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Chapter 3 Summary and Review

Study Guide

Key Terms and Concepts	Example
Section 3.1: The Complex Numbers
The number `i` is defined such that i=−1−−−√ $i = \sqrt{−1}$ and i2=−1 $i^{2} = −1$ .	Express each number in terms of `i`. $-5 - - - \sqrt = -1 \cdot 5 - - - - - \sqrt = -1 - - - \sqrt \cdot 5 - \sqrt = i 5 - \sqrt, or 5 - \sqrt i$ $\sqrt{−5} = \sqrt{−1 \cdot 5} = \sqrt{−1} \cdot \sqrt{5} = i \sqrt{5}, or \sqrt{5} i$ ; $- -36 - - - - \sqrt = - -1 \cdot 36 - - - - - - \sqrt = - -1 - - - \sqrt \cdot 36 - - \sqrt = - i \cdot 6 = -6 i$ $- \sqrt{−36} = - \sqrt{−1 \cdot 36} = - \sqrt{−1} \cdot \sqrt{36} = - i \cdot 6 = −6 i$
A complex number is a number of the form a+bi $a + b i$ , where `a` and `b` are real numbers. The number `a` is said to be the real part of a+bi $a + b i$ , and the number `b` is said to be the imaginary part of a+bi $a + b i$ . To add or subtract complex numbers, we add or subtract the real parts, and we add or subtract the imaginary parts.	Add or subtract. $(-3 + 4 i) + (5 - 8 i) (6 - 7 i) - (10 + 3 i) = = = = (-3 + 5) + (4 i - 8 i) 2 - 4 i; (6 - 10) + (-7 i - 3 i) -4 -10 i$ $\begin{array}{rcl} (−3 + 4 i) + (5 - 8 i) & = & (−3 + 5) + (4 i - 8 i) \\ = & 2 - 4 i; \\ (6 - 7 i) - (10 + 3 i) & = & (6 - 10) + (−7 i - 3 i) \\ = & −4 −10 i \end{array}$
When we multiply complex numbers, we must keep in mind the fact that i2=−1 $i^{2} = −1$ . Note that a−−√⋅b√≠ab−−√ $\sqrt{a} \cdot \sqrt{b} \neq \sqrt{a b}$ when a−−√ $\sqrt{a}$ and b√ $\sqrt{b}$ are not real numbers.	Multiply. $-4 - - - \sqrt \cdot -100 - - - - \sqrt (2 - 5 i) (3 + i) = = = = = = = = = -1 - - - \sqrt \cdot 4 - \sqrt \cdot -1 - - - \sqrt \cdot 100 - - - \sqrt i \cdot 2 \cdot i \cdot 10 i 2 \cdot 20 -1 \cdot 20 i 2 = - 1 -20; 6 + 2 i - 15 i - 5 i 2 6 - 13 i - 5 (-1) 6 - 13 i + 5 11 - 13 i$ $\begin{array}{rcl} \sqrt{−4} \cdot \sqrt{−100} & = & \sqrt{−1} \cdot \sqrt{4} \cdot \sqrt{−1} \cdot \sqrt{100} \\ = & i \cdot 2 \cdot i \cdot 10 \\ = & i^{2} \cdot 20 \\ = & −1 \cdot 20 i^{2} = - 1 \\ = & −20; \\ (2 - 5 i) (3 + i) & = & 6 + 2 i - 15 i - 5 i^{2} \\ = & 6 - 13 i - 5 (−1) \\ = & 6 - 13 i + 5 \\ = & 11 - 13 i \end{array}$
The conjugate of a complex number a+bi $a + b i$ is a−bi $a - b i$ . The numbers a+bi $a + b i$ and a−bi $a - b i$ are complex conjugates. Conjugates are used when we divide complex numbers.	Divide. $5 - 2 i 3 + i = = = = = 5 - 2 i 3 + i \cdot 3 - i 3 - i 15 - 5 i - 6 i + 2 i 2 9 - i 2 15 - 11 i - 2 9 + 1 13 - 11 i 10 13 10 - 11 10 i 3 - i i s t h e c o n j u g a t e o f t h e d i v i s o r, 3 + i . i 2 = - 1$ $\begin{array}{rcll} \frac{5 - 2 i}{3 + i} & = & \frac{5 - 2 i}{3 + i} \cdot \frac{3 - i}{3 - i} & \begin{array}{l} 3 - i is the conjugate \\ of the divisor, 3 + i . \end{array} \\ = & \frac{15 - 5 i - 6 i + 2 i^{2}}{9 - i^{2}} \\ = & \frac{15 - 11 i - 2}{9 + 1} & i^{2} = - 1 \\ = & \frac{13 - 11 i}{10} \\ = & \frac{13}{10} - \frac{11}{10} i \end{array}$
Section 3.2: Quadratic Equations, Functions, Zeros, and Models
A quadratic equation is an equation that can be written in the form $a x 2 + b x + c = 0, a \neq 0,$ $a x^{2} + b x + c = 0, a \neq 0,$ where `a`, `b`, and `c` are real numbers. A quadratic function `f` is a function that can be written in the form $f (x) = a x 2 + b x + c, a \neq 0,$ $f (x) = a x^{2} + b x + c, a \neq 0,$ where `a`, `b`, and `c` are real numbers. The zeros of a quadratic function f(x)= $f (x) =$ ax2+bx+c $a x^{2} + b x + c$ are the solutions of the associated quadratic equation ax2+bx+c=0 $a x^{2} + b x + c = 0$ .	3x2−2x+4=0 $3 x^{2} - 2 x + 4 = 0$ and 5−4x=x2 $5 - 4 x = x^{2}$ are examples of quadratic equations. The equation 3x2−2x+4=0 $3 x^{2} - 2 x + 4 = 0$ is written in standard form. The functions f(x)=2x2+x+1 $f (x) = 2 x^{2} + x + 1$ and f(x)=5x2−4 $f (x) = 5 x^{2} - 4$ are examples of quadratic functions.
The Principle of Zero Products If ab=0 $a b = 0$ is true, then a=0 $a = 0$ or b=0 $b = 0$ , and if a=0 $a = 0$ or b=0 $b = 0$ , then ab=0 $a b = 0$ .	Solve: 3x2−4=11x $3 x^{2} - 4 = 11 x$ . $3 x 2 - 4 3 x 2 - 11 x - 4 (3 x + 1) (x - 4) = = = 11 x 00 S u b t r a c t i n g 11 x o n b o t h s i d e s t o g e t 0 o n o n e s i d e o f t h e e q u a t i o n F a c t o r i n g 3 x + 1 3 x x = = = 0 -1 - 1 3 o r o r o r x - 4 x x = = = 044 U s i n g t h e p r i n c i p l e o f z e r o p r o d u c t s$ $\begin{array}{l} \begin{array}{rcll} 3 x^{2} - 4 & = & 11 x \\ 3 x^{2} - 11 x - 4 & = & 0 & Subtracting 11 x \\ on both sides to get \\ 0 on one side of the \\ equation \\ (3 x + 1) (x - 4) & = & 0 & Factoring \end{array} \\ \begin{array}{rclcrcll} 3 x + 1 & = & 0 & or & x - 4 & = & 0 & Using the principle \\ of zero products \\ 3 x & = & −1 & or & x & = & 4 \\ x & = & - \frac{1}{3} & or & x & = & 4 \end{array} \end{array}$ The solutions are −13 $- \frac{1}{3}$ and 4.
