College AlgebraSixth EditionJames Stewart Lothar Redlin Saleem Watson

Equations and

Inequalities1

SOLVING QUADRATIC

EQUATIONS1.6

Linear Equations vs. Quadratic Equations

Linear equations are first-degree equations,

such as:

2x + 1 = 5 or 4 – 3x = 2

Quadratic equations are second-degree

equations, such as:

x2 + 2x – 3 = 0 or 2x2 + 3 = 5x

Quadratic Equation—Definition

A quadratic equation is an equation of the

form

ax2 + bx + c = 0

where a, b, and c are real numbers with a ≠ 0.

Solving Quadratic Equations

by Factoring

Solving Quadratic Equations

Some quadratic equations can be solved

by factoring and using the following

basic property of real numbers.

Zero-Product Property

AB = 0 if and only if

A = 0 or B = 0

• This means that, if we can factor the LHS of a quadratic (or other) equation, then we can solve it by setting each factor equal to 0 in turn.

• This method works only when the RHS is 0.

E.g. 1—Solving a Quadratic Equation by Factoring

Solve the equation

x2 + 5x = 24

• We must first rewrite the equation so that the RHS is 0.

E.g. 1—Solving a Quadratic Equation by Factoring

x2 + 5x = 24

x2 + 5x – 24 = 0 (Subtract 24)

(x – 3)(x + 8) = 0 (Factor)

x – 3 = 0 or x + 8 = 0 (Zero-Product Property)

x = 3 x = –8 (Solve)

Solving a Quadratic Equation by Factoring

Do you see why one side of the equation

must be 0 in Example 1?

• Factoring the equation as x(x + 5) = 24 does not help us find the solutions.

• 24 can be factored in infinitely many ways, such as 6 . 4, ½ . 48, (–2/5) . (–60), and so on.

Solving Quadratic Equations by

Completing the Square

Solving Simple Quadratics

As we saw in Section P.8, Example 5(b), if a

quadratic equation is of the form (x ± a)2 = c,

we can solve it by taking the square root of

each side.

• In an equation of this form, the LHS is a perfect square: the square of a linear expression in x.

Completing the Square

So, if a quadratic equation does not factor

readily, we can solve it using the technique

of completing the square.

Completing the Square

To make x2 + bx a perfect square, add ,

the square of half the coefficient of x.

This gives the perfect

square

2

2

b

2 22

2 2

b bx bx x

Completing the Square

To compete the square, we add a constant to

an expression to make it a perfect square.

• For example, to make

x2 + 6x

a perfect square, we must add (6/2)2 = 9. Then

x2 + 6x + 9 = (x + 3)2

is a perfect square.

Completing the Square

The table gives some more examples of

completing the square.

E.g. Solving Quadratic Equations by Completing the Square

Solve each equation.

(a) x2 – 8x + 13 = 0

(b) 3x2 – 12x + 6 = 0

E.g. 2—Completing the Square

x2 – 8x + 13 = 0

x2 – 8x = –13 (Subtract 13)

x2 – 8x + 16 = –13 + 16 (Complete the square)

(x – 4)2 = 3 (Perfect square)

x – 4 = ± (Take square root)

x = 4 ± (Add 4)

Example (a)

3

3

E.g. 2—Completing the Square

First, we subtract 6 from each side.

Then, we factor the coefficient of x2 (the 3)

from the left side.

• This puts the equation in the correct form for completing the square.

Example (b)

E.g. 2—Completing the Square

3x2 – 12x + 6 = 0

3x2 – 12x = –6 (Subtract 6)

3(x2 – 4x) = –6 (Factor 3 from LHS)

E.g. 2—Completing the Square

Now, we complete the square by adding

(–2)2 = 4 inside the parentheses.

• Since everything inside the parentheses is multiplied by 3, this means that we are actually adding 3 . 4 = 12 to the left side of the equation.

• Thus, we must add 12 to the right side as well.

E.g. 2—Completing the Square

3(x2 – 4x + 4) = –6 + 3 . 4 (Complete the square)

3(x – 2)2 = 6 (Perfect square)

(x – 2)2 = 2 (Divide by 3)

(Take square root)

(Add 2)

2 = 2x

= 2 2x

The Quadratic Formula

Deriving a Formula for the Roots

We can use the technique of completing

the square to derive a formula for the roots

of the general quadratic equation

ax2 + bx + c = 0

The Quadratic Formula

The roots of the quadratic equation

ax2 + bx + c = 0

where a ≠ 0, are:

2 4

2

b b acx

a

The Quadratic Formula—Proof

First, we divide each side of the equation

by a and move the constant to the right

side, giving:

2 b cx x

a a

The Quadratic Formula—Proof

2 22

2 2

2

2

2

(Complete the square)

(Perfect square)

(Take square root)

(Subtract )2

2 2

4

2 4

4

2 2

4

2

b

a

b b c bx x

a a a a

b ac bx

a a

b b acx

a a

b b acx

a

We now complete the square by adding

(b/2a)2 to each side of the equation.

