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June 3rd, 2014, 06:10 PM  #1 
Newbie Joined: Apr 2014 From: canada Posts: 1 Thanks: 0  Solving Radical Equations
Hi, I'm having a little trouble with these two questions, and it would be great if anyone could help me, thanks I've attached the picture with the questions and any help would be appreciated! 1. Determine the roots of the following equations. Show your work to check if any of the roots are extraneous. 
June 3rd, 2014, 06:36 PM  #2 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,943 Thanks: 985 
Unclear. Square root of what? a) SQRT(2w)+ 8 or SQRT(2w + 8)? b) clarify... 
June 3rd, 2014, 10:11 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,598 Thanks: 2583 Math Focus: Mainly analysis and algebra 
Assuming \begin{align*} \frac{\sqrt{2w+8}}{w} &= 1 \\[6pt] \sqrt{2w+8} &= w \\ \text{Square the equation (possibly adding solutions):} \qquad 2w+8 &= w^2 \\ w^2 2w 8 &= 0 \\ \left(w + 2\right) \left(w  4\right) &= 0 \\ \end{align*} So we have solutions $w = 2$ and $w = 4$, but we need to verify them because we may have added solutions by squaring the equation. $$w = 2 \qquad \Rightarrow \frac{\sqrt{2w+8}}{w} = \frac{\sqrt{4}}{2} = \frac{2}{2} = 1 \ne 1 \qquad \text{So $w = 2$ is not a solution} \\ w = 4 \qquad \Rightarrow \frac{\sqrt{2w+8}}{w} = \frac{\sqrt{16}}{4} = \frac{4}{4} = 1 \qquad \text{So $w = 4$ is a solution} \\ $$ Thus, the answer to the question is $$w=4$$ Assuming \begin{align*} \sqrt{p+5}p &= 3 \\[6pt] \sqrt{p+5} &= p+3 \\ \text{Square the equation (possibly adding solutions):} \qquad p + 5 &= p^2 + 6p + 9 \\ p^2 + 5p + 4 &= 0 \\ \left(p + 1\right)\left(p + 4\right) &= 0 \\ \end{align*} So we have solutions $p = 1$ and $p = 4$ to check. $$p = 1 \qquad \Rightarrow \sqrt{p+5}p = \sqrt{4} + 1 = 3 \qquad \text{So $p = 1$ is a solution} \\ p = 4 \qquad \Rightarrow \sqrt{p+5}p = \sqrt{1} + 4 \ne 3 \qquad \text{(because $\sqrt{1} \ne 1$). So $p = 4$ is not a solution}$$ So the answer to the question is $$p = 1$$ 

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