Question

simplify 3-radical6/5-2radical6

Answer 1

\[3-\sqrt{6}\div5-2\sqrt{6}\]

Answer 2

\[\large \frac{3-\sqrt6}{5-2\sqrt6}\]Like this, yes?
And they want us to simplify it?

Answer 3

yes.

Answer 4

So the idea is, we want to `rationalize` this expression.
We want to get the `irrational number` out of the denominator.

In this case sqrt6 is our irrational number.

Do you remember what conjugates are?
Here is the basic rule:
\(\large (a-b)(a+b)=a^2-b^2\)

Multiplying conjugates, `the bracket thingies`, gives us the difference of squares.
Does that look familiar?

Answer 5

yes.
the conjugate is 5+2radical 6, yeah?

Answer 6

Yes good :)

Answer 7

So according to the rule, multiplying these conjugates should give us something like this, right?
\[\large 5^2-(2\sqrt6)^2\]

Answer 8

Oh I should clarify, we'll want to multiply the `bottom` AND `top` by the conjugate of the bottom.

\[\large \frac{3-\sqrt6}{5-2\sqrt6}\color{royalblue}{\left(\frac{5+2\sqrt6}{5+2\sqrt6}\right)}\]This blue fraction is equivalent to 1, since it has the same value on top and bottom. So it won't change the value of our expression, it will just change the way it looks. Which is important.

I just wanted to mention that in case there was any confusion.

Answer 9

got it.

Answer 10

I've had trouble simplifying the top part.

Answer 11

Ok so did you make it to this part?
\[\large \frac{(3-\sqrt6)(5+2\sqrt6)}{1}\]

Answer 12

yeah i got that.

Answer 13

We need to expand out these brackets.
Lemme show you how we do that, just in case you're a little rusty on it.

Answer 14

We multiply each term in the first set of brackets with each term in the second set of brackets. Then we add up all of those terms. You'll end up with 4 in all.