To rationalize then simplified: two squareroot of x minus three squareroot of y all divided by squareroot of x plus square root of y Thank:D

2root x - 3root y / root x + root y rational denominator by multuplying by (rootx - rooty) =(2root x - 3root y) (root x - root y) / (x-y) =(2x -2root x * root y - 3 root y * root x + 3 y / (x-y) =(2x +6xy + 3y) / (x - y)

not quite sure of this just rechecking

Wait it works Thanks

no sorry! the last line is wrong

other way around?

how do u mean it works

Well, i have this online course you see, and this answer fits the amount of blocks it gives

here look at this

-2root x * root y - 3 root y * root x is not = 6xy it = 6 root(xy)

thats the one (2x +6root(xy) + 3y) / (x - y)

yep that checks out

Thanks:D

yw

How do you know how to do this? Everybody in my group is inadeuquate to solve these

'root6' and similar are called surds and u need to know how to manipulate them. the wolfram math sites will have info about them

\[\left(2 \sqrt{x}-3 \sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)=2 x-5 \sqrt{x} \sqrt{y}+3 y \]Is there an error at untitled.jpg?

What do you mean error?

yes u are rob right its - 5 root ( xy) not 6 root (xy)

Here is another one if you want?

the jpg is wrong

the first one i know it is i fixed it

ok in the next one you take out the surd (sqrt 7 + 2) by multiplying by its complement this is (sqrt 7 - 2) (sqrt 7 + 2) * (sqrt 7 - 2) = 7 + 2 sqrt 7 - 2 sqrt 7 - 4 = 3 you also have to multiply the top by (sqrt 7 + 2) so we get 12(sqrt 7 - 2) / 3 = 4(sqrt7 - 2) or 4sqrt 7 - 8

hope i havent made another boob !

Oh i gotcha, thanks

oh oh!

Jimmyrep's answer looks ok.\[\left(\frac{12}{\sqrt{7}+2}\right)\left(\frac{\sqrt{7}-2}{\sqrt{7}-2}\right)=\frac{12\left(\sqrt{7}-2\right)}{3}=4 \sqrt{7}-8 \]

tell me rob how do u do the maths symbols using Equation box or other?

I use Mathematic 8 for solving problems on this site. In the program there is an option to copy answers in LaTex form. So it is simply a matter of a copy and paste into this sites "Sigma" Equation box.

right - ill look at that thanx

Your welcome.

Sorry. The program's name is Mathematica 8, not Mathematic 8.

ok

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