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Section 1.3 Quadratic Equations

Before Starting this Section, Review

1 Factoring (Section P.4 )
2 Square roots (Section P.6 , page 61)
3 Rationalizing (Section P.6 , page 67)
4 Linear equations (Section 1.1 , page 83)

Objectives

1 Solve a quadratic equation by factoring.
2 Solve a quadratic equation by the square root method.
3 Solve a quadratic equation by completing the square.
4 Solve a quadratic equation by using the quadratic formula.
5 Solve applied problems.

The Golden Rectangle

The ancient Greeks, in about the sixth century B.C., sought unifying principles of beauty and perfection, which they believed could be described mathematically. In their study of beauty, the Greeks used the term golden ratio. The golden ratio is frequently referred to as “phi,” after the Greek sculptor Phidias, and is symbolized by the Greek letter Φ $Φ$ (phi). A geometric figure that is commonly associated with phi is the golden rectangle. (See Example 11.) This rectangle has sides of lengths a and b that are in proportion to the golden ratio. It has been said that the golden rectangle is the most pleasing rectangle to the eye. Numerous examples of art and architecture have employed the golden rectangle. Here are three of them:

The exterior dimensions of the Parthenon in Athens, built about 440 B.C., form a perfect golden rectangle.
Leonardo Da Vinci called the golden ratio the “divine proportion” and featured it in many of his paintings, including the famous Mona Lisa. (Try drawing a rectangle around her face.)
The Great Pyramid of Giza is believed to be 4600 years old, a time long before the Greeks. Its dimensions are also based on the golden ratio.

In Example 11, we show that the numerical value of the golden ratio can be found by solving the quadratic equation x2−x−1=0. $x^{2} - x - 1 = 0 .$

A quadratic equation written in the form ax2+bx+c=0 $a x^{2} + b x + c = 0$ is said to be in standard form. Here are some examples of quadratic equations that are not in standard form.

2 x 2 - 5 x + 7 3 y 2 + 4 y + 1 x 2 - 2 x = = = x 2 + 3 x - 1 6 y - 5 3

$\begin{array}{rcl} 2 x^{2} - 5 x + 7 & = & x^{2} + 3 x - 1 \\ 3 y^{2} + 4 y + 1 & = & 6 y - 5 \\ x^{2} - 2 x & = & 3 \end{array}$

We discuss four methods for solving quadratic equations: (1) by factoring, (2) by taking square roots, (3) by completing the square, and (4) by using the quadratic formula.

We begin by studying real number solutions of quadratic equations.

Factoring Method

1 Solve a quadratic equation by factoring.

Some quadratic equations in the standard form ax2+bx+c=0 $a x^{2} + b x + c = 0$ can be solved by factoring and using the zero-product property.

For a product of real numbers, a similar property was stated in Section P.1.

Example 1 Solving a Quadratic Equation by Factoring

Solve by factoring: 2x2+5x=3 $2 x^{2} + 5 x = 3$

Solution

2 x 2 + 5 x 2 x 2 + 5 x - 3 (2 x - 1) (x + 3) = = = 300 Original equation Subtract 3 from both sides to get 0 on one side . Factor the left side .

$\begin{array}{rcll} 2 x^{2} + 5 x & = & 3 & Original equation \\ 2 x^{2} + 5 x - 3 & = & 0 & Subtract 3 from both sides to get 0 on one side . \\ (2 x - 1) (x + 3) & = & 0 & Factor the left side . \end{array}$

We now set each factor equal to 0 and solve the resulting linear equations. The vertical line segment separates the computations leading to the two solutions.

2 x - 1 2 x x = = = 01 1 2 x + 3 x x = = = 0 - 3 - 3 Zero-product property Isolate the x term on one side . Solve for x .

$\begin{array}{rclrcll} 2 x - 1 & = & 0 & x + 3 & = & 0 & Zero-product property \\ 2 x & = & 1 & x & = & - 3 & Isolate the x term on one side . \\ x & = & \frac{1}{2} & x & = & - 3 & Solve for x . \end{array}$

Check: You should check the solutions x=12 $x = \frac{1}{2}$ and x=−3 $x = - 3$ in the original equation.

The solution set is {12, −3}. ${\frac{1}{2}, - 3} .$

Practice Problem 1

Solve by factoring: x2+25x=−84 $x^{2} + 25 x = - 84$

Side Note

To use the zero-product property to solve an equation, you must make sure that one side of the equation is 0.

Example 2 Solving a Quadratic Equation by Factoring

Solve by factoring: 3t2=2t $3 t^{2} = 2 t$

Solution

3 t 2 3 t 2 - 2 t t (3 t - 2) = = = 2 t 00 Original equation Subtract 2 t from both sides . Factor . t = 0 3 t - 2 3 t t = = = 02 2 3 Zero-product property Isolate the t terms on one side . Solve for t .

$\begin{array}{l} \begin{array}{rcll} 3 t^{2} & = & 2 t & Original equation \\ 3 t^{2} - 2 t & = & 0 & Subtract 2 t from both sides . \\ t (3 t - 2) & = & 0 & Factor . \end{array} \\ \begin{array}{rclrcll} t & = & 0 & 3 t - 2 & = & 0 & Zero-product property \\ 3 t & = & 2 & Isolate the t terms on one side . \\ t & = & \frac{2}{3} & Solve for t . \end{array} \end{array}$

Check: You should check the solutions t=0 $t = 0$ and t=23 $t = \frac{2}{3}$ in the original equation.

The solution set is {0, 23}. ${0, \frac{2}{3}} .$

Practice Problem 2

Solve by factoring: 2m2=5m $2 m^{2} = 5 m$

Warning

The factoring method of solving an equation works only when one of the sides of the equation is 0. If neither side of an equation is 0, as in (x+1)(x−3)=10, $(x + 1) (x - 3) = 10,$ we cannot know the value of either factor. For example, we could have x+1=5 $x + 1 = 5$ and x−3=2, $x - 3 = 2,$ or x+1=0.1 $x + 1 = 0.1$ and x−3=100. $x - 3 = 100 .$ There are countless possibilities for two factors whose product is 10.

