Targeted Review Solutions

Write answers in simplified radical form. Use absolute value bars when necessary.
$\sqrt[4]{162 a^{7} b^{12} c^{4}}$

$\sqrt[4]{3^{4} \cdot 2 a^{7} b^{12} c^{4}} 3^{\frac{4}{4}} \cdot 2^{\frac{1}{4}} a^{\frac{7}{4}} b^{\frac{12}{4}} c^{\frac{4}{4}} 3^{1} \cdot 2^{\frac{1}{4}} a^{1 \frac{3}{4}} b^{3} c^{1}$

$3 | a b^{3} c | \sqrt[4]{2 a^{3}}$

Rationalize the denominator.

$\frac{2 x - 5}{x + \sqrt{3}}$

$\frac{2 x - 5}{x + \sqrt{3}} (\frac{x - \sqrt{3}}{x - \sqrt{3}})$

$\frac{2 x^{2} - 2 x \sqrt{3} - 5 x + 5 \sqrt{3}}{x^{2} - 3}$

Solve. Check your work.
$\sqrt{x + 8} = \sqrt{2 x - 1}$

$\begin{array}{rcl} {(\sqrt{x + 8})}^{2} & = & {(\sqrt{2 x - 1})}^{2} \\ x + 8 & = & 2 x - 1 \\ x & = & 9 \\ Check \sqrt{9 + 8} & = & \sqrt{2 (9) - 1} \\ \sqrt{17} & = & \sqrt{18 - 1} \end{array}$

$x = 9$

Solve.
$\sqrt{4 x + 5} > \sqrt{4 x + 1} + 2$

$\begin{array}{rcl} \begin{matrix} Restrictions & Solve \\ 4 x + 5 \geq 0 & 4 x + 1 \geq 0 & {(\sqrt{4 x + 5})}^{2} > {(\sqrt{4 x + 1} + 2)}^{2} \\ x \geq - \frac{5}{4} & x \geq - \frac{1}{4} & 4 x + 5 > 4 x + 1 + 4 \sqrt{4 x + 1} + 4 \\ 0 > 4 \sqrt{4 x + 1} \\ 0 > \sqrt{4 x + 1} \\ {(0)}^{2} > {(\sqrt{4 x + 1})}^{2} 0 > 4 x + 1 - 1 > 4 x \\ x < - \frac{1}{4} \end{matrix} \end{array}$

No solution.

Note

A solution cannot be both greater than or equal to and less than the same number.

Simplify. Then classify by all sets to which it belongs: real, pure imaginary, complex.
$i (\sqrt{- 12} + 3 i) - 2 \sqrt{3} + i \sqrt{- 49}$

$i (2 i \sqrt{3} + 3 i) - 2 \sqrt{3} + 7 i^{2} 2 i^{2} \sqrt{3} + 3 i^{2} - 2 \sqrt{3} + 7 i^{2} 2 (- 1) \sqrt{3} + 3 (- 1) - 2 \sqrt{3} + 7 (- 1) - 2 \sqrt{3} - 3 - 2 \sqrt{3} - 7$

$- 10 - 4 \sqrt{3} real, complex$

Simplify.
$i^{67}$

$67 \div 4 = 16 R 3 {(i^{4})}^{16} \cdot i^{3} 1 (- i)$

–i

Write the equation for each parent function in vertex form.

a cubic $y = a {(x - h)}^{3} + k$
absolute value $y = a |x - h| + k$
square root $y = a \sqrt{(x - h)} + k$
quadratic $y = a {(x - h)}^{2} + k$
rational $y = \frac{a}{x - h} + k, x \neq h$

Two students were asked to describe the transformation from the parent function using a, h, and k. Explain the errors and if either student is correct. State the correct answer if both students are incorrect. $y = - \frac{1}{4} \sqrt{x + 14} + 33$

Sample: Student A did not correctly identify the value of h under the radical. Student B changed all of the signs of a, h, and k to the opposite. The correct values of a, h, and k are $a = - \frac{1}{4}, h = - 14, k = 33$

Name domain restrictions to represent the function.

$\{x | x \in ℝ, x > 3\}$
$\{x | x \in ℝ, x \geq 3\}$
$\{x | x \in ℝ, x \geq 1\}$
$\{x | x \in ℝ, x \geq 6\}$

$a = 1, h = 3, k = 1$

This image has an empty alt attribute; its file name is image-5-1024x785.png

Note

This option does not include the initial point at (3, 1).

This option is the range restriction.

This option is the value of the graph if the scale of the graph is ignored.

Solve.
$\sqrt{2 x + 2} = 3 \sqrt{2 x - 4}$

$\frac{19}{8}$
$\frac{7}{2}$
$\frac{3}{8}$
no solution

$\begin{array}{rcl} {(\sqrt{2 x + 2})}^{2} & = & {(3 \sqrt{2 x - 4})}^{2} \\ 2 x + 2 & = & 9 (2 x - 4) \\ 2 x + 2 & = & 18 x - 36 \\ 38 & = & 16 x \\ x & = & \frac{38}{16} = \frac{19}{8} \end{array}$

Note

This option is the solution when the coefficient is not squared

This option is the solution when the coefficient is not distributed across the terms on the right side of the equation

This option is the solution if the terms are combined without adhering to order of operations

Name the equation that represents the graph of the transformed parent function.

$y = - |x + 3|$
$y = 3 |x + 1|$
$y = - 3 |x + 1|$
$y = - 3 |x - 1|$

Vertex $(- 1, 0)$ so $h = - 1$

The next whole number point is $(0, - 3)$ so $a = - 3$

Note

$h = - 1$ on the graph, not –3

This equation does not show a reflected absolute value graph

The value of $h = 1$ which represents $(1, 0)$ but the graphs shows $(- 1, 0)$

Select the graph that matches the description of the end behavior.
$As x \to + \infty, f (x) \to + \infty, and as x \to - \infty, f (x) \to + \infty$

This graph shows that all values of y are increasing.

Note

The y-values are negative when the x-values are negative.
The y-values are negative when the x-values are positive.
This graph only contains negative x-values.

Problem	1	2	3	4	5	6	7	8	9	10	11	12
Origin	L11	L12	L13	L14	L16	L15	L17	L18	L18	L13	L18	L18

L = Lesson in this level, A1 = Algebra 1: Principles of Secondary Mathematics, FD = Foundational Knowledge