The Quadratic Formula

The quadratic formula could be used to

solve the equations in Examples 1 and 2.

• You should carry out the details of these calculations.

E.g. 3—Using the Quadratic Formula

Find all solutions of each equation.

(a) 3x2 – 5x – 1 = 0

(b) 4x2 + 12x + 9 = 0

(c) x2 + 2x + 2 = 0

E.g. 3—Using Quadratic Formula

In 3x2 – 5x – 1 = 0,

a = 3 b = –5 c = –1

By the quadratic formula,

Example (a)

25 5 4 3 1 5 37

2 3 6x

E.g. 3—Using Quadratic Formula

If approximations are desired, we can use

a calculator to obtain:

5 371.8471

6

5 370.1805

6

x

x

Example (a)

E.g. 3—Using Quadratic Formula

Using the quadratic formula

with a = 4, b = 12, and c = 9 gives:

• This equation has only one solution, x = –3/2.

Example (b)

212 12 4 4 9 12 0 3

2 4 8 2x

E.g. 3—Using Quadratic Formula

Using the quadratic formula

with a = 1, b = 2, and c = 2 gives:

• Since the square of a real number is nonnegative, is undefined in the real number system.

• The equation has no real solution.

Example (c)

22 2 4 2 2 4 2 2 1

2 2 2

1 1

x

1

The Discriminant

Discriminant

The quantity b2 – 4ac that appears under

the square root sign in the Quadratic Formula

is called the discriminant of the equation

ax2 + bx + c = 0.

• It is given the symbol D.

D < 0

If D < 0, then is undefined.

• The quadratic equation has no real solution—as in Example 3(c).

2 4b ac

D = 0 and D > 0

If D = 0, then the equation has only one real

solution—as in Example 3(b).

Finally, if D > 0, then the equation has two

distinct real solutions—as in Example 3(a).

Discriminant

The following box summarizes these

observations.

E.g. 4—Using the Discriminant

Use the discriminant to determine how many

real solutions each equation has.

(a) x2 + 4x – 1 = 0

(b) 4x2 – 12x + 9 = 0

(c) 1/3x2 – 2x + 4 = 0

E.g. 4—Using the Discriminant

x2 + 4x – 1 = 0

The discriminant is:

D = 42 – 4(1)(–1) = 20 > 0

• So, the equation has two distinct real solutions.

Example (a)

E.g. 4—Using the Discriminant

4x2 – 12x + 9 = 0

The discriminant is:

D = (–12)2 – 4 . 4 . 9 = 0

• So, the equation has exactly one real solution.

Example (b)

E.g. 4—Using the Discriminant

1/3x2 – 2x + 4 = 0

The discriminant is:

D = (–2)2 – 4(1/3)4 = –4/3 < 0

• So, the equation has no real solution.

Example (c)

Modeling with Quadratic

Equations

Quadratic Equations in Real Life

Let’s consider a real-life situation that

can be modeled by a quadratic

equation.

• The principles discussed in Section 1.5 for setting up equations as models are useful here as well.

E.g. 6—Dimensions of a Building Lot

A rectangular building lot is 8 ft

longer than it is wide and has an area

of 2900 ft2.

• Find the dimensions of the lot.

E.g. 6—Dimensions of a Building Lot

We are asked to find the width and

length of the lot.

• So, let: w = width of lot

E.g. 6—Dimensions of a Building Lot

Then, we translate

the information

in the problem

into the language

of algebra.

In Words In Algebra

Width of lot w

Length of lot w + 8

E.g. 6—Dimensions of a Building Lot

Now, we set up the model.

Width of Lot . Length of Lot

= Area of Lot

E.g. 6—Dimensions of a Building Lot

w(w + 8) = 2900

w2 + 8w = 2900 (Expand)

w2 + 8w – 2900 = 0

(w – 50)(w + 58) = 0 (Factor)

w = 50 or w = –58 (Zero-Product Property)

E.g. 6—Dimensions of a Building Lot

Since the width of the lot must be a positive

number, we conclude that:

w = 50 ft

The length of the lot is:

w + 8 = 50 + 8 = 58 ft

E.g. 7—A Distance-Speed-Time Problem

A jet flew from New York (NY) to Los Angeles

(LA), a distance of 4,200 km.

