Question

3/3sqrt2-sqrt3 rationalize the denominator and simplify can i get an explanation please

Answer 1

Is your original expression:

a) \(\frac{3}{3\sqrt{2} - \sqrt{3}}\)
b) \(\frac{3}{3\sqrt{2}} - \sqrt{3}\)

Answer 2

a

Answer 3

Ok, so multiply top and bottom by the conjugate \(3\sqrt{2} + \sqrt{3}\)

Answer 4

top and bottom??

Answer 5

yes, anything you do to the denominator you must do to the numerator to keep the ratio the same.

Answer 6

Anything you multiply the denominator by that is.

\[\frac{4}{3} = \frac{4*2}{3*2} = \frac{8}{6}\]

Answer 7

im still confused i wrote this \[3/3\sqrt2-\sqrt3 *3\sqrt2+\sqrt3/3\sqrt2+\sqrt3\]

Answer 8

\[3/3\sqrt2-\sqrt3 * 3\sqrt2+\sqrt3/3\sqrt2+\sqrt3\]i mean

Answer 9

You need to use parentheses

Answer 10

\[\frac{3}{3\sqrt{2} - \sqrt{3}} = \frac{3(3\sqrt{2} + \sqrt{3})}{(3\sqrt{2} - \sqrt{3})(3\sqrt{2} + \sqrt{3})}\]

Answer 11

And on bottom we get some nice simplification.

Answer 12

i dont understand how that simplifies

Answer 13

Foil it out..
\[(3\sqrt{2} - \sqrt{3})(3\sqrt{2} + \sqrt{3})\]
\[=3\sqrt{2}*3\sqrt{2} - 3\sqrt{2}\sqrt{3} +3\sqrt{2}\sqrt{3} -\sqrt{3}*\sqrt{3}\]
\[=9(2) + 0 - 3 = 15\]

Answer 14

Which cancels with the 3 up top

Answer 15

to give you
\[\frac{3\sqrt{2} + \sqrt{3}}{5}\]

Answer 16

i am not grasping this

Answer 17

i dont understand how you got 15

Answer 18

i dont understand how you got 15

Answer 19

This is what you had in the denominator right?

\[(3\sqrt{2} - \sqrt{3})(3\sqrt{2} + \sqrt{3})\]
This is just like
\[(a - b)(a+b) = a^2 -ba + ab - b^2\]
The middle two terms cancel each other out and you are left with
\[a^2 - b^2\]
Where in your case, \(a=3\sqrt{2}\) and \(b=\sqrt{3}\)
So:
\[a^2 - b^2 \]\[= (3\sqrt{2})^2 - (\sqrt{3})^2 \]\[= 3^2\sqrt{2}^2 - \sqrt{3}^2 \]\[=9(2) - 3 = 18-3 = 15\]

Answer 20

i understand the foil method better thankyou i understand now