Example 3 Solving a Quadratic Equation with One Solution

Solve by factoring: x2+16=8x $x^{2} + 16 = 8 x$

Solution

Step 1 Write the equation in standard form.

$x 2 + 16 x 2 + 16 - 8 x x 2 - 8 x + 16 = = = 8 x 00 Original equation Subtract 8 x from both sides . Standard form$ $\begin{array}{rcll} x^{2} + 16 & = & 8 x & Original equation \\ x^{2} + 16 - 8 x & = & 0 & Subtract 8 x from both sides . \\ x^{2} - 8 x + 16 & = & 0 & Standard form \end{array}$
Step 2 Factor the left side of the equation.

$(x - 4) (x - 4) = 0 Perfect-square trinomial$ $\begin{array}{l} (x - 4) (x - 4) = 0 & Perfect-square trinomial \end{array}$
Steps 3 and 4 Set each factor equal to 0 and solve each resulting equation.

$x - 4 x = = 04 x - 4 x = = 04 Zero-product property$ $\begin{array}{rclrcll} x - 4 & = & 0 & x - 4 & = & 0 & Zero-product property \\ x & = & 4 & x & = & 4 \end{array}$
Step 5 Check the solution in the original equation.

Check: x=4 $x = 4$ because x=4 $x = 4$ is the only possible solution.

$x 2 + 16 (4) 2 + 16 32 = = ? = 8 x 8 (4) 32 ✓ Original equation Replace x with 4 .$ $\begin{array}{rcll} x^{2} + 16 & = & 8 x & Original equation \\ {(4)}^{2} + 16 & \overset{?}{=} & 8 (4) & Replace x with 4 . \\ 32 & = & 32 ✓ \end{array}$

Thus, 4 is the only solution of x2+16=8x $x^{2} + 16 = 8 x$ and the solution set is {4}.

Practice Problem 3

Solve by factoring: x2−6x=−9 $x^{2} - 6 x = - 9$

In Example 3, because the factor (x−4) $(x - 4)$ appears twice in the solution, the number 4 is called a double root, or a root of multiplicity 2, of the given equation.

The Square Root Method

2 Solve a quadratic equation by the square root method.

The solutions of an equation of the type x2=d $x^{2} = d$ are obtained by the square root method. Suppose we are interested in solving the equation x2=3, $x^{2} = 3,$ which is equivalent to the equation x2−3=0. $x^{2} - 3 = 0 .$ Applying the factoring method but removing the restriction that coefficients and constants represent only integers, we have

x 2 - 3 x 2 - (3 - \sqrt) 2 (x + 3 - \sqrt) (x - 3 - \sqrt) = = = 000 x + 3 - \sqrt x = = 0 - 3 - \sqrt or or x - 3 - \sqrt x = = 0 3 - \sqrt 3 = (3 - \sqrt) 2 a 2 - b 2 = (a - b) (a + b) Set each factor equal to zero . Solve for x .

$\begin{array}{l} \begin{array}{rcl} x^{2} - 3 & = & 0 \\ x^{2} - {(\sqrt{3})}^{2} & = & 0 \\ (x + \sqrt{3}) (x - \sqrt{3}) & = & 0 \end{array} & \begin{array}{l} 3 = {(\sqrt{3})}^{2} \\ a^{2} - b^{2} = (a - b) (a + b) \end{array} \\ \begin{array}{rcllrcl} x + \sqrt{3} & = & 0 & or & x - \sqrt{3} & = & 0 \\ x & = & - \sqrt{3} & or & x & = & \sqrt{3} \end{array} & \begin{array}{l} Set each factor equal to zero . \\ Solve for x . \end{array} \end{array}$

Thus, the solutions of the equation x2=3 $x^{2} = 3$ are −3–√ $- \sqrt{3}$ and 3–√. $\sqrt{3} .$ Because the solutions differ only in sign, we use the notation ±3–√ $\pm \sqrt{3}$ to write the two numbers 3–√ $\sqrt{3}$ and −3–√. $- \sqrt{3} .$ Similarly, for any nonnegative real number d, the solutions of the quadratic equation x2=d $x^{2} = d$ are ±d−−√. $\pm \sqrt{d} .$

Side Note

If d<0 $d < 0$ , then the equation u2=d $u^{2} = d$ has no real solutions.

Example 4 Solving an Equation by the Square Root Method

Solve: (x−3)2=5 ${(x - 3)}^{2} = 5$

Solution

(x - 3) 2 x - 3 x = = = 5 \pm 5 - \sqrt 3 \pm 5 - \sqrt Original equation Square root property Add 3 to both sides .

$\begin{array}{rcll} {(x - 3)}^{2} & = & 5 & Original equation \\ x - 3 & = & \pm \sqrt{5} & Square root property \\ x & = & 3 \pm \sqrt{5} & Add 3 to both sides . \end{array}$

The solution set is {3−5–√, 3+5–√}. ${3 - \sqrt{5}, 3 + \sqrt{5}} .$

Practice Problem 4

Solve: (x+2)2=5 ${(x + 2)}^{2} = 5$

Completing the Square

3 Solve a quadratic equation by completing the square.

We use a method called completing the square to solve quadratic equations that cannot be solved by factoring and are not in the correct form to use the square root method. The method requires two steps. First, we write a given quadratic equation in the form (x+k)2=d. ${(x + k)}^{2} = d .$ Then we solve the equation (x+k)2=d ${(x + k)}^{2} = d$ by the square root method.

Recall that

(x + k) 2 = x 2 + 2 k x + k 2 .

${(x + k)}^{2} = x^{2} + 2 k x + k^{2} .$

Notice that the constant term, k2, $k^{2},$ in the trinomial x2+2kx+k2 $x^{2} + 2 k x + k^{2}$ is the square of one-half the coefficient of x That is, k2=[12(2k)]2. $k^{2} = {[\frac{1}{2} (2 k)]}^{2} .$

To convert the expression x2+bx $x^{2} + b x$ to a perfect square, we add a constant term that results in a perfect-square trinomial.