The Principle of Square Roots If x2=k $x^{2} = k$ , then x=k−−√ $x = \sqrt{k}$ or x=−k−−√ $x = - \sqrt{k}$ .	Solve: 3x2−18=0 $3 x^{2} - 18 = 0$ . $3 x 2 - 18 3 x 2 x 2 = = = 0186 A d d i n g 18 o n b o t h s i d e s D i v i d i n g b y 3 o n b o t h s i d e s x = 6 - \sqrt o r x = - 6 - \sqrt U s i n g t h e p r i n c i p l e o f s q u a r e r o o t s$ $\begin{array}{l} \begin{array}{rcll} 3 x^{2} - 18 & = & 0 \\ 3 x^{2} & = & 18 & Adding 18 on \\ both sides \\ x^{2} & = & 6 & Dividing by 3 on \\ both sides \end{array} \\ \begin{array}{l} x = \sqrt{6} & or & x = - \sqrt{6} & Using the principle of square roots \end{array} \end{array}$ The solutions are $\sqrt{6}$ and $- \sqrt{6}$ , or $\pm \sqrt{6}$ .
To solve a quadratic equation by completing the square: Isolate the terms with variables on one side of the equation and arrange them in descending order. Divide by the coefficient of the squared term if that coefficient is not 1. Complete the square by taking half the coefficient of the first-degree term and adding its square on both sides of the equation. Express one side of the equation as the square of a binomial. Use the principle of square roots. Solve for the variable.	Solve: $2 x^{2} - 3 = 6 x$ . $\begin{array}{rcll} 2 x^{2} - 3 & = & 6 x \\ 2 x^{2} - 6 x - 3 & = & 0 & Subtracting 6 x \\ 2 x^{2} - 6 x & = & 3 & Adding 3 \\ x^{2} - 3 x & = & \frac{3}{2} & \begin{array}{l} Dividing by 2 to make the \\ x^{2} -coefficient 1 \end{array} \\ x^{2} - 3 x + \frac{9}{4} & = & \frac{3}{2} + \frac{9}{4} & \begin{array}{l} Completing the square : \\ \frac{1}{2} (- 3) = - \frac{3}{2} and \\ {(- \frac{3}{2})}^{2} = \frac{9}{4}; adding \frac{9}{4} \end{array} \\ {(x - \frac{3}{2})}^{2} & = & \frac{15}{4} & Factoring and simplifying \\ x - \frac{3}{2} & = & \pm \frac{\sqrt{15}}{2} & \begin{array}{l} Using the principle of square \\ roots and the quotient rule \\ for radicals \end{array} \\ x & = & \frac{3}{2} \pm \frac{\sqrt{15}}{2} = \frac{3 \pm \sqrt{15}}{2} \end{array}$ The solutions are $\frac{3 + \sqrt{15}}{2}$ and $\frac{\sqrt{3 - \sqrt{15}}}{2}$ , or $\frac{3 \pm \sqrt{15}}{2}$ .
The solutions of $a x^{2} + b x + c = 0, a \neq 0$ , can be found using the quadratic formula: $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} .$	Solve: $x^{2} - 6 = 3 x$ . $\begin{array}{rcll} x^{2} - 6 & = & 3 x \\ x^{2} - 3 x - 6 & = & 0 Standard form \\ a & = & 1, b = −3, c = −6 \\ x & = & \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ = & \frac{- (−3) \pm \sqrt{{(−3)}^{2} - 4 (1) (−6)}}{2 \cdot 1} \\ = & \frac{3 \pm \sqrt{9 + 24}}{2} \\ = & \frac{3 \pm \sqrt{33}}{2} Exact solutions \end{array}$ Using a calculator, we approximate the solutions to be 4.372 and −1.372.
Discriminant For $a x^{2} + b x + c = 0$ , where `a`, `b`, and `c` are real numbers: $\begin{array}{rcl} b^{2} - 4 a c = 0 & \to & One real-number \\ solution; \\ b^{2} - 4 a c > 0 & \to & Two different \\ real-number solutions; \\ b^{2} - 4 a c < 0 & \to & Two different \\ imaginary-number \\ solutions, complex \\ conjugates . \end{array}$	For the equation above, $x^{2} - 6 = 3 x$ , we see that $b^{2} - 4 a c$ is 33. Since 33 is positive, there are two different real-number solutions. For $2 x^{2} - x + 4 = 0$ , with $a = 2, b = −1$ , and $c = 4$ , the discriminant, ${(−1)}^{2} - 4 \cdot 2 \cdot 4 = 1 - 32 = −31$ , is negative, so there are two different imaginary-number (or nonreal) solutions. For $x^{2} - 6 x + 9 = 0$ , with $a = 1, b = −6$ , and $c = 9$ , the discriminant, ${(−6)}^{2} - 4 \cdot 1 \cdot 9 = 36 - 36 = 0$ , is 0 so there is one real-number solution.
Equations reducible to quadratic, or quadratic in form, can be treated as quadratic equations if a suitable substitution is made.	Solve: $x^{4} - x^{2} - 12 = 0$ . $\begin{array}{l} \begin{array}{rcll} x^{4} - x^{2} - 12 & = & 0 & Let u = x^{2} . Then u^{2} = {(x^{2})}^{2} = x^{4} . \\ u^{2} - u - 12 & = & 0 & Substituting \\ (u - 4) (u + 3) & = & 0 \end{array} \\ \begin{array}{rclcrcll} u - 4 & = & 0 & or & u + 3 & = & 0 \\ u & = & 4 & or & u & = & −3 & Solving for u \\ x^{2} & = & 4 & or & x^{2} & = & −3 \\ x & = & \pm 2 & or & x & = & \pm \sqrt{3} i & Solving for x \end{array} \end{array}$ The solutions are 2, $−2, \sqrt{3} i$ , and $- \sqrt{3} i$ .
Section 3.3: Analyzing Graphs of Quadratic Functions
