Question

unit 5 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
(9^1/2)^2 | |
(9^1/2)^2(1/2) | |
9^1/2 + 9^1/2 | |
(9^1/2)(9^1/2) | |

1. circle the expressions that are equivalent to: -√16/√64

a) (16)^-1/4(64)^-1/2 b) - 16
c) - 2/0 d) - 4
e) -(16)^1/4(64)^1/2 f) -(16)^1/4(64)^-1/2

1. simplify to exponential form:

a) (√(x^2y^2z^3))(x^2√z) b) (y^1/3x^2)(xy^1/2)

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

a) (x^4y^-3z^2)/(x^-2y^3) b) ((x^1/2y^-1/3)/(z^2))^-2 c) ((x^2y^-3z^1/2)/(x^-1y^2z^-3))^3

1. rewrite as a radical expression:

a) (x^2/3)/(y^1/3) b) (x^3/4)/(y^1/4)

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

5 + √(x - 5)=0
2√(x + 5)+8 = 18
√(3x + 2)-√(2x + 7)=0

Explanation:

1. Refuting the student's belief

Step1: Simplify \((9^{\frac{1}{2}})^2\)

Using the power - of - a - power rule \((a^m)^n=a^{mn}\), we have \((9^{\frac{1}{2}})^2 = 9^{\frac{1}{2}\times2}=9^1 = 9\)

Step2: Simplify \((9^{\frac{1}{2}})^2(\frac{1}{2})\)

First, \((9^{\frac{1}{2}})^2 = 9\), then \(9\times\frac{1}{2}=\frac{9}{2}
eq9\)

Step3: Simplify \(9^{\frac{1}{2}}+9^{\frac{1}{2}}\)

Since \(9^{\frac{1}{2}}=\sqrt{9} = 3\), then \(9^{\frac{1}{2}}+9^{\frac{1}{2}}=3 + 3=6
eq9\)

Step4: Simplify \((9^{\frac{1}{2}})(9^{\frac{1}{2}})\)

Using the product rule \(a^m\times a^n=a^{m + n}\), we get \((9^{\frac{1}{2}})(9^{\frac{1}{2}})=9^{\frac{1}{2}+\frac{1}{2}}=9^1 = 9\)

Step1: Simplify \(\sqrt[4]{16}\)

We know that \(16 = 2^4\), so \(\sqrt[4]{16}=2\)

Step2: Simplify \(\sqrt{64}\)

\(\sqrt{64}=8\)

Step3: Calculate \(-\frac{\sqrt[4]{16}}{\sqrt{64}}\)

\(-\frac{2}{8}=-\frac{1}{4}\)

Step4: Check each option

Option a: \((16)^{-\frac{1}{4}}(64)^{-\frac{1}{2}}\)

\((16)^{-\frac{1}{4}}=(2^4)^{-\frac{1}{4}}=2^{-1}=\frac{1}{2}\), \((64)^{-\frac{1}{2}}=(2^6)^{-\frac{1}{2}}=2^{-3}=\frac{1}{8}\), then \((16)^{-\frac{1}{4}}(64)^{-\frac{1}{2}}=\frac{1}{2}\times\frac{1}{8}=\frac{1}{16}
eq-\frac{1}{4}\)

Option b: \(- 16

eq-\frac{1}{4}\)

Option c: \(-\frac{1}{4}\) (matches)

Option d: \(-4

eq-\frac{1}{4}\)

Option e: \(-(16)^{\frac{1}{4}}(64)^{\frac{1}{2}}\)

\((16)^{\frac{1}{4}} = 2\), \((64)^{\frac{1}{2}}=8\), then \(-(16)^{\frac{1}{4}}(64)^{\frac{1}{2}}=-2\times8=-16
eq-\frac{1}{4}\)

Option f: \(-(16)^{\frac{1}{4}}(64)^{-\frac{1}{2}}\)

\((16)^{\frac{1}{4}} = 2\), \((64)^{-\frac{1}{2}}=\frac{1}{8}\), then \(-(16)^{\frac{1}{4}}(64)^{-\frac{1}{2}}=-2\times\frac{1}{8}=-\frac{1}{4}\) (matches)

Step1: Rewrite square - roots as exponents

\(\sqrt{x^{\frac{1}{2}}y^{\frac{1}{3}}}=(x^{\frac{1}{2}}y^{\frac{1}{3}})^{\frac{1}{2}}=x^{\frac{1}{4}}y^{\frac{1}{6}}\) and \(\sqrt{z}=z^{\frac{1}{2}}\)

Step2: Multiply the expressions

\((x^{\frac{1}{4}}y^{\frac{1}{6}}z^{\frac{1}{4}})(x^{\frac{1}{2}}z^{\frac{1}{2}})=x^{\frac{1}{4}+\frac{1}{2}}y^{\frac{1}{6}}z^{\frac{1}{4}+\frac{1}{2}}=x^{\frac{3}{4}}y^{\frac{1}{6}}z^{\frac{3}{4}}\)

b) \((y^{\frac{1}{3}}x^{\frac{1}{2}})(xy^{\frac{1}{2}})\)

Answer:

Expression	Yes	No
\((9^{\frac{1}{2}})^2(\frac{1}{2})\)		√
\(9^{\frac{1}{2}}+9^{\frac{1}{2}}\)		√
\((9^{\frac{1}{2}})(9^{\frac{1}{2}})\)	√

2. Equivalent expressions

First, simplify \(-\frac{\sqrt[4]{16}}{\sqrt{64}}\)