Here are some examples that show the number that should be added to x2+bx $x^{2} + b x$ to create a perfect-square trinomial. Note that b may be negative.

x2+bx $x^{2} + b x$	`b`	b2 $\frac{b}{2}$	Add (b2)2 ${(\frac{b}{2})}^{2}$	Result is (x+b2)2 ${(x + \frac{b}{2})}^{2}$
x2+6x $x^{2} + 6 x$	6	3	32=9 $3^{2} = 9$	x2+6x+9=(x+3)2 $x^{2} + 6 x + 9 = {(x + 3)}^{2}$
x2−4x $x^{2} - 4 x$	−4 $- 4$	−2 $- 2$	(−2)2=4 ${(- 2)}^{2} = 4$	x2−4x+4=(x−2)2 $x^{2} - 4 x + 4 = {(x - 2)}^{2}$
x2+3x $x^{2} + 3 x$	3	32 $\frac{3}{2}$	(32)2=94 ${(\frac{3}{2})}^{2} = \frac{9}{4}$	x2+3x+94=(x+32)2 $x^{2} + 3 x + \frac{9}{4} = {(x + \frac{3}{2})}^{2}$
x2−x $x^{2} - x$	−1 $- 1$	−12 $- \frac{1}{2}$	(−12)2=14 ${(- \frac{1}{2})}^{2} = \frac{1}{4}$	x2−x+14=(x−12)2 $x^{2} - x + \frac{1}{4} = {(x - \frac{1}{2})}^{2}$

Thus, to make x2+bx $x^{2} + b x$ into a perfect square, we need to add

[1 2 (coefficient of x)] 2 = [1 2 (b)] 2 = b 2 4

${[\frac{1}{2} (coefficient of x)]}^{2} = {[\frac{1}{2} (b)]}^{2} = \frac{b^{2}}{4}$

so that

x 2 + b x + b 2 4 = (x + b 2) 2 .

$x^{2} + b x + \frac{b^{2}}{4} = {(x + \frac{b}{2})}^{2} .$

We say that b24 $\frac{b^{2}}{4}$ was added to x2+bx $x^{2} + b x$ to complete the square. See Figure 1.2.

Figure 1.2

Area of a square=(x+b2)2 $Area of a square = {(x + \frac{b}{2})}^{2}$

Example 5 Solving a Quadratic Equation by Completing the Square

Solve by completing the square: x2+10x+8=0 $x^{2} + 10 x + 8 = 0$

Solution

First, isolate the constant on the right side.

x 2 + 10 x + 8 x 2 + 10 x = = 0 - 8 Original equation Subtract 8 from both sides .

$\begin{array}{rcll} x^{2} + 10 x + 8 & = & 0 & Original equation \\ x^{2} + 10 x & = & - 8 & Subtract 8 from both sides . \end{array}$

Next, add [12(10)]2=52=25 ${[\frac{1}{2} (10)]}^{2} = 5^{2} = 25$ to both sides to complete the square on the left side.

x 2 + 10 x + 25 (x + 5) 2 x + 5 x = = = = - 8 + 25 17 \pm 17 - - \sqrt - 5 \pm 17 - - \sqrt Add 25 to each side . x 2 + 10 x + 25 = (x + 5) 2 Square root property Subtract 5 from each side .

$\begin{array}{rcll} x^{2} + 10 x + 25 & = & - 8 + 25 & Add 25 to each side . \\ {(x + 5)}^{2} & = & 17 & x^{2} + 10 x + 25 = {(x + 5)}^{2} \\ x + 5 & = & \pm \sqrt{17} & Square root property \\ x & = & - 5 \pm \sqrt{17} & Subtract 5 from each side . \end{array}$

Thus, x=−5+17−−√ $x = - 5 + \sqrt{17}$ or x=−5−17−−√, $x = - 5 - \sqrt{17},$ and the solution set is {−5+17−−√, ${- 5 + \sqrt{17},$ −5−17−−√}. $- 5 - \sqrt{17}} .$

Practice Problem 5

Solve by completing the square: x2−6x+7=0 $x^{2} - 6 x + 7 = 0$

Example 6 Solving a Quadratic Equation by Completing the Square

Solve by completing the square: 3x2−4x−1=0 $3 x^{2} - 4 x - 1 = 0$

Solution

Step 1 3x2−4x−13x2−4x==01Note that the coefficient of x2 is 3.Add 1 to both sides. $\begin{array}{rcll} 3 x^{2} - 4 x - 1 & = & 0 & Note that the coefficient of x^{2} is 3 . \\ 3 x^{2} - 4 x & = & 1 & Add 1 to both sides . \end{array}$
Step 2 x2−43x=13Divide both sides by 3. $\begin{array}{l} x^{2} - \frac{4}{3} x = \frac{1}{3} & Divide both sides by 3 . \end{array}$
Step 3 x2−43x+(−23)2=13+ (−23)2Add [12(−43)]2=(−23)2 to both sides. $\begin{array}{l} x^{2} - \frac{4}{3} x + {(- \frac{2}{3})}^{2} = \frac{1}{3} + {(- \frac{2}{3})}^{2} & Add {[\frac{1}{2} (- \frac{4}{3})]}^{2} = {(- \frac{2}{3})}^{2} to both sides . \end{array}$
Step 4 (x−23)2=79x2−43x+(−23)2=(x−23)2 $\begin{array}{l} {(x - \frac{2}{3})}^{2} = \frac{7}{9} & x^{2} - \frac{4}{3} x + {(- \frac{2}{3})}^{2} = {(x - \frac{2}{3})}^{2} \end{array}$
Step 5 x−23x−23==±79−−√±7–√3Take the square root of both sides.79−−√=7–√9–√=7–√3 $\begin{array}{l} x - \frac{2}{3} & = & \pm \sqrt{\frac{7}{9}} & Take the square root of both sides . \\ x - \frac{2}{3} & = & \pm \frac{\sqrt{7}}{3} & \sqrt{\frac{7}{9}} = \frac{\sqrt{7}}{\sqrt{9}} = \frac{\sqrt{7}}{3} \end{array}$
Step 6 xx==23±7–√32±7–√3Add 23 to both sides.Add fractions. $\begin{array}{rcll} x & = & \frac{2}{3} \pm \frac{\sqrt{7}}{3} & Add \frac{2}{3} to both sides . \\ x & = & \frac{2 \pm \sqrt{7}}{3} & Add fractions . \end{array}$