Graphing Quadratic Equations The graph of the function $f (x) = a {(x - h)}^{2} + k$ is a parabola that: opens up if $a > 0$ and down if $a < 0;$ has (`h`, `k`) as the vertex; has $x = h$ as the axis of symmetry; has `k` as a minimum value (output) if $a > 0;$ has `k` as a maximum value if $a < 0$ . We can use a modification of the technique of completing the square as an aid in analyzing and graphing quadratic functions.	Find the vertex, the axis of symmetry, and the maximum or minimum value of $f (x) = 2 x^{2} + 12 x + 12$ . $\begin{array}{l} f (x) & = & 2 x^{2} + 12 x + 12 \\ = & 2 (x^{2} + 6 x) + 12 & Note that 9 completes \\ the square for \\ x^{2} + 6 x . \\ = & 2 (x^{2} + 6 x + 9 - 9) + 12 & Adding 9 - 9, \\ or 0, inside the \\ parentheses \\ = & 2 (x^{2} + 6 x + 9) −2 \cdot 9 + 12 & Using the \\ distributive law \\ to remove - 9 \\ from within the \\ parentheses \\ = & 2 {(x + 3)}^{2} - 6 \\ = & 2 {[x - (−3)]}^{2} + (−6) \end{array}$ The function is now written in the form $f (x) = a {(x - h)}^{2} + k$ with $a = 2$ , $h = −3$ , and $k = −6$ . Because $a > 0$ , we know the graph opens up and thus the function has a minimum value. We also know the following: Vertex $(h, k) : (−3, −6);$ Axis of symmetry $x = h : x = −3;$ Minimum value of the function $k : −6$ . To graph the function, we first plot the vertex and then find several points on either side of it. We plot these points and connect them with a smooth curve.
The Vertex of a Parabola The vertex of the graph of $f (x) = a x^{2} + b x + c$ is	Find the vertex of the function $f (x) = −3 x^{2} + 6 x + 1$ . $\begin{array}{rcl} - \frac{b}{2 a} & = & - \frac{6}{2 (−3)} = 1 \\ f (1) & = & −3 \cdot 1^{2} + 6 \cdot 1 + 1 = 4 \end{array}$ The vertex is $(1, 4)$ .
Some applied problems can be solved by finding the maximum or minimum value of a quadratic function.	See Examples 5–7 on pp. 194–197.
Section 3.4: Solving Rational Equations and Radical Equations
A rational equation is an equation containing one or more rational expressions. When we solve a rational equation, we usually first multiply by the least common denominator (LCD) of all the rational expressions to clear the fractions. CAUTION! When we multiply by an expression containing a variable, we might not obtain an equation equivalent to the original equation, so we must check the possible solutions obtained by substituting them in the original equation.	Solve. $\begin{array}{rcll} \frac{5}{x + 2} - \frac{4}{x^{2} - 4} & = & \frac{x - 3}{x - 2} \\ \frac{5}{x + 2} - \frac{4}{(x + 2) (x - 2)} & = & \frac{x - 3}{x - 2} & \begin{array}{l} The LCD is \\ (x + 2) (x - 2) . \end{array} \end{array}$ $(x + 2) (x - 2) (\frac{5}{x + 2} - \frac{4}{(x + 2) (x - 2)})$ $\begin{array}{l} \begin{array}{rcl} = & (x + 2) (x - 2) \cdot \frac{x - 3}{x - 2} \\ 5 (x - 2) −4 & = & (x + 2) (x - 3) \\ 5 x - 10 - 4 & = & x^{2} - x - 6 \\ 5 x - 14 & = & x^{2} - x - 6 \\ 0 & = & x^{2} - 6 x + 8 \\ 0 & = & (x - 2) (x - 4) \end{array} \\ \begin{array}{rclcrcll} x - 2 & = & 0 & or & x - 4 & = & 0 \\ x & = & 2 & or & x & = & 4 \end{array} \end{array}$ The number 2 does not check, but 4 does. The solution is 4.
A radical equation is an equation that contains one or more radicals. We use the principle of powers to solve radical equations. For any positive integer `n`: If $a = b$ is true, then $a^{n} = b^{n}$ is true. CAUTION! If $a^{n} = b^{n}$ is true, it is not necessarily true that $a = b$ , so we must check the possible solutions obtained by substituting them in the original equation.	Solve: $\sqrt{x + 2} + \sqrt{x - 1} = 3$ . $\begin{array}{rcll} \sqrt{x + 2} + \sqrt{x - 1} & = & 3 \\ \sqrt{x + 2} & = & 3 - \sqrt{x - 1} & Isolating one radical \\ {(\sqrt{x + 2})}^{2} & = & {(3 - \sqrt{x - 1})}^{2} \\ x + 2 & = & 9 - 6 \sqrt{x - 1} + (x - 1) \\ x + 2 & = & 8 - 6 \sqrt{x - 1} + x \\ −6 & = & −6 \sqrt{x - 1} & Isolating the \\ remaining radical \\ 1 & = & \sqrt{x - 1} & Dividing by - 6 \\ 1^{2} & = & {(\sqrt{x - 1})}^{2} \\ 1 & = & x - 1 \\ 2 & = & x \end{array}$ The number 2 checks. It is the solution.
Section 3.5: Solving Equations and Inequalities with Absolute Value
We use the following property to solve equations with absolute value. For $a > 0$ and an algebraic expression `X`: $\| X \| = a is equivalent to X = - a or x = a$	Solve: $\| x + 1 \| = 4$ . $\begin{array}{l} \| x + 1 \| = 4 \\ \begin{array}{rclcrcl} x + 1 & = & −4 & or & x + 1 & = & 4 \\ x & = & −5 & or & x & = & 3 \end{array} \end{array}$ Both numbers check. The solutions are −5 and 3.
The following properties are used to solve inequalities with absolute value. For $a > 0$ and an algebraic expression `X`: $\| X \| < a$ is equivalent to $- a < X < a$ . $\| X \| > a$ is equivalent to $X < - a or X > a$ . Similar statements hold for $\| X \| \leq a and \| X \| \geq a$ .	Solve: $\| x - 2 \| < 3$ . $\begin{array}{rcll} \| x - 2 \| & < & 3 \\ −3 & < & x - 2 & < & 3 \\ −1 & < & x < 5 & Adding 2 \end{array}$ The solution set is ${x \| −1 < x < 5}$ , or (−1, 5). $\begin{array}{l} \| 3 x \| \geq 6 \\ \begin{array}{rclcrcll} 3 x & \leq & −6 & or & 3 x & \geq & 6 \\ x & \leq & −2 & or & x & \geq & 2 & Dividing by 3 \end{array} \end{array}$ The solution set is ${x \| x \leq −2 or x \geq 2}$ , or $(- \infty, −2] \cup [2, \infty)$ .

