Simplify a (complex) Radical equation help!!

Mar 2013
142
0
Vancouver
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.

IMG_0001.jpg

Any help in how to even begin would be wonderful.
Thanks.
 
Feb 2015
2,255
510
Ottawa Ontario
That ain't no equation...
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
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.
 
Last edited:
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Mar 2013
142
0
Vancouver
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?
 
Mar 2013
142
0
Vancouver
I'm sorry i found out, Thanks skeeter
 
Mar 2013
142
0
Vancouver
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 ?
 
Last edited:

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
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 ...

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