The solution set is {2−7–√3, 2+7–√3}. ${\frac{2 - \sqrt{7}}{3}, \frac{2 + \sqrt{7}}{3}} .$

Practice Problem 6

Solve by completing the square: 4x2−24x+25=0 $4 x^{2} - 24 x + 25 = 0$

The Quadratic Formula

4 Solve a quadratic equation by using the quadratic formula.

We can generalize the method of completing the square to derive a formula that gives a solution of any quadratic equation.

We solve a quadratic equation in standard form by completing the square.

a x 2 + b x + c = 0, a \neq 0

$a x^{2} + b x + c = 0, a \neq 0$

Step 1 ax2+bx=−cIsolate the constant on the right side. $\begin{array}{l} a x^{2} + b x = - c & Isolate the constant on the right side . \end{array}$
Step 2 x2+bax=−caDivide both sides by a. $\begin{array}{l} x^{2} + \frac{b}{a} x = - \frac{c}{a} & Divide both sides by a . \end{array}$
Step 3 x2+bax+(b2a)2=−ca+(b2a)2Add the square of one-half thecoefficient of x to both sides. $\begin{array}{l} x^{2} + \frac{b}{a} x + {(\frac{b}{2 a})}^{2} = - \frac{c}{a} + {(\frac{b}{2 a})}^{2} & \begin{array}{l} Add the square of one-half the \\ coefficient of x to both sides . \end{array} \end{array}$
Step 4 (x+b2a)2(x+b2a)2==−ca+b24a2b2−4ac4a2The left side in Step 3 is a perfectsquare.Rearrange and combine fractions onthe right side. $\begin{array}{rcll} {(x + \frac{b}{2 a})}^{2} & = & - \frac{c}{a} + \frac{b^{2}}{4 a^{2}} & \begin{array}{l} The left side in Step 3 is a perfect \\ square . \end{array} \\ {(x + \frac{b}{2 a})}^{2} & = & \frac{b^{2} - 4 a c}{4 a^{2}} & \begin{array}{l} Rearrange and combine fractions on \\ the right side . \end{array} \end{array}$
Step 5 x+b2a=±b2−4ac4a2−−−−−−−√Take the square roots of both sides. $\begin{array}{l} x + \frac{b}{2 a} = \pm \sqrt{\frac{b^{2} - 4 a c}{4 a^{2}}} & Take the square roots of both sides . \end{array}$
Step 6 Solve the equations in Step 5.

$x+b2axxx====±b2−4ac4a2−−−−−−−√−b2a±b2−4ac−−−−−−−√2|a|−b2a±b2−4ac−−−−−−−√2a−b±b2−4ac−−−−−−−√2aEquations from Step 5Add −b2a to both sides; 4a2−−−√=4–√a2−−√=2|a|.±2|a|=±2a because |a|=a or |a|=−a.Combine fractions.$ $\begin{array}{rcll} x + \frac{b}{2 a} & = & \pm \sqrt{\frac{b^{2} - 4 a c}{4 a^{2}}} & Equations from Step 5 \\ x & = & - \frac{b}{2 a} \pm \frac{\sqrt{b^{2} - 4 a c}}{2 | a |} & Add - \frac{b}{2 a} to both sides; \sqrt{4 a^{2}} = \sqrt{4} \sqrt{a^{2}} = 2 | a | . \\ x & = & - \frac{b}{2 a} \pm \frac{\sqrt{b^{2} - 4 a c}}{2 a} & \pm 2 | a | = \pm 2 a because | a | = a or | a | = - a . \\ x & = & \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} & Combine fractions . \end{array}$

The last equation in Step 6 is called the quadratic formula.

Side Note

The quadratic formula shows that a quadratic equation can have, at most, two distinct solutions, and for any solution to be real we must have b2−4ac≥0. $b^{2} - 4 a c \geq 0 .$

Warning

To use the quadratic formula, write the given quadratic equation in standard form. Then determine the values of a (coefficient of x2 $x^{2}$ ), b (coefficient of x), and c (constant term).

Example 7 Solving a Quadratic Equation by Using the Quadratic Formula

Solve by using the quadratic formula: 3x2=5x+2 $3 x^{2} = 5 x + 2$

Solution

The equation 3x2=5x+2 $3 x^{2} = 5 x + 2$ must be rewritten in the standard form.

3 x 2 - 5 x - 2 = 0 3 x 2 + (- 5) x + (- 2) = 0 ↑ ↑ ↑ a b c Subtract 5 x + 2 from both sides of 3 x 2 = 5 x + 2 . Identify values of a, b, and c to be used in the quadratic formula .

$\begin{array}{r} 3 x^{2} - 5 x - 2 = 0 & Subtract 5 x + 2 from both sides of 3 x^{2} = 5 x + 2 . \\ \begin{array}{l} 3 x^{2} + (- 5) x + (- 2) = 0 \\ ↑ ↑ ↑ \\ a b c \end{array} & \begin{array}{l} Identify values of a, b, and c to be used in the \\ quadratic formula . \end{array} \end{array}$

Then a=3, b=−5, $a = 3, b = - 5,$ and c=−2, $c = - 2,$ and we have

x x = = = = - b \pm b 2 - 4 a c - - - - - - - \sqrt 2 a - ( - 5 ) \pm ( - 5 ) 2 - 4 ( 3 ) ( - 2 ) - - - - - - - - - - - - - - \sqrt 2 ( 3 ) 5 \pm 25 + 24 - - - - - - \sqrt 6 5 \pm 49 - - \sqrt 6 = 5 \pm 7 6 Quadratic formula Substitute 3 for a, - 5 for b, and - 2 for c . Simplify . Simplify .

