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I teach math for Schools. How can Help me in the following past Olympiad question?

If $y,z$ be two negative distinct number and $x$ and $y$ be negate of each other, how we can calculate $ \displaystyle\frac {\sqrt {x^2} + \sqrt {y^2} }{2 \sqrt {xyz}}$?

1) $\frac {\sqrt{x}}{x}$

2)$\frac {\sqrt{-y}}{y}$

3) $\frac {\sqrt{z}}{z}$

4) $\frac {\sqrt{-z}}{-z}$

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    $\begingroup$ The answer is $\frac1{\sqrt{-z}}$ no? $\endgroup$ Feb 16, 2016 at 20:57
  • $\begingroup$ $\sqrt{a^2}=|a|$. $\endgroup$
    – Anurag A
    Feb 16, 2016 at 20:58
  • $\begingroup$ @GregoryGrant is not in the choice $\endgroup$ Feb 16, 2016 at 20:59
  • $\begingroup$ Note that $\sqrt{x^2}=|x|$. So since $y=-x$ the top is $2|x|$ and the denominator is $2\sqrt{-x^2z}$ which equals $2|x|\sqrt{-z}$. So everything cancels but the $\sqrt{-z}$ in the denominator. Where's the mistake? $\endgroup$ Feb 16, 2016 at 21:00
  • $\begingroup$ @LoveMathContest What are the choices? $\endgroup$
    – egreg
    Feb 16, 2016 at 21:02

2 Answers 2

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Note that $\sqrt{x^2}=|x|$. So since $y=-x$ the top is $2|x|$ and the denominator is $2\sqrt{-x^2z}$ which equals $2|x|\sqrt{-z}$. So everything cancels but the $\sqrt{-z}$ in the denominator. Now multiply top and bottom by $\sqrt{-z}$ to get $\frac{\sqrt{-z}}{-z}$

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  • $\begingroup$ see edited questions $\endgroup$ Feb 16, 2016 at 21:04
  • $\begingroup$ @LoveMathContest I edited my answer, should be complete now. $\endgroup$ Feb 16, 2016 at 21:05
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$$\frac{\sqrt{x^2}+\sqrt{y^2}}{2\sqrt{xyz}}=\frac{2\sqrt{x^2}}{2\sqrt{-x^2z}}$$ using $y=-x$. This yields: $$\frac{2\sqrt{x^2}}{2\sqrt{x^2}\sqrt{-z}}=\frac{1}{\sqrt{-z}}$$. Note that $z<0$, so $-z>0$, so $\sqrt{-z}$ is well-defined.

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  • $\begingroup$ I said the same thing but he says that's not one of the choices. I've asked for clarification. $\endgroup$ Feb 16, 2016 at 21:01
  • $\begingroup$ @GregoryGrant Ah, I missed the discussion while writing my answer. Thanks for the comment. I agree with your answer. $\endgroup$ Feb 16, 2016 at 21:02
  • $\begingroup$ He edited the question, turns out we're both right. I'll upvote your answer. $\endgroup$ Feb 16, 2016 at 21:05
  • $\begingroup$ @GregoryGrant Likewise, since we have the same thing. $\endgroup$ Feb 16, 2016 at 21:05

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