My Math Forum Solving Radical Equations

 Algebra Pre-Algebra and Basic Algebra Math Forum

June 3rd, 2014, 05:10 PM   #1
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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.
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 June 3rd, 2014, 05:36 PM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Unclear. Square root of what? a) SQRT(2w)+ 8 or SQRT(2w + 8)? b) clarify...
 June 3rd, 2014, 09:11 PM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 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 $$p = -1$$ Thanks from xoxhannahc