Search in book...
Toggle Font Controls
Create new playlist

Name your new playlist

Playlist description (optional)
Sign In

Email address

Password

Forgot Password?

or

Continue with Facebook

Continue with Google
Sign Up

Full Name

Email address

Confirm Email Address

Password

or

Continue with Facebook

Continue with Google

Chapter 1 Review and Tests

Review

Definitions, Concepts, and Formulas

Examples

1.1 Linear Equations in One Variable

An equation is a statement that two mathematical expressions are equal.
The solutions or roots of an equation are those values (if any) of the variable that satisfy the equation.
Equivalent equations are equations that have the same solution set.
You can generate equivalent equations using the operations given on page 83.
A formula is an equation that expresses a relationship between two or more variables.
The standard form of a conditional linear equation in x is $a x + b = 0, a \neq 0 .$ $a x + b = 0, a \neq 0 .$

Solve the equation $5 x + 3 = 2 (x + 1) + 7 .$ $5 x + 3 = 2 (x + 1) + 7 .$

Solution

\begin{array}{rcll} 5 x + 3 & = & 2 x + 2 + 7 & Distributive property \\ 5 x + 3 & = & 2 x + 9 & Simplify . \\ 5 x + 3 - 3 & = & 2 x + 9 - 3 & Subtract 3 from both sides . \\ 5 x & = & 2 x + 6 & Simplify . \\ 5 x - 2 x & = & 2 x + 6 - 2 x & Subtract 2 x from both sides . \\ 3 x & = & 6 & Simplify . \\ x & = & 2 & Divide both sides by 3 . \end{array}

$\begin{array}{rcll} 5 x + 3 & = & 2 x + 2 + 7 & Distributive property \\ 5 x + 3 & = & 2 x + 9 & Simplify . \\ 5 x + 3 - 3 & = & 2 x + 9 - 3 & Subtract 3 from both sides . \\ 5 x & = & 2 x + 6 & Simplify . \\ 5 x - 2 x & = & 2 x + 6 - 2 x & Subtract 2 x from both sides . \\ 3 x & = & 6 & Simplify . \\ x & = & 2 & Divide both sides by 3 . \end{array}$

Each equation above is equivalent to the preceding equation, and so to one another.

1.2 Applications of Linear Equations

Modeling and Solving an Applied Problem

Step 1 Read the problem.
Step 2 Assign a variable.
Step 3 Write an equation.
Step 4 Solve the equation.
Step 5 Answer the original question.
Step 6 Check the answer using the information in the original problem.

A bus leaves Houston heading for New Orleans traveling at 40 mph. Three hours later a motorcycle leaves Houston heading for New Orleans traveling at 55 mph on the same route as the bus. How far is the motorcycle from Houston when it overtakes the bus?

Solution

Use the $d = r t$ $d = r t$ formula and let t represent the time it takes the motorcycle to overtake the bus. We need this time to find the distance.

Organize the given information.

Object	`r`, mph	`t`, hours	$d = r t$ $d = r t$
Motorcycle	55	`t`	55`t`
Bus	40	$t + 3$ $t + 3$	$40 (t + 3)$ $40 (t + 3)$

\begin{array}{rcll} 55 t & = & 40 (t + 3) & Both traveled the same \\ distance over time t . \\ 55 t & = & 40 t + 120 & Distributive property \\ 55 t & - & 40 t = 120 & Subtract 40 t from both \\ sides and simplify \\ 15 t & = & 120 & Simplify . \\ t & = & 8 & Divide both sides by 15 . \end{array}

$\begin{array}{rcll} 55 t & = & 40 (t + 3) & Both traveled the same \\ distance over time t . \\ 55 t & = & 40 t + 120 & Distributive property \\ 55 t & - & 40 t = 120 & Subtract 40 t from both \\ sides and simplify \\ 15 t & = & 120 & Simplify . \\ t & = & 8 & Divide both sides by 15 . \end{array}$

It takes the motorcycle 8 hours to overtake the bus. Since the motorcycle travels at 55 mph, it travels $8 \cdot 55 = 440$ $8 \cdot 55 = 440$ miles and overtakes the bus 440 miles from Houston.

The bus travels $t + 3 = 8 + 3 = 11$ $t + 3 = 8 + 3 = 11$ hours at 40 mph, so it travels $11 \cdot 40 = 440$ $11 \cdot 40 = 440$ miles also.