$\begin{array}{l} x & = & \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} & Quadratic formula \\ x & = & \frac{- (- 5) \pm \sqrt{{(- 5)}^{2} - 4 (3) (- 2)}}{2 (3)} & \begin{array}{l} Substitute 3 for a, - 5 \\ for b, and - 2 for c . \end{array} \\ = & \frac{5 \pm \sqrt{25 + 24}}{6} & Simplify . \\ = & \frac{5 \pm \sqrt{49}}{6} = \frac{5 \pm 7}{6} & Simplify . \end{array}$

Then $x = \frac{5 + 7}{6} = \frac{12}{6} = 2$ or $x = \frac{5 - 7}{6} = \frac{- 2}{6} = - \frac{1}{3},$ and the solution set is ${- \frac{1}{3}, 2}$ .

Now you should solve $3 x^{2} = 5 x + 2$ by factoring to confirm that you get the same result.

Practice Problem 7

Solve $6 x^{2} - x - 2 = 0$ by using the quadratic formula.

The Discriminant

The expression $b^{2} - 4 a c$ , under the radical in the quadratic formula, is called the discriminant. If the coefficients, a, b, and c are all integers, the number and type of solutions of $a x^{2} + b x + c = 0$ are determined by the value of the discriminant.

Solutions of $a x^{2} + b x + c = 0,$ where `a`, `b`, and `c` are All integers
Discriminant	Number of Real Solutions	Type of Solutions
Positive, perfect square	Two	Rational
Positive, but not a perfect square	Two	Irrational
Zero	One—a double root	Rational
Negative	None	No real solutions

Example 8 Using the Discriminant

Evaluate the discriminant for each equation. Then use the discriminant to determine the number and type of solutions.

$2 x^{2} - 3 x - 7 = 0$
$x^{2} + 9 = - 6 x$
$x^{2} = - 2 x - 3$

Solution

For $2 x^{2} - 3 x - 7 = 0,$ we have $a = 2, b = - 3,$ and $c = - 7,$ so

$\begin{array}{l} b^{2} - 4 a c = {(- 3)}^{2} - 4 (2) (- 7) = 65 & discriminant = 65 \end{array}$

Because the discriminant, 65, is not a perfect square, there are two distinct irrational solutions.
First, rewrite $x^{2} + 9 = - 6 x$ in standard form: $x^{2} + 6 x + 9 = 0 .$ Then, $a = 1, b = 6,$ and $c = 9,$ so $\begin{array}{l} b^{2} - 4 a c = {(6)}^{2} - 4 (1) (9) = 0 & discriminant = 0 \end{array}$

Because the discriminant is zero there is only one rational solution.
First, rewrite $x^{2} = - 2 x - 3$ in standard form: $x^{2} + 2 x + 3 = 0 .$ Then, $a = 1, b = 2,$ and $c = 3,$ so $\begin{array}{l} b^{2} - 4 a c = {(2)}^{2} - 4 (1) (3) = - 8 & discriminant = - 8 \end{array}$

Because the discriminant, $- 8$ , is negative there are no real solutions.

Practice Problem 8

Evaluate the discriminant for each equation. Then use the discriminant to determine the number and type of solutions.
1. $3 x^{2} - 2 x - 5 = 0$
2. $x^{2} + 4 = 4 x$
3. $3 x^{2} + 6 x = - 6$

Applications

5 Solve applied problems.

Example 9 Dimension of an Electronic Component

A circular lithium battery is housed on a square surface, as shown in the figure in the margin. If the area of the square is 90.25 sq mm, what is the radius of the battery?

Solution

Let $\begin{array}{l} r = battery radius . & This is the value you want to find . \end{array}$

Then $2 r = width of the square,$ so ${(2 r)}^{2} = area of the square .$

We have $\begin{array}{rcll} {(2 r)}^{2} & = & 90.25 & Area of the square = 90.25 sq mm \\ 2 r & = & \pm \sqrt{90.25} = \pm 9.5 & Take the square root of both sides . \\ r & = & \pm 4.75 mm & Divide both sides by 2 . \end{array}$

We reject $- 4.75$ because the radius must be positive.

The radius of the battery is 4.75 mm.

Practice Problem 9

A circular lithium battery is housed on a square surface as in Example 9 . If the area of the battery is $64 π$ sq mm, find the area of the portion of the square not covered by the battery. Round your answer to the nearest square millimeter.

Example 10 Partitioning a Building

A rectangular building whose depth (from the front of the building) is three times its frontage is divided into two parts by a partition that is 45 feet from and parallel to the front wall. Assuming that the rear portion of the building contains 2100 square feet, find the dimensions of the building.

Solution

In solving problems of this type, it is advisable to draw a diagram such as Figure 1.3. We let

$\begin{array}{lrcl} x & = & the frontage of the building, in feet . \\ Then & 3 x & = & the depth of the building, in feet, \\ and & 3 x - 45 & = & the depth of the rear portion, in feet . \end{array}$

The area of the rear portion of the building is $x (3 x - 45)$ square feet. This area is 2100 square feet, so

$\begin{array}{l} \begin{array}{rcl} x (3 x - 45) & = & 2100 \\ 3 x^{2} - 45 x & = & 2100 \\ 3 x^{2} - 45 x - 2100 & = & 0 \\ x^{2} - 15 x - 700 & = & 0 \\ (x - 35) (x + 20) & = & 0 \end{array} & \begin{array}{l} Distributive property \\ Subtract 2100 from both sides . \\ Divide both sides by 3 . \\ Factor . \end{array} \\ \begin{array}{rcllrcl} x - 35 & = & 0 & or & x + 20 & = & 0 \\ x & = & 35 & or & x & = & - 20 \end{array} & \begin{array}{l} Zero-product property \\ Solve for x . \end{array} \end{array}$

Because the dimensions of the building are positive numbers, we reject $x = - 20 .$ Thus,

$\begin{array}{rcll} x & = & 35 & Frontage in feet \\ 3 x & = & 105 & Depth in feet \end{array}$

Check: $3 x - 45 = 3 (35) - 45 = 105 - 45 = 60$ is the depth of the rear portion, and the area of the rear portion is $60 \cdot 35 = 2100$ square feet.

