Help me with simplifying radicals? I'll medal!!!

Question

camzzzie · Answer

(square root 2 +square root 5)/(square root 2 -square root 5)

anonymous · Answer

$$
\frac{\sqrt2+\sqrt 5}{\sqrt 2 - \sqrt 5}
$$

campbell_st · Answer

you need to rationalize the denominator by multiplying 

$$\frac{(\sqrt{2} + \sqrt{5})}{(\sqrt{2} - \sqrt{5})} \times \frac{(\sqrt{2}+\sqrt{5})}{(\sqrt{2} + \sqrt{5})}$$

this will allow you to get a rational number in the denominator and the 2 binomials are the difference of 2 squares. 

Hope it helps

anonymous · Answer

First take the conjugate of the bottom.   You get the conjugate by changing from addition to subtraction or vice versa.  In this case it is: $
\sqrt 2+\sqrt 5
$.
Then you multiply top and bottom by it: $$
\frac{\sqrt 2+\sqrt 5}{\sqrt 2-\sqrt 5}\times \frac{\sqrt 2+\sqrt 5}{\sqrt 2+\sqrt 5}
$$

anonymous · Answer

The conjugate results in difference of squares, that is why we use it: $$
(\sqrt 2-\sqrt 5)(\sqrt 2+\sqrt 5) = 2-5 = -3
$$

anonymous · Answer

When you do difference of square, the square roots go away.

camzzzie · Answer

do you have to foil it when youre at that point

campbell_st · Answer

well you do for the numerator...

camzzzie · Answer

and then the denominator cancels out?

campbell_st · Answer

the denominator becomes a rational number

look at what @wio posted

camzzzie · Answer

so is the finale answer -3?

anonymous · Answer

No.

anonymous · Answer

$$
\frac{(\sqrt2+\sqrt5)(\sqrt2+\sqrt5)}{(\sqrt2-\sqrt5)(\sqrt2+\sqrt5)} = \frac{(\sqrt2+\sqrt5)(\sqrt2+\sqrt5)}{-3}
$$You still have to foil the top.

camzzzie · Answer

Oh ok