1.3 Quadratic Equations

The equation $a x^{2} + b x + c = 0, a \neq 0,$ $a x^{2} + b x + c = 0, a \neq 0,$ is the standard form of the quadratic equation in x.
Zero-product property: Let A and B be two algebraic expressions. Then $A B = 0$ $A B = 0$ if and only if $A = 0$ $A = 0$ or $B = 0 .$ $B = 0 .$
Square root method: If $u^{2} = d,$ $u^{2} = d,$ then $u = \pm \sqrt{d} .$ $u = \pm \sqrt{d} .$
Method for completing the square:

Add ${(\frac{1}{2} b)}^{2}$ ${(\frac{1}{2} b)}^{2}$ to $x^{2} + b x$ $x^{2} + b x$ to get the perfect square ${(x + \frac{1}{2} b)}^{2} .$ ${(x + \frac{1}{2} b)}^{2} .$ See page 110.
Quadratic formula: $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

Solve by using the zero-product property:

Solution

\begin{array}{rcll} x^{2} + 8 x + 15 & = & 0 \\ (x + 5) \cdot (x + 3) & = & 0 & Factor x^{2} + 8 x + 15 . \\ x + 5 = 0 or x + 3 & = & 0 & Set each factor equal to 0 . \\ x = - 5 or x & = & - 3 & Solve each equation . \end{array}

$\begin{array}{rcll} x^{2} + 8 x + 15 & = & 0 \\ (x + 5) \cdot (x + 3) & = & 0 & Factor x^{2} + 8 x + 15 . \\ x + 5 = 0 or x + 3 & = & 0 & Set each factor equal to 0 . \\ x = - 5 or x & = & - 3 & Solve each equation . \end{array}$

The solution set is ${- 5, - 3} .$ ${- 5, - 3} .$

Solve $x^{2} + 10 x - 2 = 0$ $x^{2} + 10 x - 2 = 0$ by completing the square.

Solution

\begin{array}{l} \begin{array}{rcll} x^{2} + 10 x & = & 2 & Add 2 to both sides; \\ simplify . \\ x^{2} + 10 x + 5^{2} & = & 2 + 5^{2} & Add {[\frac{1}{2} (10)]}^{2} = 5^{2} to \\ both sides . \\ {(x + 5)}^{2} & = & 27 & Factor the left side; \\ simplify . \\ x + 5 = \pm \sqrt{27} & = & \pm 3 \sqrt{3} & Take the square root of \\ both sides . \sqrt{27} = \sqrt{9 \cdot 3} \end{array} \\ \begin{array}{ll} x + 5 = 3 \sqrt{3} or x + 5 = - 3 \sqrt{3} & Rewrite \\ x + 5 = \pm 3 \sqrt{3} \\ x = - 5 + 3 \sqrt{3} or x = - 5 - 3 \sqrt{3} & Subtract 5 from \\ each equation . \end{array} \end{array}

$\begin{array}{l} \begin{array}{rcll} x^{2} + 10 x & = & 2 & Add 2 to both sides; \\ simplify . \\ x^{2} + 10 x + 5^{2} & = & 2 + 5^{2} & Add {[\frac{1}{2} (10)]}^{2} = 5^{2} to \\ both sides . \\ {(x + 5)}^{2} & = & 27 & Factor the left side; \\ simplify . \\ x + 5 = \pm \sqrt{27} & = & \pm 3 \sqrt{3} & Take the square root of \\ both sides . \sqrt{27} = \sqrt{9 \cdot 3} \end{array} \\ \begin{array}{ll} x + 5 = 3 \sqrt{3} or x + 5 = - 3 \sqrt{3} & Rewrite \\ x + 5 = \pm 3 \sqrt{3} \\ x = - 5 + 3 \sqrt{3} or x = - 5 - 3 \sqrt{3} & Subtract 5 from \\ each equation . \end{array} \end{array}$

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

1.4 Complex Numbers: Quadratic Equations with Complex Solutions

Complex numbers are of the form $a + b i,$ $a + b i,$ where a and b are real numbers and $i = \sqrt{- 1}; i^{2} = - 1 .$ $i = \sqrt{- 1}; i^{2} = - 1 .$
The number $a - b i$ $a - b i$ is called the complex conjugate of $a + b i$ $a + b i$ .
Operations with complex numbers can be performed as if they are binomials with the variable i. Set $i^{2} = - 1$ $i^{2} = - 1$ to simplify.