Key Terms and Concepts

Example

Section 3.1: The Complex Numbers

The number i is defined such that i=−1−−−√ $i = \sqrt{−1}$ and i2=−1 $i^{2} = −1$ .

Express each number in terms of i.

-5 - - - \sqrt = -1 \cdot 5 - - - - - \sqrt = -1 - - - \sqrt \cdot 5 - \sqrt = i 5 - \sqrt, or 5 - \sqrt i

$\sqrt{−5} = \sqrt{−1 \cdot 5} = \sqrt{−1} \cdot \sqrt{5} = i \sqrt{5}, or \sqrt{5} i$ ;

- -36 - - - - \sqrt = - -1 \cdot 36 - - - - - - \sqrt = - -1 - - - \sqrt \cdot 36 - - \sqrt = - i \cdot 6 = -6 i

$- \sqrt{−36} = - \sqrt{−1 \cdot 36} = - \sqrt{−1} \cdot \sqrt{36} = - i \cdot 6 = −6 i$

A complex number is a number of the form a+bi $a + b i$ , where a and b are real numbers. The number a is said to be the real part of a+bi $a + b i$ , and the number b is said to be the imaginary part of a+bi $a + b i$ . To add or subtract complex numbers, we add or subtract the real parts, and we add or subtract the imaginary parts.

Add or subtract.

(-3 + 4 i) + (5 - 8 i) (6 - 7 i) - (10 + 3 i) = = = = (-3 + 5) + (4 i - 8 i) 2 - 4 i; (6 - 10) + (-7 i - 3 i) -4 -10 i

$\begin{array}{rcl} (−3 + 4 i) + (5 - 8 i) & = & (−3 + 5) + (4 i - 8 i) \\ = & 2 - 4 i; \\ (6 - 7 i) - (10 + 3 i) & = & (6 - 10) + (−7 i - 3 i) \\ = & −4 −10 i \end{array}$

When we multiply complex numbers, we must keep in mind the fact that i2=−1 $i^{2} = −1$ .

Note that a−−√⋅b√≠ab−−√ $\sqrt{a} \cdot \sqrt{b} \neq \sqrt{a b}$ when a−−√ $\sqrt{a}$ and b√ $\sqrt{b}$ are not real numbers.

Multiply.

-4 - - - \sqrt \cdot -100 - - - - \sqrt (2 - 5 i) (3 + i) = = = = = = = = = -1 - - - \sqrt \cdot 4 - \sqrt \cdot -1 - - - \sqrt \cdot 100 - - - \sqrt i \cdot 2 \cdot i \cdot 10 i 2 \cdot 20 -1 \cdot 20 i 2 = - 1 -20; 6 + 2 i - 15 i - 5 i 2 6 - 13 i - 5 (-1) 6 - 13 i + 5 11 - 13 i

$\begin{array}{rcl} \sqrt{−4} \cdot \sqrt{−100} & = & \sqrt{−1} \cdot \sqrt{4} \cdot \sqrt{−1} \cdot \sqrt{100} \\ = & i \cdot 2 \cdot i \cdot 10 \\ = & i^{2} \cdot 20 \\ = & −1 \cdot 20 i^{2} = - 1 \\ = & −20; \\ (2 - 5 i) (3 + i) & = & 6 + 2 i - 15 i - 5 i^{2} \\ = & 6 - 13 i - 5 (−1) \\ = & 6 - 13 i + 5 \\ = & 11 - 13 i \end{array}$

The conjugate of a complex number a+bi $a + b i$ is a−bi $a - b i$ . The numbers a+bi $a + b i$ and a−bi $a - b i$ are complex conjugates.