Practice Problem 10

In Example 10 , find approximate dimensions of the building assuming that its depth is five times its frontage.

Side Note

Recall that two rectangles are similar if the lengths of corresponding sides are proportional.

Golden Rectangle

A rectangle of length p and width q, with $p > q,$ is called a golden rectangle if you can divide the rectangle into a square with side of length q and a smaller rectangle that is similar to the original one. See Figure 1.4. The ratio of the longer side to the shorter side of a golden rectangle is called the golden ratio. In Figure 1.4, the golden ratio is $\frac{p}{q} .$

Example 11 Calculating the Golden Ratio

Use the golden rectangle in Figure 1.4 to calculate the golden ratio.

Solution

The rectangle A in Figure 1.4 is a golden rectangle. This means that we can divide rectangle A into a square S and a smaller rectangle B such that

$\frac{Longer side of A}{Shorter side of A} = \frac{Longer side of B}{Shorter side of B} .$

Thus, because rectangle B is the smaller rectangle, $p - q < q,$ and we have

$\begin{array}{l} \begin{array}{rcll} \frac{p}{q} & = & \frac{q}{p - q} \\ p (p - q) & = & q \cdot q & Multiply both sides by q (p - q); simplify . \\ p^{2} - p q & = & q^{2} & Distributive property \\ {(\frac{p}{q})}^{2} - \frac{p}{q} & = & 1 & Divide both sides by q^{2}; simplify . \\ Φ^{2} - Φ & = & 1 & Replace \frac{p}{q} with Φ, the golden ratio . \\ Φ^{2} - Φ - 1 & = & 0 & Write in standard form, treating Φ as the variable . \end{array} \\ \begin{array}{rcll} Φ & = & \frac{- (- 1) \pm \sqrt{{(- 1)}^{2} - 4 (1) (- 1)}}{2 (1)} & \begin{array}{l} Substitute 1 for a and - 1 \\ both b and c in the quadratic \\ formula . \end{array} \\ = & \frac{1 \pm \sqrt{5}}{2} & Simplify . \end{array} \end{array}$

Because the ratio of two positive numbers is positive, we reject the negative root $\frac{1 - \sqrt{5}}{2} .$ Therefore, the golden ratio is

$Φ = \frac{1 + \sqrt{5}}{2} \approx 1.618 .$

Practice Problem 11

Find the length of a golden rectangle whose width is 36 feet.

Section 1.3 Exercises

Concepts and Vocabulary

Any equation of the form $a x^{2} + b x + c = 0$ with $a \neq 0$ is called a(n) equation.
If P(x), D(x), and Q(x) are polynomials and $P (x) = D (x) Q (x),$ then the solutions of $P (x) = 0$ are the solutions of $Q (x) = 0$ together with the solutions of $\underline{} = 0.$
From the square root property, we know that if $u^{2} = 5,$ then $u = \underline{}$ .
If you complete the square in the quadratic equation $a x^{2} + b x + c = 0,$ you get the quadratic formula for the solutions $x = \underline{}$ .
True or False. Every quadratic equation has, at most, two distinct solutions.
True or False. If we are given $x^{2} + k x,$ we form a perfect square by adding $k^{2} .$
True or False. The quadratic formula can be used to solve any quadratic equation.
True or False. To solve $(x + 1) (x - 3) = 1,$ we solve $x + 1 = 0$ and $x - 3 = 0 .$

Building Skills

In Exercises 9–16, show by substitution whether the number r is a solution of the corresponding quadratic equation.

$x^{2} + 4 x - 12 = 0; r = - 6$
$x^{2} - 8 x - 9 = 0; r = 9$
$3 x^{2} + 7 x - 6 = 0; r = \frac{2}{3}$
$2 x^{2} - 5 x - 3 = 0; r = - \frac{1}{2}$
$x^{2} - 4 x + 1 = 0; r = 2 - \sqrt{3}$
$x^{2} - 6 x + 1 = 0; r = 3 + 2 \sqrt{2}$
$4 x^{2} - 8 x + 13 = 0; r = 2 + \sqrt{3}$
$x^{2} - 6 x + 13 = 5; r = 5 - \sqrt{2}$
Find k, assuming that $x = 1$ is a solution of the equation $k x^{2} + x - 3 = 0 .$
Find k, assuming that $x = \sqrt{7}$ is a solution of the equation $k x^{2} + x - 3 = 0 .$

In Exercises 19–24, solve each equation by factoring.

$x^{2} - 5 x = 0$
$x^{2} - 5 x + 4 = 0$
$x^{2} + 5 x = 14$
$x^{2} - 11 x = 12$
$x^{2} = 5 x + 6$
$x = x^{2} - 12$

In Exercises 25–30, solve each equation by the square root method.

$3 x^{2} = 48$
$2 x^{2} = 50$
$x^{2} + 1 = 5$
$2 x^{2} - 1 = 17$
${(x - 1)}^{2} = 16$
${(2 x - 3)}^{2} = 25$

In Exercises 31–40, add a constant term to the expression to convert it to a perfect square.

$x^{2} + 4 x$
$y^{2} + 10 y$
$x^{2} + 6 x$
$y^{2} - 8 y$
$x^{2} - 7 x$
$x^{2} - 3 x$
$x^{2} + \frac{1}{3} x$
$x^{2} - \frac{3}{2} x$
$x^{2} + a x$
$x^{2} - \frac{2 a}{3} x$

In Exercises 41–46, solve each equation by completing the square.

