Question

Help?

Answer 1

To rationalize the denominator of fractions with roots, we multiply by the conjugate (We change the sign between the real number and the root). That is:
\[\frac{1}{a+\sqrt{b}} \frac{a-\sqrt{b}}{a-\sqrt{b}}\]

This will get rid of the roots in the denominator. (Notice, we're multiplying by the same thing on the top and the bottom, so we're really just multiplying by 1, so we don't change the value at all)

Answer 2

Okay so... Confused lol.

Answer 3

In your problem, you have:
\[\frac{6}{6+\sqrt{5}}\]
We'll multiply, top and bottom, by the conjugate:
\[\frac{6}{6+\sqrt{5}}\frac{6-\sqrt{5}}{6-\sqrt{5}}=\frac{6(6-\sqrt{5})}{(6+\sqrt{5})(6-\sqrt{5})}\]
Can you simplify that?

Answer 4

The square roots would cancel out one another... So would it be 36?

Answer 5

The square roots don't cancel. When we multiply, we have to distribute everything.
\[(a+b)(a-b) = a^2 -ab + ab - b^2 = a^2 - b^2\]

Answer 6

I'm lost...

Answer 7

\[(6+\sqrt{5})(6-\sqrt{5}) = 6^2 -6 \sqrt{5} + 6 \sqrt{5} - \sqrt{5}\sqrt{5} = 6^2 - \sqrt{5}\sqrt{5} = 36 - 5 = 31\]

Answer 8

Am I right?

Answer 9

No. Look up a few posts. I actually did EXACTLY that problem as a demonstration.

Answer 10

So D?

Answer 11

Yep

Answer 12

Right?

Answer 13

Try doing it with something that has two roots.
\[\frac{1}{\sqrt{a}+\sqrt{b}}\]
See if it works

Answer 14

Nope, it's false lol