Conjugates are used when we divide complex numbers.

Divide.

5 - 2 i 3 + i = = = = = 5 - 2 i 3 + i \cdot 3 - i 3 - i 15 - 5 i - 6 i + 2 i 2 9 - i 2 15 - 11 i - 2 9 + 1 13 - 11 i 10 13 10 - 11 10 i 3 - i i s t h e c o n j u g a t e o f t h e d i v i s o r, 3 + i . i 2 = - 1

$\begin{array}{rcll} \frac{5 - 2 i}{3 + i} & = & \frac{5 - 2 i}{3 + i} \cdot \frac{3 - i}{3 - i} & \begin{array}{l} 3 - i is the conjugate \\ of the divisor, 3 + i . \end{array} \\ = & \frac{15 - 5 i - 6 i + 2 i^{2}}{9 - i^{2}} \\ = & \frac{15 - 11 i - 2}{9 + 1} & i^{2} = - 1 \\ = & \frac{13 - 11 i}{10} \\ = & \frac{13}{10} - \frac{11}{10} i \end{array}$

Section 3.2: Quadratic Equations, Functions, Zeros, and Models

A quadratic equation is an equation that can be written in the form

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

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

where a, b, and c are real numbers.

A quadratic function f is a function that can be written in the form

f (x) = a x 2 + b x + c, a \neq 0,

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

where a, b, and c are real numbers.

The zeros of a quadratic function f(x)= $f (x) =$ ax2+bx+c $a x^{2} + b x + c$ are the solutions of the associated quadratic equation ax2+bx+c=0 $a x^{2} + b x + c = 0$ .

3x2−2x+4=0 $3 x^{2} - 2 x + 4 = 0$ and 5−4x=x2 $5 - 4 x = x^{2}$ are examples of quadratic equations. The equation 3x2−2x+4=0 $3 x^{2} - 2 x + 4 = 0$ is written in standard form.

The functions f(x)=2x2+x+1 $f (x) = 2 x^{2} + x + 1$ and f(x)=5x2−4 $f (x) = 5 x^{2} - 4$ are examples of quadratic functions.

The Principle of Zero Products

If ab=0 $a b = 0$ is true, then a=0 $a = 0$ or b=0 $b = 0$ , and if a=0 $a = 0$ or b=0 $b = 0$ , then ab=0 $a b = 0$ .

Solve: 3x2−4=11x $3 x^{2} - 4 = 11 x$ .

3 x 2 - 4 3 x 2 - 11 x - 4 (3 x + 1) (x - 4) = = = 11 x 00 S u b t r a c t i n g 11 x o n b o t h s i d e s t o g e t 0 o n o n e s i d e o f t h e e q u a t i o n F a c t o r i n g 3 x + 1 3 x x = = = 0 -1 - 1 3 o r o r o r x - 4 x x = = = 044 U s i n g t h e p r i n c i p l e o f z e r o p r o d u c t s

$\begin{array}{l} \begin{array}{rcll} 3 x^{2} - 4 & = & 11 x \\ 3 x^{2} - 11 x - 4 & = & 0 & Subtracting 11 x \\ on both sides to get \\ 0 on one side of the \\ equation \\ (3 x + 1) (x - 4) & = & 0 & Factoring \end{array} \\ \begin{array}{rclcrcll} 3 x + 1 & = & 0 & or & x - 4 & = & 0 & Using the principle \\ of zero products \\ 3 x & = & −1 & or & x & = & 4 \\ x & = & - \frac{1}{3} & or & x & = & 4 \end{array} \end{array}$

The solutions are −13 $- \frac{1}{3}$ and 4.

The Principle of Square Roots

If x2=k $x^{2} = k$ , then x=k−−√ $x = \sqrt{k}$ or x=−k−−√ $x = - \sqrt{k}$ .

Solve: 3x2−18=0 $3 x^{2} - 18 = 0$ .

3 x 2 - 18 3 x 2 x 2 = = = 0186 A d d i n g 18 o n b o t h s i d e s D i v i d i n g b y 3 o n b o t h s i d e s x = 6 - \sqrt o r x = - 6 - \sqrt U s i n g t h e p r i n c i p l e o f s q u a r e r o o t s

$\begin{array}{l} \begin{array}{rcll} 3 x^{2} - 18 & = & 0 \\ 3 x^{2} & = & 18 & Adding 18 on \\ both sides \\ x^{2} & = & 6 & Dividing by 3 on \\ both sides \end{array} \\ \begin{array}{l} x = \sqrt{6} & or & x = - \sqrt{6} & Using the principle of square roots \end{array} \end{array}$

The solutions are $\sqrt{6}$ and $- \sqrt{6}$ , or $\pm \sqrt{6}$ .

To solve a quadratic equation by completing the square:

Isolate the terms with variables on one side of the equation and arrange them in descending order.
Divide by the coefficient of the squared term if that coefficient is not 1.
Complete the square by taking half the coefficient of the first-degree term and adding its square on both sides of the equation.
Express one side of the equation as the square of a binomial.
Use the principle of square roots.
Solve for the variable.

Solve: $2 x^{2} - 3 = 6 x$ .

$\begin{array}{rcll} 2 x^{2} - 3 & = & 6 x \\ 2 x^{2} - 6 x - 3 & = & 0 & Subtracting 6 x \\ 2 x^{2} - 6 x & = & 3 & Adding 3 \\ x^{2} - 3 x & = & \frac{3}{2} & \begin{array}{l} Dividing by 2 to make the \\ x^{2} -coefficient 1 \end{array} \\ x^{2} - 3 x + \frac{9}{4} & = & \frac{3}{2} + \frac{9}{4} & \begin{array}{l} Completing the square : \\ \frac{1}{2} (- 3) = - \frac{3}{2} and \\ {(- \frac{3}{2})}^{2} = \frac{9}{4}; adding \frac{9}{4} \end{array} \\ {(x - \frac{3}{2})}^{2} & = & \frac{15}{4} & Factoring and simplifying \\ x - \frac{3}{2} & = & \pm \frac{\sqrt{15}}{2} & \begin{array}{l} Using the principle of square \\ roots and the quotient rule \\ for radicals \end{array} \\ x & = & \frac{3}{2} \pm \frac{\sqrt{15}}{2} = \frac{3 \pm \sqrt{15}}{2} \end{array}$

The solutions are $\frac{3 + \sqrt{15}}{2}$ and $\frac{\sqrt{3 - \sqrt{15}}}{2}$ , or $\frac{3 \pm \sqrt{15}}{2}$ .

