Question

Who can rationalize

g^3/4 h^5/4 over square root of 3r^3/4 ?

Answer 1

mutiply both the numerator and the denominator with the denominator

Answer 2

\[\LARGE {{g^{1 \over 4} * g^{1 \over 4} * g^{1 \over 4} * h * h^{1 \over 4}} \over 3r^{3 \over 2} / 4} \]

Answer 3

@ParthKohli . Eh? Elaborate pls.

Answer 4

You can see my tutorial, you'll understand everything.. If you don't then turn over to me.

Answer 5

\[{g^{{3}\over{4}}\times h^{{5}\over{4}}}\over{\sqrt{3r^{{3}\over{4}}}}\]
\[={{g^{{3}\over{4}}\times h^{{5}\over{4}}}\over\sqrt{3}r^{{3}\over{8}}}\]
\[={{g^{{3}\over{4}}\times h^{{5}\over{4}}}\times r^{{3}\over{8}}\over\sqrt{3}}\]

Answer 6

I don't understand. What the procedure? Like the steps?

Answer 7

@sheg square root of 3 is not rationalized. Simplify it further..

Answer 8

Ask sheg, he'll tell you.

Answer 9

How can the radical be removed?

Answer 10

@sheg ?

Answer 11

Multiply both the numerator and denominator by square root of 3.

Answer 12

\[{g^{{3}\over{4}}\times h^{{5}\over{4}}}\over{\sqrt{3r^{{3}\over{4}}}}\]
\[={{g^{{3}\over{4}}\times h^{{5}\over{4}}}\over\sqrt{3}r^{{3}\over{8}}}\]
\[={{g^{{3}\over{4}}\times h^{{5}\over{4}}}\times r^{{3}\over{8}}\over\sqrt{3}}\]
\[={3^{{-1}\over{2}}}\times {{g^{{3}\over{4}}\times h^{{5}\over{4}}}\times r^{{3}\over{8}}}\]

Answer 13

@sheg Is it like the final answer?

Answer 14

on simplification you will get
3^(-1/2)g^(3/4)h^(5/4)r^(-3/8)
hey in above i forgot to put minus sign in the power of r @chol

Answer 15

Ok. Thanks so much @sheg

Answer 16

did u understood how to solve this one

Answer 17

Ahm. I kind of understand it now :)

Answer 18

see sqrt of 3 can be written as\[\sqrt{3} = 3^{{1}\over{2}}\]

Answer 19

Yeah. I know that thing.

Answer 20

ok so in case of sqrt(r^(3/4)) = r^(3/8)
and as it is in the denominator so it can be written in the numerator as r^(-3/8)

Answer 21

Oh. That's kind of helpful. Thank you.

Answer 22

@sheg :)