Question

divide. rationalize all denominators.
3+square root 18/ 1+square root 8.
teacher put on the equation multiply the denominators conjugate with numerators and the denominator. how do i do this?

Answer 1

@eliassaab

Answer 2

i thought you did that but i wasnt sure

Answer 3

\[
\frac {3+\sqrt{18}}
{1+\sqrt 8}=\frac {3+\sqrt{18}}
{1+\sqrt 8}\frac {1-\sqrt 8}{1-\sqrt 8}
\]

Answer 4

There was a misprint in the older post. Look at the new one

Answer 5

yeah i was about to ask why it would be like that.

Answer 6

\[
(a-b)(a+b)=a^2-b^2
\]

Answer 7

ok now im confused. what do we plug into that equation

Answer 8

and how do we multiply the 3+square root 18 by 1- square root 8

Answer 9

\[
\frac {3+\sqrt{18}}
{1+\sqrt 8}=\frac {3+\sqrt{18}}
{1+\sqrt 8}\frac {1-\sqrt 8}{1-\sqrt 8}=
\frac{(3+ \sqrt{18})(1-\sqrt 8)}{1^2 - (\sqrt 8)^2}=\\
\frac{(3+ \sqrt{18})(1-\sqrt 8)}{-7}\\


\]

Answer 10

Expand the numerator and you are done

Answer 11

so would we use foil to expand the numerator

Answer 12

You see now, you do not have any radical in the denominator

Answer 13

Yes

Answer 14

give me one minute to work it out and if it is wrong can you help me

Answer 15

um i think i got the first part but the last im not sure.
3+3square root 8+1square root 18- square root 7.07

Answer 16

\[3+3\sqrt{8}+1\sqrt{18}-\sqrt{7.07}\]

Answer 17

is it correct or i think the last part is wrong

Answer 18

Leave the numeratro in terms of sqrt

Answer 19

i lost internet sorry its taking so long my internet is being very funny today

Answer 20

\[
\left(\sqrt{18}+2\right)
\left(1-\sqrt{8}\right)=-10-\sqrt{2}
\]

Answer 21

ok so we factor it out as far down as it will go. and that is are answer

Answer 22

\[
\frac{ -10-\sqrt 2}{-7}=\frac{ 10+\sqrt 2}{7}
\]

Answer 23

Yes

Answer 24

ok is that our answer.

Answer 25

I used the fact that
\[
\sqrt{18} = 3 \sqrt 2\\
\sqrt{8} = 2 \sqrt 2
\]

Answer 26

Yes

Answer 27

ok so it is \[10+\sqrt{2}\div 7\]

Answer 28

that is our answer correct

Answer 29

@eliassaab can you help with my new problem i have posted