The solutions of $a x^{2} + b x + c = 0, a \neq 0$ , can be found using the quadratic formula:

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} .$

Solve: $x^{2} - 6 = 3 x$ .

$\begin{array}{rcll} x^{2} - 6 & = & 3 x \\ x^{2} - 3 x - 6 & = & 0 Standard form \\ a & = & 1, b = −3, c = −6 \\ x & = & \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ = & \frac{- (−3) \pm \sqrt{{(−3)}^{2} - 4 (1) (−6)}}{2 \cdot 1} \\ = & \frac{3 \pm \sqrt{9 + 24}}{2} \\ = & \frac{3 \pm \sqrt{33}}{2} Exact solutions \end{array}$

Using a calculator, we approximate the solutions to be 4.372 and −1.372.

Discriminant

For $a x^{2} + b x + c = 0$ , where a, b, and c are real numbers:

$\begin{array}{rcl} b^{2} - 4 a c = 0 & \to & One real-number \\ solution; \\ b^{2} - 4 a c > 0 & \to & Two different \\ real-number solutions; \\ b^{2} - 4 a c < 0 & \to & Two different \\ imaginary-number \\ solutions, complex \\ conjugates . \end{array}$

For the equation above, $x^{2} - 6 = 3 x$ , we see that $b^{2} - 4 a c$ is 33. Since 33 is positive, there are two different real-number solutions.

For $2 x^{2} - x + 4 = 0$ , with $a = 2, b = −1$ , and $c = 4$ , the discriminant, ${(−1)}^{2} - 4 \cdot 2 \cdot 4 = 1 - 32 = −31$ , is negative, so there are two different imaginary-number (or nonreal) solutions.

For $x^{2} - 6 x + 9 = 0$ , with $a = 1, b = −6$ , and $c = 9$ , the discriminant, ${(−6)}^{2} - 4 \cdot 1 \cdot 9 = 36 - 36 = 0$ , is 0 so there is one real-number solution.

Equations reducible to quadratic, or quadratic in form, can be treated as quadratic equations if a suitable substitution is made.

Solve: $x^{4} - x^{2} - 12 = 0$ .

$\begin{array}{l} \begin{array}{rcll} x^{4} - x^{2} - 12 & = & 0 & Let u = x^{2} . Then u^{2} = {(x^{2})}^{2} = x^{4} . \\ u^{2} - u - 12 & = & 0 & Substituting \\ (u - 4) (u + 3) & = & 0 \end{array} \\ \begin{array}{rclcrcll} u - 4 & = & 0 & or & u + 3 & = & 0 \\ u & = & 4 & or & u & = & −3 & Solving for u \\ x^{2} & = & 4 & or & x^{2} & = & −3 \\ x & = & \pm 2 & or & x & = & \pm \sqrt{3} i & Solving for x \end{array} \end{array}$

The solutions are 2, $−2, \sqrt{3} i$ , and $- \sqrt{3} i$ .

Section 3.3: Analyzing Graphs of Quadratic Functions

Graphing Quadratic Equations

The graph of the function $f (x) = a {(x - h)}^{2} + k$ is a parabola that:

opens up if $a > 0$ and down if $a < 0;$
has (h, k) as the vertex;
has $x = h$ as the axis of symmetry;
has k as a minimum value (output) if $a > 0;$
has k as a maximum value if $a < 0$ .

We can use a modification of the technique of completing the square as an aid in analyzing and graphing quadratic functions.

Find the vertex, the axis of symmetry, and the maximum or minimum value of $f (x) = 2 x^{2} + 12 x + 12$ .

$\begin{array}{l} f (x) & = & 2 x^{2} + 12 x + 12 \\ = & 2 (x^{2} + 6 x) + 12 & Note that 9 completes \\ the square for \\ x^{2} + 6 x . \\ = & 2 (x^{2} + 6 x + 9 - 9) + 12 & Adding 9 - 9, \\ or 0, inside the \\ parentheses \\ = & 2 (x^{2} + 6 x + 9) −2 \cdot 9 + 12 & Using the \\ distributive law \\ to remove - 9 \\ from within the \\ parentheses \\ = & 2 {(x + 3)}^{2} - 6 \\ = & 2 {[x - (−3)]}^{2} + (−6) \end{array}$

The function is now written in the form $f (x) = a {(x - h)}^{2} + k$ with $a = 2$ , $h = −3$ , and $k = −6$ . Because $a > 0$ , we know the graph opens up and thus the function has a minimum value. We also know the following:

Vertex $(h, k) : (−3, −6);$
Axis of symmetry $x = h : x = −3;$
Minimum value of the function $k : −6$ .

To graph the function, we first plot the vertex and then find several points on either side of it. We plot these points and connect them with a smooth curve.

The Vertex of a Parabola

The vertex of the graph of $f (x) = a x^{2} + b x + c$ is

Find the vertex of the function $f (x) = −3 x^{2} + 6 x + 1$ .

$\begin{array}{rcl} - \frac{b}{2 a} & = & - \frac{6}{2 (−3)} = 1 \\ f (1) & = & −3 \cdot 1^{2} + 6 \cdot 1 + 1 = 4 \end{array}$

The vertex is $(1, 4)$ .

Some applied problems can be solved by finding the maximum or minimum value of a quadratic function.

See Examples 5–7 on pp. 194–197.

