How is $\displaystyle \sqrt{50}y=100$ solved? The answer in my book is $\displaystyle 2\sqrt{50}$

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- Apr 8th 2010, 04:15 PMMike9182Help solving equation
How is $\displaystyle \sqrt{50}y=100$ solved? The answer in my book is $\displaystyle 2\sqrt{50}$

- Apr 8th 2010, 04:21 PMpickslides
- Apr 8th 2010, 04:27 PMMike9182
How did $\displaystyle \frac{20}{\sqrt{2}}$ become $\displaystyle 10\sqrt{2}$?

- Apr 8th 2010, 04:32 PMharish21
$\displaystyle \frac{20}{\sqrt{2}} = \frac{20}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{20 \times \sqrt{2}}{2} = 10 \times \sqrt{2}$

Note: The answer you have written in the question is $\displaystyle 2\sqrt{50} = 2 \sqrt{5 \times 5 \times 2} = 2 \times 5 \times \sqrt{2} = 10 \times \sqrt{2}$ - Apr 8th 2010, 04:37 PMpickslides
$\displaystyle

\frac{20}{\sqrt{2}}=\frac{20}{\sqrt{2}}\times {\color{red}\frac{\sqrt{2}}{\sqrt{2}}}= \frac{20\sqrt{2}}{\sqrt{2}\times \sqrt{2}}=\frac{20\sqrt{2}}{2}=10\sqrt{2}

$ - Apr 8th 2010, 04:46 PMMike9182
Thanks, I understand now.

- Apr 8th 2010, 04:49 PMMike9182
I have another equation I need help solving, the correct answer is 1/8.

$\displaystyle \frac{2}{3}x^{-1/3}-\frac{1}{3}x^{-2/3}=0$ - Apr 8th 2010, 05:06 PMharish21
Let $\displaystyle u = x^{-1/3} $.

then the given equation becomes:

$\displaystyle \frac{2u}{3}-\frac{u^2}{3}=0$

$\displaystyle \frac{2u - u^2}{3} = 0$

$\displaystyle 2u - u^{2} = 0$

$\displaystyle u^{2} - 2u = 0$

$\displaystyle u(u-2) = 0$

So, $\displaystyle u = 0$ OR $\displaystyle u = 2$

substitute $\displaystyle u= x^{\frac{-1}{3}}$, you get

$\displaystyle x^{\frac{-1}{3}} = 2$

cube both sides to get:

$\displaystyle x^{-1} = 8$

that is

$\displaystyle \frac{1}{x} = 8$

therefore,

$\displaystyle x = \frac{1}{8}$ - Apr 8th 2010, 05:23 PMbjhopperanother question