simplest radical form with the denominator rationalized

[synapsis]

Having a bit of trouble with this one- i keep getting the answer sq root of 3 + 4
Here is the problem- i wish i had some kind of program to express the problem as is, but i hope my description works...

sq root of 8 / sq root of 6 - sq root of 2

According to my math book, so far as looking at similar problems in the examples, i need to multiply the numerator and denominator by the conjugate of the denominator, which is sq root of 6 + sq root of 2

thank you for your help, if you could show all your steps i would appretiate it.

 

[Denis]

Having a bit of trouble with this one- i keep getting the answer sq root of 3 + 4
Here is the problem- i wish i had some kind of program to express the problem as is, but i hope my description works...

sq root of 8 / sq root of 6 - sq root of 2

According to my math book, so far as looking at similar problems in the examples, i need to multiply the numerator and denominator by the conjugate of the denominator, which is sq root of 6 + sq root of 2

thank you for your help, if you could show all your steps i would appretiate it.

"sq root of 8 / sq root of 6 - sq root of 2"
Program or not, you need a set of brackets:
sq root of 8 / (sq root of 6 - sq root of 2) : do you understand why?

Next time, show sq root of n as sqrt(n); applied to your expression:
sqrt(8) / (sqrt(6) - sqrt(2))

I don't see the need to multiply as you state, since:
sqrt(8) = 2sqrt(2), and
sqrt(6) - sqrt(2) = sqrt(2) * (sqrt(3) - 1)

So we now have:
2sqrt(2) / sqrt(2)(sqrt(3)-1)
= 2 / (sqrt(3) - 1) : since sqrt(2) cancels out

 

[soroban]

Hello, synapsis!

Having a bit of trouble with this one - i keep getting the answer sq root of 3 + 4
\(\displaystyle \;\;\) Are you cancelling illegally?

\(\displaystyle \L\:\frac{\sqrt{8}}{sqrt{6}\,-\,\sqrt{2}}\)

We have: \(\displaystyle \L\:\frac{\sqrt{8}}{\sqrt{6}\,-\,\sqrt{2}}\,\cdot\,\frac{\sqrt{6}\,+\,\sqrt{2}}{\sqrt{6}\,+\,\sqrt{2}}\)

The numerator is: \(\displaystyle \,\sqrt{8}(\sqrt{6}\,+\,\sqrt{2})\:=\:\sqrt{48}\,+\,\sqrt{16}\:=\:4\sqrt{3}\,+\,4\:=\:4(\sqrt{3}\,+\,1)\)

The denominator is: \(\displaystyle \,(\sqrt{6}\,-\,\sqrt{2})(\sqrt{6}\,+\,\sqrt{2})\:=\\sqrt{6})^2\,-\,(\sqrt{2})^2\:=\:6\,-\,2\:=\:4\)

The fraction becomes: \(\displaystyle \L\,\frac{\not{4}(\sqrt{3}\,+\,1)}{\not{4}}\:=\:\sqrt{3}\,+\,1\)

 

[synapsis]

AHHHH i didn't treat (4)sqrt3 + (4) as i should of- i didn't factor it- dumb dum dumb- thx for showing your work that was easier than i thought.

much appretiated

-Syn