Question

Can anyone help me with my algebra 2 semester 2 e2020 class? Specifically, the pretests for Adding and Subtracting Radicals, Multiplying Radicals, Dividing Radicals, and Radical Equations and Extraneous Roots?

Answer 1

$\bf (8x-8)^{\frac{3}{2}}=64~~ \begin{cases} 64=2\cdot 2\cdot 2\\ \qquad 2^3 \end{cases}\implies (8x-8)^{\frac{3}{2}}=2^3 \\\\\\ \textit{then we raise both sides by }\frac{2}{3}\implies \left[ (8x-8)^{\frac{3}{2}} \right]^{\frac{2}{3}}=(2^3)^{\frac{2}{3}} \\\\\\ (8x-8)^1=2^2\implies 8x-8=4\implies 8x=12 \\\\\\ x=\cfrac{12}{8}\implies x=\cfrac{3}{2}$


now onto the second one,


$\bf (45-3x)^{\frac{1}{2}}=x-9\impliedby \textit{we'll raise both by }^2 \\\\\\ \left[ (45-3x)^{\frac{1}{2}} \right]^2=(x-9)^2\implies (45-3x)^1=x^2-18x+81 \\\\\\ 45-3x=x^2-18x+81\implies 0=x^2-15x+36 \\\\\\ 0=(x-12)(x-3)\implies x= \begin{cases} 12\\ 3 \end{cases}$

now, the extraneous part, I don't see either as extraneous, 3 and 12 do work, however, I take it is referring to the root, so if we nevermind the root has ± versions, and say only assuming the value coming from the root is say positive, then

$\bf (45-3x)^{\frac{1}{2}}=x-9\implies \sqrt{45-3x}=x-9\implies \stackrel{x=3}{\sqrt{45-3(3)}=3-9} \\\\\\ \sqrt{36}=-6\implies \stackrel{\textit{extraneous}}{6\ne -6}$