1. ## Help solving equation

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

2. Originally Posted by Mike9182
How is $\sqrt{50}y=100$ solved? The answer in my book is $2\sqrt{50}$
I get

$\sqrt{50}y=100$

$5\sqrt{2}y=100$

$y=\frac{100}{5\sqrt{2}}$

$y=\frac{20}{\sqrt{2}}=10\sqrt{2}$

3. How did $\frac{20}{\sqrt{2}}$ become $10\sqrt{2}$?

4. Originally Posted by Mike9182
How did $\frac{20}{\sqrt{2}}$ become $10\sqrt{2}$?
$\frac{20}{\sqrt{2}} = \frac{20}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{20 \times \sqrt{2}}{2} = 10 \times \sqrt{2}$

Note: The answer you have written in the question is $2\sqrt{50} = 2 \sqrt{5 \times 5 \times 2} = 2 \times 5 \times \sqrt{2} = 10 \times \sqrt{2}$

5. $
\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}
$

6. Thanks, I understand now.

7. I have another equation I need help solving, the correct answer is 1/8.

$\frac{2}{3}x^{-1/3}-\frac{1}{3}x^{-2/3}=0$

8. Originally Posted by Mike9182
I have another equation I need help solving, the correct answer is 1/8.

$\frac{2}{3}x^{-1/3}-\frac{1}{3}x^{-2/3}=0$
Let $u = x^{-1/3}$.

then the given equation becomes:

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

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

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

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

$u(u-2) = 0$

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

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

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

cube both sides to get:

$x^{-1} = 8$

that is

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

therefore,

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

9. ## another question

Originally Posted by harish21
Let $u = x^{-1/3}$.

then the given equation becomes:

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

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

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

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

$u(u-2) = 0$

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

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

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

cube both sides to get:

$x^{-1} = 8$

that is

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

therefore,

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

A direct solution if a-b =0 a=b Simplify the equals to find x^1/3 =1/2 so x=1/8

bjh