$x^{2} + 2 x - 5 = 0$
$x^{2} + 6 x = - 7$
$x^{2} - 3 x - 1 = 0$
$x^{2} - x - 3 = 0$
$2 r^{2} + 3 r = 9$
$3 k^{2} - 5 k + 1 = 0$

In Exercises 47–52, solve each equation by using the quadratic formula.

$x^{2} + 2 x - 4 = 0$
$m^{2} + 3 m + 2 = 0$
$6 x^{2} = 7 x + 5$
$t^{2} - 7 = 4 t$
$3 z^{2} - 2 z = 7$
$6 y^{2} + 11 y = 10$

In Exercises 53–78, solve each equation by any method.

$2 x^{2} + 5 x - 3 = 0$
$2 x^{2} - 9 x + 10 = 0$
${(3 x - 2)}^{2} - 16 = 0$
${(4 x + 1)}^{2} - 25 = 0$
$5 x^{2} - 6 x = 4 x^{2} + 6 x - 3$
$x^{2} + 7 x - 5 = x - x^{2}$
$3 p^{2} + 8 p + 4 = 0$
$x^{2} = 5 (x - 1)$
$3 y^{2} + 5 y + 2 = 0$
$6 x^{2} + 11 x + 4 = 0$
$5 x^{2} + 12 x + 4 = 0$
$3 x^{2} - 2 x - 5 = 0$
$5 y^{2} + 10 y + 4 = 2 y^{2} + 3 y + 1$
$3 x^{2} - 1 = 5 x^{2} - 3 x - 5$
$2 x^{2} + x = 15$
$6 x^{2} = 1 - x$
$12 x^{2} - 10 x = 12$
$- x^{2} + 10 x + 1200 = 0$
$(x + 13) (x + 5) = - 2$
$3 (x^{2} + 1) = 2 x^{2} + 4 x + 1$
$18 x^{2} - 45 x = - 7$
$18 x^{2} + 57 x + 45 = 0$
$2 t^{2} - 5 = 0$
$3 k^{2} - 48 = 0$
$4 x^{2} - 10 x - 750 = 0$
$12 x^{2} + 43 x + 36 = 0$

In Exercises 79–86, evaluate the discriminant for each equation. Then use the discriminant to determine the number and type of solutions. Do not solve the equation.

$x^{2} + 4 x - 2 = 0$
$2 x^{2} + 18 = 12 x$
$- 3 x^{2} = 4$
$x^{2} + 8 = 0$
$6 x^{2} - 3 x - 2 = 0$
$12 x^{2} + 5 x - 3 = 0$
$2 x^{2} - 3 x + 1 = 0$
$4 x^{2} + 3 = 10 x$

In Exercises 87–90, write your answer rounded to two decimal places.

Find the length of the golden rectangle whose width is 14.72 in.
Find the length of the golden rectangle whose width is 18.63 ft.
Find the width of the golden rectangle whose length is 8.46 cm.
Find the width of the golden rectangle whose length is 4.68 m.

Applying the Concepts

Dimensions. The length of a rectangular plot is three times its width. Assuming that the area of the plot is 10,800 square feet, find the dimensions of the plot.
Dimensions. The diagonal of a square tile is 4 inches longer than its side. Find the length of the side.
Finding integers. Find two numbers whose sum is 28 and whose product is 147.
Finding integers. Find an integer such that the sum of twice the square of the integer and the integer itself is 55.
Interpreting a model. The percent, p, of persons age 25‌ and older who obtained bachelor’s degrees from an elite high school four years after graduating can be approximated by $p = 0.003 x^{2} + 0.2573 x + 83.975,$ where x is the number of years after 2000 (so, $x = 0$ corresponds to the year 2000). Based on this model (rounded to the nearest percent)
1. what percent obtained bachelor’s degrees in 2014?
2. what percent will obtain bachelor’s degrees in 2020?
Interpreting a model. For several years beginning in 2011, the percent, p, of Americans who owned smartphones can be approximated by $p = - 0.0893 x^{2} + 8.3893 x + 36.179,$ where x is the number of years after 2011 (so, $x = 0$ corresponds to the year 2011). Based on this model
1. what percent (rounded to the nearest percent) owned smartphones in 2016?
2. how do you know this model is inaccurate for the year 2020?
  
  (Source: Pew Research Center)
Geometry. The length of a rectangle is 5 centimeters more than its width. The area of the rectangle is 500 square centimeters. Find the dimensions of the rectangle.
Geometry. The sides of a rectangle are in the ratio 3:2. The area of the rectangle is 216 square centimeters. Find the dimensions of the rectangle.
Cutting a wire. A wire of length 16 inches is to be cut into two pieces; then each piece will be bent to form a square. Find the length of the two pieces assuming that the sum of the areas of the two squares is 10 square inches.
Cutting a wire. A piece of wire is 38 inches long. The wire is cut into two pieces; then each piece is bent to form a square. Find the length of each piece if three times the area of the larger square exceeds the area of the smaller square by 95.75 square inches.
Manufacturing. The surface area A of a cylinder with height h and radius r is given by the equation $A = 2 π r h + 2 π r^{2} .$ A company makes soup cans by using $32 π$ square inches of aluminum sheet for each can. Assuming that the height of the can is 6 inches, find the radius of the can.
Concrete patio. To make a rectangular concrete patio, a homeowner used 70 feet of perimeter, into which she poured 138 cubic feet of concrete to form a slab 6 inches thick. What were the dimensions of the patio?
Making a box. The volume of a box is given by the equation $V = length \cdot width \cdot height .$ A topless box is to be made from a square sheet of tin by cutting 5-inch squares from each of the four corners and then shaping the tin into an open box by turning up the sides. Assuming that the box is to hold 480 cubic inches, find the size of the piece of tin to be used.
Making a box. A 2-inch square is cut from each corner of a rectangular piece of cardboard whose length exceeds the width by 4 inches. The sides are then turned up to form an open box. Assuming that the volume of the box is 64 cubic inches, find the dimensions of the box.
Bus travel. Two buses leave Atlanta at the same time, one traveling west at an average of 52 miles per hour and the other traveling north at an average of 39 miles per hour. When will they be 390 miles apart?
Plane travel. Two planes left San Francisco at 2 p.m. One flew east at a certain speed, and the other flew south at a speed that was 100 kilometers per hour more than the speed of the plane heading east. The planes were 2100 kilometers apart at 5 p.m. How fast was each plane flying?
Gardening. You want to expand your present 25-foot by 15-foot garden by planting a border of flowers. The border is to be of the same width around the entire garden. The flowers you bought will fill an area of 624 square feet. How wide should the border be?
Sky divers. Sky divers are in free fall from the time they jump out of a plane until they open their parachutes. A sky diver jumps from 5000 feet. The diver’s height h above the ground t seconds after the jump is described by the equation $h = - 16 t^{2} + 5000 .$ Find the time during which the diver is in free fall, assuming that the parachute opens at 1000 feet.
Physics. Suppose you throw a ball straight up from the ground with a velocity of 112 feet per second. As the ball moves up, gravity slows it. Eventually, the ball begins to fall back to the ground. The height h of the ball after t seconds in the air is given by the equation $h = - 16 t^{2} + 112 t .$
1. Find the height of the ball after two seconds.
2. How long will it take the ball to reach a height of 96 feet?
3. How long after you throw the ball will it return to the ground?
Sound velocity. The speed of sound v(in ft/sec) depends on the temperature $T °$ Celsius and is given by