Division is performed by first multiplying the numerator and the denominator by the complex conjugate of the denominator.
The quantity $b^{2} - 4 a c$ $b^{2} - 4 a c$ in the quadratic formula is called the discriminant $(= D)$ $(= D)$ .
If $D > 0,$ $D > 0,$ the quadratic equation has two distinct real solutions.

If $D = 0,$ $D = 0,$ there is one real solution.

If $D < 0,$ $D < 0,$ there are two nonreal complex solutions, and they are conjugates.

The complex number $- 2 + 13 i,$ $- 2 + 13 i,$ has real part $- 2$ $- 2$ and imaginary part 13.

The complex conjugate of $- 2 + 13 i$ $- 2 + 13 i$ is $- 2 - 13 i$ $- 2 - 13 i$ ,

$\begin{array}{l} (7 + 3 i) + (2 + i) = (7 + 2) + (3 + 1) i = 9 + 4 i \\ (4 - 5 i) - (2 + 3 i) = (4 - 2) + (- 5 - 3) i = 2 - 8 i \end{array}$ $\begin{array}{l} (7 + 3 i) + (2 + i) = (7 + 2) + (3 + 1) i = 9 + 4 i \\ (4 - 5 i) - (2 + 3 i) = (4 - 2) + (- 5 - 3) i = 2 - 8 i \end{array}$

$\begin{array}{l} (2 + 3 i) (4 + 5 i) = 8 + 10 i + 12 i + 15 i^{2} = \\ 8 + 10 i + 12 i + 15 (- 1) = - 7 + 22 i \end{array}$ $\begin{array}{l} (2 + 3 i) (4 + 5 i) = 8 + 10 i + 12 i + 15 i^{2} = \\ 8 + 10 i + 12 i + 15 (- 1) = - 7 + 22 i \end{array}$

$\begin{array}{l} \frac{2 + 3 i}{4 - i} = \frac{(2 + 3 i) (4 + i)}{(4 - i) (4 + i)} \\ \begin{array}{l} = & \frac{8 + 2 i + 12 i + 3 i^{2}}{16 + 4 i - 4 i - i^{2}} = \frac{8 + 2 i + 12 i - 3}{16 + 4 i - 4 i + 1} \\ = & \frac{5 + 14 i}{17} = \frac{5}{17} + \frac{14}{17} i \end{array} \end{array}$ $\begin{array}{l} \frac{2 + 3 i}{4 - i} = \frac{(2 + 3 i) (4 + i)}{(4 - i) (4 + i)} \\ \begin{array}{l} = & \frac{8 + 2 i + 12 i + 3 i^{2}}{16 + 4 i - 4 i - i^{2}} = \frac{8 + 2 i + 12 i - 3}{16 + 4 i - 4 i + 1} \\ = & \frac{5 + 14 i}{17} = \frac{5}{17} + \frac{14}{17} i \end{array} \end{array}$

For $4 x^{2} + 4 x - 7 = 0,$ $4 x^{2} + 4 x - 7 = 0,$ we have $a = 4, b = 4,$ $a = 4, b = 4,$ and $c = - 7$ $c = - 7$ . Then $D = b^{2} - 4 a c = 4^{2} - 4 \cdot 4 (- 7) = 128 > 0,$ $D = b^{2} - 4 a c = 4^{2} - 4 \cdot 4 (- 7) = 128 > 0,$ so $4 x^{2} + 4 x - 7 = 0$ $4 x^{2} + 4 x - 7 = 0$ has two distinct real solutions.

1.5 Solving Other Types of Equations

Some equations can be recognized as quadratic in form by replacing a recurrent expression by a variable, solving the resulting quadratic equation, replacing the variable by the recurrent expression, and solving the equations that result.

Many types of equations can be solved by factoring. However, it is possible to introduce extraneous solutions when solving rational equations, equations involving radicals, or equations involving rational exponents. Remember to check the solutions.

Extraneous solutions may be introduced when you:

Multiply both sides by an LCD that contains a variable.
Raise both sides to an even power.

