# Simplify a (complex) Radical equation help!!

#### Gurp925

Hi all, i am stuck on how to even begin with this simplification so i know that i need to start within the brackets but that is about it.

Any help in how to even begin would be wonderful.
Thanks.

#### DenisB

That ain't no equation...

#### skeeter

MHF Helper
for future reference, this is an expression (equations have equal signs) ...

order of operations; start by getting a common denominator and combining the two terms in parentheses ...

$\dfrac{\sqrt{y}(x+\sqrt{xy})+\sqrt{y}(x-\sqrt{xy})}{x^2-xy} = \dfrac{2x\sqrt{y}}{x(x-y)} = \dfrac{2\sqrt{y}}{x-y}$

do the multiplication ...

$\dfrac{\sqrt{x}-\sqrt{y}}{2\sqrt{xy}} \cdot \dfrac{2\sqrt{y}}{x-y} = \dfrac{1}{\sqrt{x}(\sqrt{x}+\sqrt{y})} = \dfrac{1}{x+\sqrt{xy}}$

add this to the first term ...

$\dfrac{\sqrt{x}+\sqrt{y}-1}{x+\sqrt{xy}} + \dfrac{1}{x+\sqrt{xy}} = \dfrac{\sqrt{x}+\sqrt{y}}{x+\sqrt{xy}}=\dfrac{ \sqrt{x}+\sqrt{y}}{\sqrt{x}(\sqrt{x}+\sqrt{y})} = \dfrac{1}{\sqrt{x}}$

check it ... make sure I didn't screw something up.

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1 person

#### Gurp925

for future reference, this is an expression (equations have equal signs) ...

order of operations; start by getting a common denominator and combining the two terms in parentheses ...

$\dfrac{\sqrt{y}(x+\sqrt{xy})+\sqrt{y}(x-\sqrt{xy})}{x^2-xy} = \dfrac{2x\sqrt{y}}{x(x-y)} = \dfrac{2\sqrt{y}}{x-y}$

do the multiplication ...

$\dfrac{\sqrt{x}-\sqrt{y}}{2\sqrt{xy}} \cdot \dfrac{2\sqrt{y}}{x-y} = \dfrac{1}{\sqrt{x}(\sqrt{x}+\sqrt{y})} = \dfrac{1}{x+\sqrt{xy}}$

add this to the first term ...

$\dfrac{\sqrt{x}+\sqrt{y}-1}{x+\sqrt{xy}} + \dfrac{1}{x+\sqrt{xy}} = \dfrac{\sqrt{x}+\sqrt{y}}{x+\sqrt{xy}}=\dfrac{ \sqrt{x}+\sqrt{y}}{\sqrt{x}(\sqrt{x}+\sqrt{y})} = \dfrac{1}{\sqrt{x}}$

check it ... make sure I didn't screw something up.

Question skeeter how do you get that LCD?

#### Gurp925

I'm sorry i found out, Thanks skeeter

#### Gurp925

How do you do that multiplication step, dont know what cancels to leave you with that.

and it that last step would it not be x(rootx + rooty) = 1/x ?

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#### skeeter

MHF Helper
How do you do that multiplication step, dont know what cancels to leave you with that.
$\dfrac{\sqrt{x}-\sqrt{y}}{2\sqrt{xy}} \cdot \dfrac{2\sqrt{y}}{x-y} = \dfrac{2\sqrt{y}(\sqrt{x}-\sqrt{y})}{2\sqrt{xy}(x-y)} = \color{red}{\dfrac{2\sqrt{y}(\sqrt{x}-\sqrt{y})}{2\sqrt{x} \cdot \sqrt{y}(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})}} = \dfrac{1}{\sqrt{x}(\sqrt{x}+\sqrt{y})} = \dfrac{1}{x+\sqrt{xy}}$

and it that last step would it not be x(rootx + rooty) = 1/x ?
no ...

$\dfrac{\sqrt{x}+\sqrt{y}-1}{x+\sqrt{xy}} + \dfrac{1}{x+\sqrt{xy}} = \dfrac{\sqrt{x}+\sqrt{y}}{x+\sqrt{xy}}= \color{red}{\dfrac{ \sqrt{x}+\sqrt{y}}{\sqrt{x} \cdot \sqrt{x} + \sqrt{x} \sqrt{y}}} =\dfrac{ \sqrt{x}+\sqrt{y}}{\sqrt{x}(\sqrt{x}+\sqrt{y})} = \dfrac{1}{\sqrt{x}}$