Question

unit 3 part 1 review

1. a student believes that all of the expressions in the table below are equal to 9. refute the students belief by determining which expressions equal 9. check yes or no.

expression yes no
(9^3)^2
(9^3)^3(1/2)
9^2 + 9^3
(9^3)(9^3)

1. circle the expressions that are equivalent to: -\frac{\sqrt4{16}}{\sqrt{64}}

a) (16)^{-1/4}(64)^{-1/2} b) - 16
c) -\frac{2}{8} d) - 4
e) -(16)^{1/4}(64)^{1/2} f) -(16)^{1/4}(64)^{-1/2}

1. simplify to exponential form:

a) (\sqrt{x^3yz^2})(x^2\sqrt{z}) b) (y^{1/3}x^2)(xy^{1/2})

1. simplify the expression. use only positive exponents in your solution.

a) \frac{x^2y^{-3}z^0}{x^5y^2} b) (\frac{-2x^3}{y^2z})^{-4} c) (\frac{4x^3y^{-2}}{9x^{-2}y^{-2}})^2

1. rewrite as a radical expression:

a) \frac{x^{3/4}}{y^{1/4}} b) \frac{x^{3/2}y}{y^{1/2}}

1. determine whether each equation has one or more extraneous solutions. indicate your answer with yes or no. without calculator.

5+\sqrt{x - 5}=0
2\sqrt{x + 5}+8 = 18
\sqrt{3x + 2}-\sqrt{2x+7}=0

Explanation:

Step1: Evaluate expressions in 1)

• For $(9^{\frac{1}{2}})^2$, by power - of - a - power rule $(a^m)^n=a^{mn}$, we have $(9^{\frac{1}{2}})^2 = 9^{\frac{1}{2}\times2}=9^1 = 9$ (Yes).
• For $(9^{\frac{1}{2}})^{\frac{1}{2}}(\frac{1}{2})$, first $(9^{\frac{1}{2}})^{\frac{1}{2}}=9^{\frac{1}{2}\times\frac{1}{2}} = 9^{\frac{1}{4}}

eq9$ (No).

• For $9^{\frac{1}{2}}+9^{\frac{1}{2}}$, since $9^{\frac{1}{2}} = 3$, then $9^{\frac{1}{2}}+9^{\frac{1}{2}}=3 + 3=6

eq9$ (No).

• For $(9^{\frac{1}{2}})(9^{\frac{1}{2}})$, by product rule $a^m\times a^n=a^{m + n}$, we have $(9^{\frac{1}{2}})(9^{\frac{1}{2}})=9^{\frac{1}{2}+\frac{1}{2}}=9^1 = 9$ (Yes).

Step2: Evaluate expressions in 2)

First, simplify $-\frac{\sqrt[4]{16}}{\sqrt{64}}$. We know that $\sqrt[4]{16}=2$ and $\sqrt{64}=8$, so $-\frac{\sqrt[4]{16}}{\sqrt{64}}=-\frac{2}{8}=-\frac{1}{4}$.

• For $(16)^{-\frac{1}{4}}(64)^{-\frac{1}{2}}$, $(16)^{-\frac{1}{4}}=\frac{1}{16^{\frac{1}{4}}}=\frac{1}{2}$ and $(64)^{-\frac{1}{2}}=\frac{1}{64^{\frac{1}{2}}}=\frac{1}{8}$, then $(16)^{-\frac{1}{4}}(64)^{-\frac{1}{2}}=\frac{1}{2}\times\frac{1}{8}=\frac{1}{16}

eq-\frac{1}{4}$.

• $- 16

eq-\frac{1}{4}$.

• $-\frac{2}{8}=-\frac{1}{4}$ (c is equivalent).
• $-4

eq-\frac{1}{4}$.

• For $-(16)^{\frac{1}{4}}(64)^{\frac{1}{2}}$, $(16)^{\frac{1}{4}} = 2$ and $(64)^{\frac{1}{2}}=8$, then $-(16)^{\frac{1}{4}}(64)^{\frac{1}{2}}=-2\times8=-16

eq-\frac{1}{4}$.

• For $-(16)^{\frac{1}{4}}(64)^{-\frac{1}{2}}$, $(16)^{\frac{1}{4}} = 2$ and $(64)^{-\frac{1}{2}}=\frac{1}{8}$, then $-(16)^{\frac{1}{4}}(64)^{-\frac{1}{2}}=-2\times\frac{1}{8}=-\frac{1}{4}$ (f is equivalent).