Solve $x^{4} - 10 x^{2} + 9 = 0 .$ $x^{4} - 10 x^{2} + 9 = 0 .$

Solution

We rewrite $x^{4} - 10 x^{2} + 9 = 0$ $x^{4} - 10 x^{2} + 9 = 0$ as ${(x^{2})}^{2} - 10 x^{2} + 9 = 0$ ${(x^{2})}^{2} - 10 x^{2} + 9 = 0$ and then replace $x^{2}$ $x^{2}$ by u to get $u^{2} - 10 u + 9 = 0 .$ $u^{2} - 10 u + 9 = 0 .$

\begin{array}{ll} (u - 9) (u - 1) = 0 & Factor u^{2} - 10 u + 9 = 0 . \\ \begin{array}{rcllrcl} u - 9 & = & 0 & or & u - 1 & = & 0 \\ u & = & 9 & or & u & = & 1 \end{array} & \begin{array}{l} Set each factor equal to 0 . \\ Solve each equation . \end{array} \\ \begin{array}{rcllrcl} x^{2} & = & 9 & or & x^{2} & = & 1 \\ x & = & \pm 3 & or & x & = & \pm 1 \end{array} & \begin{array}{l} Replace u by x^{2} . \\ Extract square roots . \end{array} \end{array}

$\begin{array}{ll} (u - 9) (u - 1) = 0 & Factor u^{2} - 10 u + 9 = 0 . \\ \begin{array}{rcllrcl} u - 9 & = & 0 & or & u - 1 & = & 0 \\ u & = & 9 & or & u & = & 1 \end{array} & \begin{array}{l} Set each factor equal to 0 . \\ Solve each equation . \end{array} \\ \begin{array}{rcllrcl} x^{2} & = & 9 & or & x^{2} & = & 1 \\ x & = & \pm 3 & or & x & = & \pm 1 \end{array} & \begin{array}{l} Replace u by x^{2} . \\ Extract square roots . \end{array} \end{array}$

The solutions set is ${- 3, - 1, 1, 3} .$ ${- 3, - 1, 1, 3} .$

1.6 Inequalities

An inequality is a statement that one algebraic expression is less than or is less than or equal to another algebraic expression.
The real numbers that result in a true statement when those numbers are substituted for the variable in the inequality are called solutions of the inequality.
A linear inequality is an inequality that can be written in the form $a x + b < 0 .$ $a x + b < 0 .$ The symbol $<$ $<$ can be replaced with $\leq, >,$ $\leq, >,$ or $\geq$ $\geq$ .
Test points can be used to solve polynomial or rational inequalities. See page 151.

Solve $4 (x + 1) \geq 3 x + 7$ $4 (x + 1) \geq 3 x + 7$

Solution

\begin{array}{l} 4 x + 4 \geq 3 x + 7 & Distributive property \\ 4 x \geq 3 x + 3 & Subtract 4 from each side and simplify . \\ x \geq 3 & Subtract 3 x from each side and simplify . \end{array}

$\begin{array}{l} 4 x + 4 \geq 3 x + 7 & Distributive property \\ 4 x \geq 3 x + 3 & Subtract 4 from each side and simplify . \\ x \geq 3 & Subtract 3 x from each side and simplify . \end{array}$

The solution set is $[3, \infty) .$ $[3, \infty) .$

Solve $x^{2} + x - 12 > 0$ $x^{2} + x - 12 > 0$

Solution

\begin{array}{l} \begin{array}{l} (x + 4) (x - 3) > 0 & Factor . \end{array} \\ \begin{array}{rcllrl} x + 4 & = & 0 & or & x - 3 = 0 & Set each factor equal to 0 . \\ x & = & - 4 & or & x = 3 & Solve each equation . \end{array} \end{array}

$\begin{array}{l} \begin{array}{l} (x + 4) (x - 3) > 0 & Factor . \end{array} \\ \begin{array}{rcllrl} x + 4 & = & 0 & or & x - 3 = 0 & Set each factor equal to 0 . \\ x & = & - 4 & or & x = 3 & Solve each equation . \end{array} \end{array}$

The points $- 4$ $- 4$ and 3 determine the intervals $(- \infty, - 4), (- 4, 3),$ $(- \infty, - 4), (- 4, 3),$ and $(3, \infty) .$ $(3, \infty) .$

Choose a test point in each interval; we choose: $- 5, 0,$ $- 5, 0,$ and 4.