Section 3.4: Solving Rational Equations and Radical Equations

A rational equation is an equation containing one or more rational expressions. When we solve a rational equation, we usually first multiply by the least common denominator (LCD) of all the rational expressions to clear the fractions.

CAUTION! When we multiply by an expression containing a variable, we might not obtain an equation equivalent to the original equation, so we must check the possible solutions obtained by substituting them in the original equation.

Solve.

$\begin{array}{rcll} \frac{5}{x + 2} - \frac{4}{x^{2} - 4} & = & \frac{x - 3}{x - 2} \\ \frac{5}{x + 2} - \frac{4}{(x + 2) (x - 2)} & = & \frac{x - 3}{x - 2} & \begin{array}{l} The LCD is \\ (x + 2) (x - 2) . \end{array} \end{array}$

$(x + 2) (x - 2) (\frac{5}{x + 2} - \frac{4}{(x + 2) (x - 2)})$

$\begin{array}{l} \begin{array}{rcl} = & (x + 2) (x - 2) \cdot \frac{x - 3}{x - 2} \\ 5 (x - 2) −4 & = & (x + 2) (x - 3) \\ 5 x - 10 - 4 & = & x^{2} - x - 6 \\ 5 x - 14 & = & x^{2} - x - 6 \\ 0 & = & x^{2} - 6 x + 8 \\ 0 & = & (x - 2) (x - 4) \end{array} \\ \begin{array}{rclcrcll} x - 2 & = & 0 & or & x - 4 & = & 0 \\ x & = & 2 & or & x & = & 4 \end{array} \end{array}$

The number 2 does not check, but 4 does. The solution is 4.

A radical equation is an equation that contains one or more radicals. We use the principle of powers to solve radical equations.

For any positive integer n:

If $a = b$ is true, then $a^{n} = b^{n}$ is true.

CAUTION! If $a^{n} = b^{n}$ is true, it is not necessarily true that $a = b$ , so we must check the possible solutions obtained by substituting them in the original equation.

Solve: $\sqrt{x + 2} + \sqrt{x - 1} = 3$ .

$\begin{array}{rcll} \sqrt{x + 2} + \sqrt{x - 1} & = & 3 \\ \sqrt{x + 2} & = & 3 - \sqrt{x - 1} & Isolating one radical \\ {(\sqrt{x + 2})}^{2} & = & {(3 - \sqrt{x - 1})}^{2} \\ x + 2 & = & 9 - 6 \sqrt{x - 1} + (x - 1) \\ x + 2 & = & 8 - 6 \sqrt{x - 1} + x \\ −6 & = & −6 \sqrt{x - 1} & Isolating the \\ remaining radical \\ 1 & = & \sqrt{x - 1} & Dividing by - 6 \\ 1^{2} & = & {(\sqrt{x - 1})}^{2} \\ 1 & = & x - 1 \\ 2 & = & x \end{array}$

The number 2 checks. It is the solution.

Section 3.5: Solving Equations and Inequalities with Absolute Value

We use the following property to solve equations with absolute value.

For $a > 0$ and an algebraic expression X:

$| X | = a is equivalent to X = - a or x = a$

Solve: $| x + 1 | = 4$ .

$\begin{array}{l} | x + 1 | = 4 \\ \begin{array}{rclcrcl} x + 1 & = & −4 & or & x + 1 & = & 4 \\ x & = & −5 & or & x & = & 3 \end{array} \end{array}$

Both numbers check. The solutions are −5 and 3.

The following properties are used to solve inequalities with absolute value.

For $a > 0$ and an algebraic expression X:

$| X | < a$ is equivalent to $- a < X < a$ .
$| X | > a$ is equivalent to $X < - a or X > a$ .

Similar statements hold for $| X | \leq a and | X | \geq a$ .

Solve: $| x - 2 | < 3$ .

$\begin{array}{rcll} | x - 2 | & < & 3 \\ −3 & < & x - 2 & < & 3 \\ −1 & < & x < 5 & Adding 2 \end{array}$

The solution set is ${x | −1 < x < 5}$ , or (−1, 5).

$\begin{array}{l} | 3 x | \geq 6 \\ \begin{array}{rclcrcll} 3 x & \leq & −6 & or & 3 x & \geq & 6 \\ x & \leq & −2 & or & x & \geq & 2 & Dividing by 3 \end{array} \end{array}$

The solution set is ${x | x \leq −2 or x \geq 2}$ , or $(- \infty, −2] \cup [2, \infty)$ .

Review Exercises

Determine whether the statement is true or false.

1. We can use the quadratic formula to solve any quadratic equation. [3.2]
2. The function $f (x) = −3 {(x + 4)}^{2} - 1$ has a maximum value. [3.3]
3. For any positive integer n, if $a^{n} = b^{n}$ is true, then $a = b$ is true. [3.4]
4. An equation with absolute value cannot have two negative-number solutions. [3.5]

Solve. [3.2]

5. $(2 y + 5) (3 y - 1) = 0$
6. $x^{2} + 4 x - 5 = 0$
7. $3 x^{2} + 2 x = 8$
8. $5 x^{2} = 15$
9. $x^{2} + 10 = 0$

Find the zero(s) of the function. [3.2]

10. $f (x) = x^{2} - 2 x + 1$
11. $f (x) = x^{2} + 2 x - 15$
12. $f (x) = 2 x^{2} - x - 5$
13. $f (x) = 3 x^{2} + 2 x + 3$

Solve.