$v = \frac{1087 \sqrt{T + 273}}{16.25} .$
1. Find the speed of sound if the air temperature is $T = 20 ° C .$
2. Find the air temperature if the speed of sound is 1150 ft/sec.

Height of a projectile. The height, h, of a projectile thrown straight up from the top of a 480-foot hill t seconds after being thrown upward with initial velocity $v_{0} ft / sec$ is given by $h = - 16 t^{2} + v_{0} t + 480 .$ In Exercises 111–114, for a given value of $v_{0},$ find the time(s) when the projectile will (a) be at a height of 592 ft above the ground and (b) crash on the ground. Round your answers to two decimal places.

$v_{0} = 96$
$v_{0} = 112$
$v_{0} = 256$
$v_{0} = 64$

Beyond the Basics

In Exercises 115–118, find the values of k for which the given equation has equal roots.

$x^{2} - k x + 3 = 0$
$x^{2} + 3 k x + 8 = 0$
$2 x^{2} + k x + k = 0$
$k x^{2} + 2 x + 6 = 0$
Assuming that r and s are the roots of the quadratic equation $a x^{2} + b x + c = 0,$ show that

$r + s = - \frac{b}{a} and r \cdot s = \frac{c}{a} .$
Find the sum and the product of the roots of the following equations without solving the equation.
1. $3 x^{2} + 5 x - 5 = 0$
2. $3 x^{2} - 7 x = 1$
3. $\sqrt{3} x^{2} = 3 x + 4$
4. $(1 + \sqrt{2}) x^{2} - \sqrt{2} x - 5 = 0$

In Exercises 121 and 122, determine k so that the sum and the product of the roots are equal.

$2 x^{2} + (k - 3) x + 3 k - 5 = 0$
$5 x^{2} + (2 k - 3) x - 2 k + 3 = 0$
Suppose r and s are the roots of the quadratic equation $a x^{2} + b x + c = 0 .$ Show that you can write $a x^{2} + b x + c = a (x - r) (x - s) .$ Thus, the quadratic trinomial can be written in factored form.

[Hint: From Exercise 119, $b = - a (r + s)$ and $c = a r s .$ Substitute in $a x^{2} + b x + c$ and factor.]
Use Exercise 123 to factor each trinomial.
1. $4 x^{2} + 4 x - 5$
2. $25 x^{2} + 40 x + 11$
3. $25 x^{2} - 30 x + 4$
4. $72 x^{2} + 95 x - 1000$
If we know the roots r and s in advance, then by Exercise 123, a quadratic equation with roots r and s can be expressed in the form

$a (x - r) (x - s) = 0 .$

Form a quadratic equation that has the listed numbers as roots.
1. $- 3, 4$
2. 5,5
3. $3 + \sqrt{2}, 3 - \sqrt{2}$
How to construct a golden rectangle. Draw a square ABCD with side of length x, as in the accompanying diagram. Let M be the midpoint of DC. The segment MB sweeps out the arc BF so that F lies on the line through the segment DC. Show that the rectangle AEFD is a golden rectangle.

Critical Thinking/Discussion/Writing

If a, b, and k are real numbers, show that the solutions of the equation $(x - a) (x - b) = k^{2}$ are real.
Show that for $0 < a < 4,$ the equation $a x (1 - x) = 1$ has no real solutions.

Getting Ready for the Next Section

In Exercises 129–144, perform the indicated operations.

$(3 + \sqrt{2}) (3 - \sqrt{2})$
$(1 - 2 \sqrt{7}) (1 + 2 \sqrt{7})$
${(2 + \sqrt{5})}^{2}$
${(\sqrt{2} - \sqrt{3})}^{2}$
$\frac{5 + 5 \sqrt{20}}{5}$
$\frac{6 + \sqrt{27}}{3}$
$\frac{16 - \sqrt{100}}{2}$
$\frac{18 - \sqrt{108}}{12}$
$(3 x + 2) + (x - 7)$
$(5 x - 9) + (6 - x)$
$(9 x + 4) - (2 x + 12)$
$(6 x - 5) - (3 x + 4)$
$(3 x + 2) (x - 9)$
$(2 x - 5) (3 x + 4)$
$(x - 3) (x + 3)$
$(5 x + 2) (5 x - 2)$

In Exercises 145–148, solve each equation.

$x^{2} + 4 x + 1 = 0$
$x^{2} + 5 x + 5 = 0$
$5 x^{2} + 8 x + 2 = 0$
$- 2 x^{2} + 5 x - 1 = 0$

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Table of Contents for Section 1.3 Quadratic Equations

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Table of Contents for
Section 1.3 Quadratic Equations