Step3: Simplify 3)a

\[

$$\begin{align*} (\sqrt[3]{x^{2}yz^{2}})(x^{2}\sqrt{z})&=(x^{\frac{2}{3}}y^{\frac{1}{3}}z^{\frac{2}{3}})(x^{2}z^{\frac{1}{2}})\\ &=x^{\frac{2}{3}+2}y^{\frac{1}{3}}z^{\frac{2}{3}+\frac{1}{2}}\\ &=x^{\frac{2 + 6}{3}}y^{\frac{1}{3}}z^{\frac{4+3}{6}}\\ &=x^{\frac{8}{3}}y^{\frac{1}{3}}z^{\frac{7}{6}} \end{align*}$$

\]

Step4: Simplify 3)b

\[

$$\begin{align*} (y^{\frac{1}{3}}x^{2})(xy^{\frac{1}{2}})&=x^{2 + 1}y^{\frac{1}{3}+\frac{1}{2}}\\ &=x^{3}y^{\frac{2 + 3}{6}}\\ &=x^{3}y^{\frac{5}{6}} \end{align*}$$

\]

Step5: Simplify 4)a

\[

$$\begin{align*} \frac{x^{2}y^{-3}z^{-1}}{x^{5}yz^{3}}&=x^{2-5}y^{-3 - 1}z^{-1-3}\\ &=x^{-3}y^{-4}z^{-4}\\ &=\frac{1}{x^{3}y^{4}z^{4}} \end{align*}$$

\]

Step6: Simplify 4)b

\[

$$\begin{align*} (\frac{-3x^{2}}{y^{3}z^{2}})^{-\frac{1}{4}}&=(\frac{y^{3}z^{2}}{-3x^{2}})^{\frac{1}{4}}\\ &=\frac{y^{\frac{3}{4}}z^{\frac{1}{2}}}{(-3)^{\frac{1}{4}}x^{\frac{1}{2}}} \end{align*}$$

\]

Step7: Simplify 4)c

\[

$$\begin{align*} (\frac{4x^{3}y^{-2}}{9x^{-2}y^{-3}z^{2}})^{2}&=(\frac{4x^{3+2}y^{-2 + 3}}{9z^{2}})^{2}\\ &=(\frac{4x^{5}y}{9z^{2}})^{2}\\ &=\frac{16x^{10}y^{2}}{81z^{4}} \end{align*}$$

\]

Step8: Rewrite 5)a

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

Step9: Rewrite 5)b

$\frac{\sqrt[3]{x^{2}y}}{y^{\frac{3}{4}}}=\frac{\sqrt[3]{x^{2}y}}{\sqrt[4]{y^{3}}}$

Step10: Analyze 6)

• For $5+\sqrt{x - 5}=0$, we have $\sqrt{x - 5}=-5$. Since the square - root of a number is non - negative in the real number system, this equation has no real solutions and no extraneous solutions (No).
• For $2\sqrt{x}+5 = 18$, we get $2\sqrt{x}=13$, then $\sqrt{x}=\frac{13}{2}$, and $x=\frac{169}{4}$. Substituting $x = \frac{169}{4}$ back into the original equation, it holds, so no extraneous solutions (No).
• For $\sqrt{3x + 2}-\sqrt{2x+7}=0$, we have $\sqrt{3x + 2}=\sqrt{2x + 7}$, then $3x+2=2x + 7$, and $x = 5$. Substituting $x = 5$ back into the original equation, it holds, so no extraneous solutions (No).

Answer:

1. $(9^{\frac{1}{2}})^2$: Yes; $(9^{\frac{1}{2}})^{\frac{1}{2}}(\frac{1}{2})$: No; $9^{\frac{1}{2}}+9^{\frac{1}{2}}$: No; $(9^{\frac{1}{2}})(9^{\frac{1}{2}})$: Yes
2. Equivalent expressions: c, f
3. a) $x^{\frac{8}{3}}y^{\frac{1}{3}}z^{\frac{7}{6}}$; b) $x^{3}y^{\frac{5}{6}}$
4. a) $\frac{1}{x^{3}y^{4}z^{4}}$; b) $\frac{y^{\frac{3}{4}}z^{\frac{1}{2}}}{(-3)^{\frac{1}{4}}x^{\frac{1}{2}}}$; c) $\frac{16x^{10}y^{2}}{81z^{4}}$
5. a) $\frac{\sqrt{x^{3}}}{\sqrt[4]{y}}$; b) $\frac{\sqrt[3]{x^{2}y}}{\sqrt[4]{y^{3}}}$
6. $5+\sqrt{x - 5}=0$: No; $2\sqrt{x}+5 = 18$: No; $\sqrt{3x + 2}-\sqrt{2x+7}=0$: No