14. $\frac{5}{2 x + 3} + \frac{1}{x - 6} = 0$ [3.4]
15. $\frac{3}{8 x + 1} + \frac{8}{2 x + 5} = 1$ [3.4]
16. $\sqrt{5 x + 1} −1 = \sqrt{3 x}$ [3.4]
17. $\sqrt{x - 1} - \sqrt{x - 4} = 1$ [3.4]
18. $| x - 4 | = 3$ [3.5]
19. $| 2 y + 7 | = 9$ [3.5]

Solve and write interval notation for the solution set. Then graph the solution set. [3.5]

20. $| 5 x | \geq 15$
21. $| 3 x + 4 | < 10$
22. $| 1 - 6 x | < 5$
23. $| x + 4 | \geq 2$
24. Solve $\frac{1}{M} + \frac{1}{N} = \frac{1}{P}$ for P. [3.4]

Express in terms of i. [3.1]

25. $- \sqrt{−40}$
26. $\sqrt{−12} \cdot \sqrt{−20}$
27. $\frac{\sqrt{−49}}{- \sqrt{−64}}$

Simplify each of the following. Write the answer in the form $a + b i$ , where a and b are real numbers. [3.1]

28. $(6 + 2 i) + (- 4 - 3 i)$
29. $(3 - 5 i) - (2 - i)$
30. $(6 + 2 i) (- 4 - 3 i)$
31. $\frac{2 - 3 i}{1 - 3 i}$
32. $i^{23}$

Solve by completing the square to obtain exact solutions. Show your work. [3.2]

33. $x^{2} - 3 x = 18$
34. $3 x^{2} - 12 x - 6 = 0$

Solve. Give exact solutions. [3.2]

35. $3 x^{2} + 10 x = 8$
36. $r^{2} - 2 r + 10 = 0$
37. $x^{2} = 10 + 3 x$
38. $x = 2 \sqrt{x} - 1$
39. $y^{4} - 3 y^{2} + 1 = 0$
40. ${(x^{2} - 1)}^{2} - (x^{2} - 1) - 2 = 0$
41. $(p + 2) (3 p + 2) (p - 3) = 0$
42. $x^{3} + 5 x^{2} - 4 x - 20 = 0$

In Exercises 43 and 44, complete the square to (a) find the vertex; (b) find the axis of symmetry; (c) determine whether there is a maximum or minimum value and find that value; (d) find the range; and (e) graph the function. [3.3]

43. $f (x) = - 4 x^{2} + 3 x - 1$
44. $f (x) = 5 x^{2} - 10 x + 3$

In Exercises 45–48, match the equation with one of the figures (a)–(d) that follow. [3.3]

45. $y = {(x - 2)}^{2}$
46. $y = {(x + 3)}^{2} - 4$
47. $y = - 2 {(x + 3)}^{2} + 4$
48. $y = - \frac{1}{2} {(x - 2)}^{2} + 5$
49. Legs of a Right Triangle. The hypotenuse of a right triangle is 50 ft. One leg is 10 ft longer than the other. What are the lengths of the legs?

[3.2]
50. Bicycling Speed. Harry and Rebecca leave a campsite, Harry biking due north and Rebecca biking due east. Harry bikes 7 km/h slower than Rebecca. After 4 hr, they are 68 km apart. Find the speed of each bicyclist.
[3.2]
51. Sidewalk Width. A 60-ft by 80-ft parking lot is torn up to install a sidewalk of uniform width around its perimeter. The new area of the parking lot is two-thirds of the old area. How wide is the sidewalk?

[3.2]
52. Maximizing Volume. The Garcias have 24 ft of flexible fencing with which to build a rectangular “toy corral.” If the fencing is 2 ft high, what dimensions should the corral have in order to maximize its volume?
[3.3]
53. Dimensions of a Box. An open box is made from a 10-cm by 20-cm piece of aluminum by cutting a square from each corner and folding up the edges. The area of the resulting base is $90 {cm}^{2}$ . What is the length of the sides of the squares? [3.2]
54. Find the zeros of $f (x) = 2 x^{2} - 5 x + 1$ . [3.2]
1. $\frac{5 \pm \sqrt{17}}{2}$
2. $\frac{5 \pm \sqrt{17}}{4}$
3. $\frac{5 \pm \sqrt{33}}{4}$
4. $\frac{- 5 \pm \sqrt{17}}{4}$
55. Solve: $\sqrt{4 x + 1} + \sqrt{2 x} = 1$ . [3.4]
1. There are two solutions.
2. There is only one solution. It is less than 1.
3. There is only one solution. It is greater than 1.
4. There is no solution.
56. The graph of $f (x) = {(x - 2)}^{2} - 3$ is which of the following? [3.3]

Synthesis

Solve.

57. $\sqrt{\sqrt{\sqrt{x}}} = 2$ [3.4]
58. ${(t - 4)}^{4 / 5} = 3$ [3.4]
59. ${(x - 1)}^{2 / 3} = 4$ [3.4]
60. ${(2 y - 2)}^{2} + y - 1 = 5$ [3.2]
61. $\sqrt{x + 2} + \sqrt[4]{x + 2} - 2 = 0$ [3.2]
62. At the beginning of the year, $3500 was deposited in a savings account. One year later, $4000 was deposited in another account. The interest rate was the same for both accounts. At the end of the second year, there was a total of $8518.35 in the accounts. What was the annual interest rate? [3.2]
63. Find b such that $f (x) = - 3 x^{2} + b x - 1$ has a maximum value of 2.
[3.3]

Collaborative Discussion and Writing

64. Is the product of two imaginary numbers always an imaginary number? Explain your answer.
[3.1]
65. Is it possible for a quadratic function to have one real zero and one imaginary zero? Why or why not? [3.2]
66. If the graphs of

$f (x) = a_{1} {(x - h_{1})}^{2} + k_{1}$

and

$g (x) = a_{2} {(x - h_{2})}^{2} + k_{2}$

have the same shape, what, if anything, can you conclude about the a’s, the h’s, and the k’s? Explain your answer. [3.3]
67. Explain why it is necessary to check the possible solutions of a rational equation. [3.4]
68. Explain in your own words why it is necessary to check the possible solutions when the principle of powers is used to solve an equation. [3.4]
69. Explain why $| x | < p$ has no solution for $p \leq 0$ . [3.5]
70. Explain why all real numbers are solutions of $| x | > p$ , for $p < 0$ .
[3.5]

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Table of Contents for Chapter 3 Summary and Review

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Table of Contents for
Chapter 3 Summary and Review