Targeted Review Solutions

Simplify.

$\sqrt{35 x} \cdot \sqrt{15 x^{3}}$

$\sqrt{5 \cdot 7 \cdot 3 \cdot 5 \cdot x^{4}} 3^{\frac{1}{2}} \cdot 5^{\frac{2}{2}} \cdot 7^{\frac{1}{2}} \cdot x^{\frac{4}{2}}$

$5 x^{2} \sqrt{21}$

$(10 + \sqrt{7}) - (8 + 12 \sqrt{7})$

$10 + \sqrt{7} - 8 - 12 \sqrt{7}$

$2 - 11 \sqrt{7}$

$\frac{2 + \sqrt{5}}{10 + \sqrt{6}}$

$\frac{2 + \sqrt{5}}{10 + \sqrt{6}} (\frac{10 - \sqrt{6}}{10 - \sqrt{6}}) \frac{20 - 2 \sqrt{6} + 10 \sqrt{5} - \sqrt{30}}{100 - 6}$

$\frac{20 - 2 \sqrt{6} + 10 \sqrt{5} - \sqrt{30}}{94}$

Solve.

$\sqrt[3]{x + 3} = 3$

$\begin{array}{rcl} {(\sqrt[3]{x + 3})}^{3} & = & {(3)}^{3} \\ x + 3 & = & 27 \end{array}$

$x = 24$

$\sqrt{x} + 5 \geq 7$

$\begin{array}{rcl} \sqrt{x} & \geq & 2 \\ {(\sqrt{x})}^{2} & \geq & {(2)}^{2} \end{array}$

$x \geq 4$

$\sqrt{2 x - 3} < 7$

$\begin{array}{rcl} 2 x - 3 & \geq & 0 \\ x & \geq & 1.5 principal root \\ {(\sqrt{2 x - 3})}^{2} & < & {(7)}^{2} \\ 2 x - 3 & < & 49 \\ 2 x & < & 52 \\ x & < & 26 \end{array}$

$\begin{array}{rcl} 1.5 & \leq & x < 26 \end{array}$

Graph.

$y = \frac{4}{3} x - 2$

$m = \frac{4}{3}, b = - 2$

$y = - \frac{1}{x - 3} - 2$

$asymptotes : x = 3, y = 2 a = - 1$

Multiple Choice

Simplify. $\sqrt[3]{48 x^{8} y^{3}}$

$2 x^{2} y \sqrt[3]{6 x^{2}}$
$2 x^{2} y \sqrt[3]{3 x^{2}}$
$4 x^{4} y^{2} \sqrt[3]{3 y}$
$16 x^{2} y \sqrt[3]{x^{2}}$

$2^{\frac{4}{3}} 3^{\frac{1}{3}} x^{\frac{8}{3}} y^{\frac{3}{3}} 2^{1 \frac{1}{3}} 3^{\frac{1}{3}} x^{2 \frac{2}{3}} y^{1} 2 x^{2} y \sqrt[3]{6 x^{2}}$

Note

The coefficient of the radicand is missing a factor of 2.

This option uses an index of 2 rather than 3.

This option divides 48 by 3 rather than using the prime factorization and exponent rules.

Solve. $\sqrt{3 x - 1} - 6 = - 2$

$\frac{1}{3}$
5
$\frac{17}{3}$
no solution

$\begin{array}{rcl} \sqrt{3 x - 1} & = & 4 \\ {(\sqrt{3 x - 1})}^{2} & = & {(4)}^{2} \\ 3 x - 1 & = & 16 \\ 3 x & = & 17 \\ x & = & \frac{17}{3} \end{array}$

Note

This option represents the restriction of the principal root.

This option occurs if 1 is subtracted rather than added to both sides.

The radical needs to be isolated on one side before determining if there is no solution.

Name the range using the graph of the rational function.

$\{y | y \in ℝ, y \neq - 1\}$
$\{y | y \in ℝ, y \neq 0\}$
$\{y | y \in ℝ, y \neq 4\}$
$\{y | y \in ℝ, y \neq 5\}$

Note

–1 represents the restriction on the domain.

This option is the x-axis, and not a restriction for the range.

This option is the value of the y-intercept.

Determine the equation that best represents the linear function.

$y = \frac{1}{4} x + 2$
$y = - \frac{1}{4} x + 2$
$y = - \frac{1}{5} x + 2$
$y = \frac{1}{5} x + 2$

Note

A, D) The slope of the graph is negative.

C, D) Using $(- 4, 3)$ and $(0, 2)$ , the slope is $- \frac{1}{4}$ .

Problem	1	2	3	4	5	6	7	8	9	10	11	12
Origin	L11	L12	L12	L13	L14	L14	A1	L10	L11	L13	L10	A1

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