Test: $- 5$ $- 5$ in $(x + 4) (x - 3)$ $(x + 4) (x - 3)$ to get $(- 5 + 4) (- 5 - 3) = 8 > 0 .$ $(- 5 + 4) (- 5 - 3) = 8 > 0 .$

Test: 0 in $(x + 4) (x - 3)$ $(x + 4) (x - 3)$ to get $(0 + 4) (0 - 3) = - 12 < 0 .$ $(0 + 4) (0 - 3) = - 12 < 0 .$

Test: 4 in $(x + 4) (x - 3)$ $(x + 4) (x - 3)$ to get $(4 + 4) (4 - 3) = 8 > 0 .$ $(4 + 4) (4 - 3) = 8 > 0 .$

So $x^{2} + x - 12 > 0$ $x^{2} + x - 12 > 0$ in the intervals containing $- 5$ $- 5$ and 4. The solution set is $(- \infty, - 4) \cup (3, \infty) .$ $(- \infty, - 4) \cup (3, \infty) .$

1.7 Equations and Inequalities Involving Absolute Value

Definition; $| a | = a$ $| a | = a$ if $a \geq 0$ $a \geq 0$ and $| a | = - a$ $| a | = - a$ if $a < 0 .$ $a < 0 .$
Let u be a variable or an algebraic expression and let $a > 0 .$ $a > 0 .$

Then

$| u | = a$ $| u | = a$ if and only if $u = - a$ $u = - a$ or $u = a .$ $u = a .$
$| u | < a$ $| u | < a$ if and only if $- a < u < a .$ $- a < u < a .$
$| u | > a$ $| u | > a$ if and only if $u < - a$ $u < - a$ or $u > a .$ $u > a .$

The properties in (a) and (b) are valid if $<$ $<$ and $>$ $>$ are replaced with $\leq$ $\leq$ and $\geq,$ $\geq,$ respectively.

Solve $| 3 x - 2 | = 7$ $| 3 x - 2 | = 7$

Solution

$| 3 x - 2 | = 7$ $| 3 x - 2 | = 7$ if and only if $3 x - 2 = - 7$ $3 x - 2 = - 7$ or $3 x - 2 = 7 .$ $3 x - 2 = 7 .$

\begin{array}{rcllrcll} 3 x & = & - 5 & or & 3 x & = & 9 & Add 2 to each equation and simplify . \\ x & = & - \frac{5}{3} & or & x & = & 3 & \begin{array}{l} Divide each equation by 3 and \\ simplify . \end{array} \end{array}

$\begin{array}{rcllrcll} 3 x & = & - 5 & or & 3 x & = & 9 & Add 2 to each equation and simplify . \\ x & = & - \frac{5}{3} & or & x & = & 3 & \begin{array}{l} Divide each equation by 3 and \\ simplify . \end{array} \end{array}$

The solution set is ${- \frac{5}{3}, 3} .$ ${- \frac{5}{3}, 3} .$

Solve $| 2 x + 3 | < 15$ $| 2 x + 3 | < 15$

Solution

$| 2 x + 3 | < 15$ $| 2 x + 3 | < 15$ if and only if $- 15 < 2 x + 3 < 15 .$ $- 15 < 2 x + 3 < 15 .$

\begin{array}{l} - 18 < 2 x < 12 & Subtract 3 from each part and simplify . \\ - 9 < x < 6 & Divide each part by 2 and simplify . \end{array}

$\begin{array}{l} - 18 < 2 x < 12 & Subtract 3 from each part and simplify . \\ - 9 < x < 6 & Divide each part by 2 and simplify . \end{array}$

The solution set is $(- 9, 6) .$ $(- 9, 6) .$

Solve $| 5 x - 7 | > 8$ $| 5 x - 7 | > 8$

Solution

$| 5 x - 7 | > 8$ $| 5 x - 7 | > 8$ if and only if $5 x - 7 < - 8$ $5 x - 7 < - 8$ or $5 x - 7 > 8 .$ $5 x - 7 > 8 .$

\begin{array}{rcllrcll} 5 x & < & - 1 & or & 5 x & > & 15 & Add 7 to each part and simplify . \\ x & < & - \frac{1}{5} & or & x & > & 3 & Divide each part by 5 and simplify . \end{array}

$\begin{array}{rcllrcll} 5 x & < & - 1 & or & 5 x & > & 15 & Add 7 to each part and simplify . \\ x & < & - \frac{1}{5} & or & x & > & 3 & Divide each part by 5 and simplify . \end{array}$

The solution set is $(- \infty, - \frac{1}{5}) \cup (3, \infty) .$ $(- \infty, - \frac{1}{5}) \cup (3, \infty) .$

..................Content has been hidden....................

You can't read the all page of ebook, please click here login for view all page.

Table of Contents for Chapter 1 Review and Tests

Create new playlist

Sign In

Sign Up

Table of Contents for
Chapter 1 Review and Tests