How do I rationalize the denominator for this problem?

Question

2*sqrt(5) + sqrt(6)
---divided by------
sqrt(6) - sqrt(5)

Please help, I have no idea how to do this.

Answer 1

remember that (a+b)(a-b) = a² - b², so

2√5 + √6
------------ =
√6 - √5

(2√5 + √6)(√6 + √5)
-------------------------- =
(√6 - √5)(√6 + √5)

2√30 + 2(5) + 6 + √30
----------------------------- =
6 - 5

3√30 + 16
-------------- =
1

3√30 + 16

Answer 2

Rationalize the denominator by multiplying both numerator and denominator by the conjugate of the denominator.
The conjugate is the same expression as the denominator but with the opposite sign in the middle.

(2*sqrt(5) + sqrt(6))/(sqrt(6) - sqrt(5))
= (2*sqrt(5) + sqrt(6))*(sqrt(6)+sqrt(5))/(sqrt(6) - sqrt(5))((sqrt(6) + sqrt(5))
= (2*sqrt(5) + sqrt(6))*(sqrt(6)+sqrt(5))/(6 -5)
= (2*sqrt(5) + sqrt(6))*(sqrt(6)+sqrt(5))/1
= (2*sqrt(5) + sqrt(6))*(sqrt(6)+sqrt(5))
= 16 + 3sqrt(30)

Answer 3

I believe you multiply the entire fraction by
sqrt(6) + sqrt(5)
---divided by---
sqrt(6) + sqrt(5)

because this way you'll get rid of the radicals on the bottom, and the middle part will cancel out. Just do it and you'll see what I mean. Hopefully this helps!

Answer 4

To get a binomial radical out of the denominator, multiply top and botton by the opposite of the denominator (in this case, sqrt6+sqrt5)
FOIL it, and get:
sqrt(6*6)+sqrt(5*6)-sqrt(5*6)-sqrt(5*5)

the two sqrt(5*^)'s cancel out(one positive and one negative), leaving you with:
sqrt(6*6)-sqrt(5*5)
You can simplify these( if there is a squared in a radical, one number comes out) to get:
6-5
1
The denominator=1
Now, if you multiply the bottom by sqrt6+sqrt5, you have to do the same for the top.
[2*sqrt(5)+sqrt(6)][sqrt6+sqrt5]
FOIL it, and get:
2*sqrt(5*6) + 2*sqrt(5*5)+sqrt(6*6)+sqrt(5*6)
the sqrt(5*5) and sqrt(6*6) become whole numbers, 5 and 6, respectively (the whole number squared under a radical again)
and the sqrt(5*6) and 2*sqrt(5*6) add together to become 3*sqrt(5*6), leaving you with
3*sqrt(5*6) +6 +5
simplify:
3*sqrt(6*5)+11

E-mail or IM me if you have any other questions

Answer 5

multiply top and bottom by the negative of the bottom, or sqrt(6)+sqrt(5).

(sqrt(6)+sqrt(5))(sqrt(6)- sqrt(5))= (sqrt(6))^2-(sqrt(5))^2 by the difference of squares law,

therefore the new denominator is (6) - (5), or 1

ps. dont forget to multiply the top by sqrt(6)+sqrt(5) also

Answer 6

you solve this by multiplying the denominator by the conjugate. If you ever have a+b on the denominator, you multiply by a-b, which gets rid of cross terms. The same rule applies here.

root 6+root 5(root 6-root5)=6-5=1. Don't forget to multiply on the top as well (i'm just too lazy to do it for you)

Answer 7

suppose rt(6) - rt(5) is like a-b
to make it a^2 + b^2 you wud multiply it with a+b.
Do the same.
Multiply with
rt(6) + rt(5)

But multiply with [rt(6) + rt(5)] / [rt(6) + rt(5)] coz you can't change the equ and this fraction can be easily cancelled out.

Good now ...

[2*rt(5) + rt(6)] by rt(6) - rt(5)

multiply by

[rt(6) + rt(5)] / [rt(6) + rt(5)]

or,

[2*rt(5) + rt(6] * [rt(6) + rt(5)]
------------------- ---------------------
[rt(6] ^2 + [rt(5)] ^2

or,

2* rt (5*6) + 2* [rt(5)] ^2 + 2* [rt(6)] ^2 + rt( 5*6)
----------------------- ------------------------- ---------------------
6 + 5

or,

2* rt(30) + 2*5 + 2*6 + rt (30)
------------------- ---------------------
11

or,

3*rt(30) + 22
---------------------
11

I hope that is the answer