how do i rationalize this?

Question

6
------
√2-√3

Answer 1

Try multiplying top and bottom by √2+√3.

Then we get 6 * (√2+√3) / (2-3)
Does that help along the way?
.

Answer 2

You can't "rationalize" the whole thing (in the sense of getting rid of all surd signs), but you CAN rationalize the DENOMINATOR.

One does that in this kind of question by noting that (√2-√3) is a factor of a "difference of squares" expression.

The difference of squares is given generally by:

a^2 - b^2 = (a - b)(a + b).

In this case, the first term shows that in YOUR problem you want a = √2, and b = √3. So that means that you want to multiply both the numerator and the denominator by the corresponding factor (√2+√3):

So you have 6 (√2+√3) / [(√2-√3)(√2+√3)] = 6 (√2+√3) / (2 - 3)

= - 6 (√2+√3).

Live long and factor!

Answer 3

Multiply by (√2+√3) / (√2+√3).

Then you get : 6(√2+√3) / (2 - 3)

= 6(√2+√3) / (-1)

= -6(√2+√3)

Answer 4

6 / (√2-√3) x (√2+√3)/(√2+√3)

= 6 (√2+√3)/(√2-√3)(√2+√3)

but (√2-√3)(√2+√3) = (√2)^2 + (√2 x √3) - (√2 x √3) + (√3)^2

= 2 + 3 = 5

Solution: 6 (√2+√3)/5

Answer 5

You do this by multiplying top and bottom of the fraction by the "conjugate" of the denominator. That
means use the same binomial, but change the sign.
6 / rt2-rt3 X( rt2+rt3)/(rt2+rt3)
= 6(rt2 + rt3)/2-3
=6(rt2+rt3)/-1
=-6(rt2+rt3)
The denominator has now been "rationalized".

Answer 6

6/(sq rt 2-sq rt 3)
multiply & divide by (sq rt 2 + sq rt 3)we get
6(sq rt 2-sq rt 3)/(2-3)------(a-b)(a+b)=a^2-b^2
6(sq rt 2-sq rt 3)/(-1)
=-6(sq rt2-sq rt3) ans

Answer 7

6/(sq2 - sq rt3) x (sq rt2 + sq rt3)/(sq rt2 + sq rt3) (Multiply through by the conjugate)

6(sq2 + sq rt3)/ (sq rt2 - sq rt 3)(sq rt2 + sq rt3)
6(sq rt2 + sq rt3)/(2 - sq rt2sq rt3 + sq rt2sq rt3 - 3)
6(sq rt2 + sq rt3)/(2 - 3)
6(sq rt2 + sq rt3)/(-1)
-6(sq rt2 + sq rt3)
-6sq rt2 - 6sq rt3 = -18.87758622

Answer 8

Um, divide it ?