Question

The square root of 8x times the square root of 8x

Answer 1

\[\sqrt{8x}\sqrt{8x}\] ?

Answer 2

= 8x and x>=0

Answer 3

Can you explain better how you got that?

Answer 4

\[\sqrt a\times\sqrt a=(\sqrt a)^2=a\]

Answer 5

Actually the original problem is the square root of 5 divided by the square root of 8x. Can you show the whole process because I'm really confused

Answer 6

But i understand how the square roots cancel out

Answer 7

\[\sqrt{5}/\sqrt{8x} = \sqrt{\left(\begin{matrix}5 \\ 8x\end{matrix}\right)}\]

Answer 8

I need to rationalize the denominator

Answer 9

\[\frac{\sqrt5}{\sqrt{8x}}=\frac{\sqrt5}{\sqrt{2^3x}}\cdot\frac{\sqrt{2x}}{\sqrt{2x}}=\frac{\sqrt{10x}}{4x}\]

Answer 10

Why don't i multiply the numerator and denominator by the swuare root of 8x?? why do i have to break it up

Answer 11

\[\frac{\sqrt5}{\sqrt{8x}}=\frac{\sqrt5}{\sqrt{8x}}\cdot\frac{\sqrt{8x}}{\sqrt{8x}}=\frac{\sqrt{40x}}{\sqrt{64x^2}}=\frac{2\sqrt{10x}}{8x}=\frac{\sqrt{10x}}{4x}\]

Answer 12

wait i thought you said \[\sqrt{8x } * \sqrt{8x} = 8x\]

Answer 13

yeah it does, look at the denominator.
it winds up as 8x, then simplifies to 4x

Answer 14

and on the last step you just divided the top and bottum by 2 to get rid of it?

Answer 15

yeah

Answer 16

ok thank you!

Answer 17

no prob