Question

I have a few math questions involving dividing radicals with extra variables... Help please?!?!?!

Answer 1

?

Answer 2

\[3+2\sqrt{5}\div2\sqrt{5}-3\]
This is one example, could anyone help explain how to work this?

Answer 3

Oh That is a fraction btw...

Answer 4

i am going to make a guess that this is really
\[\frac{3+2\sqrt5}{2\sqrt5-3}\]

Answer 5

multiply top and bottom by 2*sqrt(5) +3

Answer 6

and simplify

Answer 7

remember sqrt(x)*sqrt(X) = x

Answer 8

You'll need to use the difference of two squares here: \[(a+b)(a-b)=a^2+b^2\]So we multiply the top and bottom by 2sqrt{5}+3:
\[\frac{3+2\sqrt{5}}{2\sqrt{5}-3}* \frac{2\sqrt{5}+3}{2\sqrt{5}+3} = \frac{(3+2\sqrt{5})(2\sqrt{5}+3)}{(2\sqrt{5})^2-3^2} \]

Simplify for your answer :)

Answer 9

2*sqrt(5) *2*sqrt(5) = 4*5 = 20

Answer 10

Okay, so my next question, how would I simplify?

Answer 11

3*2*sqrt(5) + 3*3 + 2sqrt(5)*2sqrt(5) + 2sqrt(5)*3

That is the top .... remember the FOIL thing

Answer 12

(x+y)(a+b) = x*a + x*b + y*a + y*b

Answer 13

I think it stood for Firsts, Outers, Inners, Lasts

Answer 14

OH, Okay.

Answer 15

6*sqrt(5) + 9 + 4*5 + 6sqrt(5)

Answer 16

12 sqrt(5) + 29

that is the numerator

Answer 17

\[(2\sqrt{5}-3)(2\sqrt{5}+3) = \]

Answer 18

4*5 +6sqrt(5) - 6 sqrt(5) - 9

Answer 19

= 20 - 9
=11

for the Denominator

Answer 20

so \[\frac{ 12\sqrt{5}+29 }{ 11 }\]

Answer 21

If you ever see a quantity like \[(\sqrt{x}+a)\] in the denominator, then
multiply both numerator and denominator by \[(\sqrt{x}-a)\]

Answer 22

It will get rid of the radical in the denominator, which is not simplified

Answer 23

OKay, thank you, would you be willing to help with a few more?

Answer 24

sure

Answer 25

can you open a new thread and tag me

Answer 26

Alright, let me close this question and open a new one.