Question

How do you divide this radical?
5/-3-3√3
I know that you need to rationalize the denominator.
My question is: how do I multiply -3-3√3 and -3-3√3?
I still end up with a radical in my denominator.

Answer 1

that is because you want to multiply by the "conjugate" of the denominator, namely
\[-3+\sqrt{3}\]

Answer 2

you get
\[(-3-\sqrt{3})(-3+\sqrt{3})=9-3=6\] and no more radical

Answer 3

^We do this because it's simply reversing the "difference of two squares" case you learned in factoring.

Answer 4

what yakeyglee said, although he/she is
making a rather broad assumption, but that is right
\[(a+b)(a-b)=a^2-b^2\]

Answer 5

Don't you use Multiply each number by FOIL?

Answer 6

Don't you use FOIL?

Answer 7

When you're multiplying the denominators?

Answer 8

Why not?

Answer 9

foil is good for wrapping turkey leftovers, not much else
a trick math teachers play on students

Answer 10

it is always true that
\[(a-b)(a+b)=a^2-b^2\] so
\[(-3+\sqrt{3})(-3-\sqrt{3})=(-3)^2-(\sqrt{3})^2=9-3=6\]

Answer 11

But -3 is not a perfect square

Answer 12

true that, but i am multiplying, not factoring

Answer 13

can you do that for any radical? I have another one too
2+5√3/-4+4√2

Answer 14

\[\frac{2+5\sqrt{3}}{-4+4\sqrt{2}}\] like that?

Answer 15

yes

Answer 16

then multiply by
\[\frac{2+5\sqrt{3}}{-4+4\sqrt{2}}\frac{(-4+4\sqrt{2})}{(-4+4\sqrt{2})}\]

Answer 17

denominator will be
\[(-4)^2-(4\sqrt{2})^2=16-32=-16\] the numerator will be a hot mess so i will let you do that on your own

Answer 18

Thank you

Answer 19

Hey i keep on messing up when i use your thing