Targeted Review Solutions

Simplify. Rationalize denominators.

$\frac{6}{\sqrt[3]{5 x^{2} y}}$

$\frac{6}{\sqrt[3]{5 x^{2} y}} (\frac{\sqrt[3]{5^{2} x y^{2}}}{\sqrt[3]{5^{2} x y^{2}}})$

$\frac{6 \sqrt[3]{25 x y^{2}}}{5 x y}$

$(i^{18}) (i^{5})$

$i^{23} = i^{20} \cdot i^{3} {(i^{4})}^{5} \cdot - i$

–i

$\frac{2 - \sqrt{3}}{6 + \sqrt{3}}$

$\frac{2 - \sqrt{3}}{6 + \sqrt{3}} (\frac{6 - \sqrt{3}}{6 - \sqrt{3}}) \frac{12 - 2 \sqrt{3} - 6 \sqrt{3} + 3}{36 - 3}$

$\frac{15 - 8 \sqrt{3}}{33}$

$\frac{8 a b}{\sqrt[4]{3 a^{2} b}}$

$\frac{8 a b}{\sqrt[4]{3 a^{2} b}} (\frac{\sqrt[4]{3^{3} a^{2} b^{3}}}{\sqrt[4]{3^{3} a^{2} b^{3}}}) \frac{8 a b \sqrt[4]{3^{3} a^{2} b^{3}}}{3 a b}$

$\frac{8 \sqrt[4]{3^{3} a^{2} b^{3}}}{3}$

$\frac{3}{2 - 5 i}$

$\frac{3}{2 - 5 i} (\frac{2 + 5 i}{2 + 5 i}) \frac{6 + 15 i}{4 - 25 i^{2}} \frac{6 + 15 i}{4 - 25 (- 1)}$

$\frac{6 + 15 i}{29}$

Solve.

$\sqrt{x + 10} = 4 - \sqrt{x + 2}$

$\begin{array}{rcl} {(\sqrt{x + 10})}^{2} & = & {(4 - \sqrt{x + 2})}^{2} \\ x + 10 & = & 16 - 8 \sqrt{x + 2} + x + 2 \\ 10 & = & 18 - 8 \sqrt{x + 2} \\ - 8 & = & - 8 \sqrt{x + 2} \\ 1 & = & \sqrt{x + 2} \\ (1^{2}) & = & {(\sqrt{x + 2})}^{2} \\ 1 & = & x + 2 \end{array}$

x = –1

${(3 x + 2)}^{\frac{1}{4}} = 3$

$\begin{array}{rcl} {({(3 x + 2)}^{\frac{1}{4}})}^{4} & = & {(3)}^{4} \\ 3 x + 2 & = & 81 \\ 3 x & = & 79 \end{array}$

$x = \frac{79}{3}$

$\sqrt{4 x - 7} > \sqrt{x - 1}$

$\begin{array}{rcl} R : \\ \begin{matrix} Restrictions & Solution \\ 4 x - 7 \geq 0 & x - 1 \geq 0 & {(\sqrt{4 x - 7})}^{2} > {(\sqrt{x - 1})}^{2} \\ 4 x \geq 7 & x \geq 1 & 4 x - 7 > x - 1 \\ x \geq \frac{7}{4} & 3 x > 6 \\ x > 2 \end{matrix} \end{array}$

x > 2

Multiple Choice

The __ test can be used to determine if a graph represents a function.

horizontal line
vertical line
domain
range

Note

By definition, the vertical line test is a visual representation that determines whether or not a graph on the coordinate plane is a function by running a vertical line across the graph. If the vertical line touches more than one point at a time, the graph is not a function.

$5 + \sqrt{2 x - 6} > 8$

$\begin{array}{rcl} Restrict \\ \begin{matrix} Restrictions : & Solution : \\ 2 x - 6 \geq 0 & \sqrt{2 x - 6} > 3 \\ 2 x \geq 6 & {(\sqrt{2 x - 6})}^{2} > {(3)}^{2} \\ x \geq 3 & 2 x - 6 > 9 \\ 2 x > 15 \\ x > \frac{15}{2} \\ x > 7.5 \end{matrix} \end{array}$

Note

This graph occurs when the direction of the inequality symbol is changed.
This graph shows the restriction and boundary point, 7.5, as an AND compound inequality.
This graph has the incorrect direction for the restriction.

Simplify ${(3 - i)}^{2} + 4 i^{3} (- i)$

4
6
$4 - 6 i$
$8 - 7 i$

$(3 - i) (3 - i) - 4 i^{4} 9 - 6 i + i^{2} - 4 (1) 9 - 6 i - 1 - 4 4 - 6 i$

Note

A–B) These answers occur when the binomial is not squared, resulting in a trinomial.

This answer occurs when $4 i^{3} (- i)$ is added rather than multiplied.

Select all classifications for the expression $8 - 3 i^{2}$

Real numbers
Rational Numbers
Imaginary Numbers
Complex Numbers

Note

Numbers cannot be imaginary and real at the same time.

$8 - 3 (- 1) = 11$

Problem	1	2	3	4	5	6	7	8	9	10	11	12
Origin	L11	L15	L12	L12	L16	L13	L13	L14	A1	L14	L16	A16

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