# how we can calculate $\frac {\sqrt {x^2} + \sqrt {y^2} }{2 \sqrt {xyz}}$? [closed]

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

• The answer is $\frac1{\sqrt{-z}}$ no? Feb 16, 2016 at 20:57
• $\sqrt{a^2}=|a|$. Feb 16, 2016 at 20:58
• @GregoryGrant is not in the choice Feb 16, 2016 at 20:59
• 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? Feb 16, 2016 at 21:00
• @LoveMathContest What are the choices? Feb 16, 2016 at 21:02

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 $\frac{\sqrt{-z}}{-z}$
$$\frac{\sqrt{x^2}+\sqrt{y^2}}{2\sqrt{xyz}}=\frac{2\sqrt{x^2}}{2\sqrt{-x^2z}}$$ 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.