = 45, c = 1

Use a calculator.

SOLUTION

Since the baton is thrown (not dropped), use the model h = –16t 2 + v 0 t + h 0 with v

0 = 45 and h 0 = 6. To determine how long the baton is in the air, find the value of

t for which h = 5.

Reject the solution – 0.022 since the baton’s time in the air cannot be negative. The baton is in the air for about 2.8 seconds.

t = – 45 ± 2089

–32

Solving Equations with the Quadratic Formula By completing the square once for the general equation a x 2 + b x + c = 0, you can develop a formula that.

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