The speed for the return trip was 100 km/h

faster than the outbound speed.

• If the total trip took 13 hours, what was the jet’s speed from NY to LA?

E.g. 7—A Distance-Speed-Time Problem

We are asked for the speed of the jet

from NY to LA.

• So, let: s = speed from NY to LA

• Then, s + 100 = speed from LA to NY

E.g. 7—A Distance-Speed-Time Problem

Now, we organize the information in a table.

First, we fill in the “Distance” column—as

we know that the cities are 4,200 km apart.

E.g. 7—A Distance-Speed-Time Problem

Then, we fill in the “Speed” column—as

we have expressed both speeds (rates)

in terms of the variable s.

E.g. 7—A Distance-Speed-Time Problem

Finally, we calculate the entries for

the “Time” column, using:distance

time = rate

E.g. 7—A Distance-Speed-Time Problem

The total trip took 13 hours.

So, we have the model:

Time from NY to LA

+ Time from LA to NY

= Total time

• This gives:4200 4200

13100s s

E.g. 7—A Distance-Speed-Time Problem

Multiplying by the common denominator,

s(s + 100), we get:

• Although this equation does factor, with numbers this large, it is probably quicker to use the quadratic formula and a calculator.

2

2

4200 100 4200 13 100

8400 420,000 13 1300

0 13 7100 420,000

s s s s

s s s

s s

E.g. 7—A Distance-Speed-Time Problem

• As s represents speed, we reject the negative answer and conclude that the jet’s speed from NY to LA was 600 km/h.

27100 7100 4 13 420,000

2 13

7100 8500

261400

600 or 53.826

s

s s

E.g. 8—The Path of a Projectile

An object thrown or fired straight upward

at an initial speed of v0 ft/s will reach a height

of h feet after t seconds, where h and t are

related by the formula

h = –16t2 + v0t

E.g. 8—The Path of a Projectile

Suppose that a bullet is shot straight

upward with an initial speed of 800 ft/s.

a) When does the bullet fall back to ground level?

b) When does it reach a height of 6,400 ft?

c) When does it reach a height of 2 mi?

d) How high is the highest point the bullet reaches?

E.g. 8—The Path of a Projectile

The initial speed is v0 = 800 ft/s.

Thus, the formula is:

h = –16t2 + 800t

E.g. 8—The Path of a Projectile

Ground level corresponds

to h = 0.

So, we must solve:

0 = –16t2 + 800t (Set h = 0)

0 = –16t(t – 50) (Factor)

• Thus, t = 0 or t = 50.• This means the bullet starts (t = 0) at ground level

and returns to ground level after 50 s.

Example (a)

E.g. 8—The Path of a Projectile

Setting h = 6400 gives:

6400 = –16t2 + 800t

(Set h = 6400)

16t2 – 800t + 6400 = 0

(All terms to LHS)

t2 – 50t + 400 = 0

(Divide by 16)

Example (b)

E.g. 8—The Path of a Projectile

(t – 10)(t – 40) = 0(Factor)

t = 10 or t = 40 (Solve)

• The bullet reaches 6400 ft after 10 s (on its ascent) and again after 40 s (on its descent to earth).

Example (b)

E.g. 8—The Path of a Projectile

Two miles is:

2 x 5,280 = 10,560 ft

10,560 = –16t2 + 800t

(Set h = 10,560)

16t2 – 800t + 10,560 = 0

(All terms to LHS)

t2 – 50t + 660 = 0 (Divide by 16)

Example (c)

E.g. 8—The Path of a Projectile

The discriminant of this

equation is:

D = (–50)2 – 4(660) = –140

• It is negative.

• Thus, the equation has no real solution.

• The bullet never reaches a height of 2 mi.

Example (c)

E.g. 8—The Path of a Projectile

Each height the bullet

reaches is attained twice—

once on its ascent and once

on its descent.

• The only exception is the highest point of its path, which is reached only once.

Example (d)

E.g. 8—The Path of a Projectile

This means that, for

the highest value of h,

the following equation has

only one solution for t:

h = –16t2 + 800t

16t2 – 800t + h = 0

• This, in turn, means that the discriminant D of the equation is 0.

Example (d)

E.g. 8—The Path of a Projectile

So,

D = (–800)2 – 4(16)h = 0

640,000 – 64h = 0

h = 10,000

• The maximum height reached is 10,000 ft.

Example (d)