# Help solving equation

• Apr 8th 2010, 04:15 PM
Mike9182
Help 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 PM
pickslides
Quote:

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

I get

$\displaystyle \sqrt{50}y=100$

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

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

$\displaystyle y=\frac{20}{\sqrt{2}}=10\sqrt{2}$
• Apr 8th 2010, 04:27 PM
Mike9182
How did $\displaystyle \frac{20}{\sqrt{2}}$ become $\displaystyle 10\sqrt{2}$?
• Apr 8th 2010, 04:32 PM
harish21
Quote:

Originally Posted by Mike9182
How did $\displaystyle \frac{20}{\sqrt{2}}$ become $\displaystyle 10\sqrt{2}$?

$\displaystyle \frac{20}{\sqrt{2}} = \frac{20}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{20 \times \sqrt{2}}{2} = 10 \times \sqrt{2}$

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 PM
pickslides
$\displaystyle \frac{20}{\sqrt{2}}=\frac{20}{\sqrt{2}}\times {\color{red}\frac{\sqrt{2}}{\sqrt{2}}}= \frac{20\sqrt{2}}{\sqrt{2}\times \sqrt{2}}=\frac{20\sqrt{2}}{2}=10\sqrt{2}$
• Apr 8th 2010, 04:46 PM
Mike9182
Thanks, I understand now.
• Apr 8th 2010, 04:49 PM
Mike9182
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 PM
harish21
Quote:

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

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

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 PM
bjhopper
another question
Quote:

Originally Posted by harish21
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,

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

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

bjh