Question

Some1 help?

Answer 1

Find the exact value of the radical expression in simplest form.\[2\sqrt{7x} + \sqrt{7x}-3\sqrt{7x}-7\]

Answer 2

If we set \( \sqrt{7x}=y\) then, \(2\sqrt{7x} + \sqrt{7x}-3\sqrt{7x}-7=2y+y-3y-7\) which equas \(0-7=-7\)

Answer 3

Find the exact value of the radical expression in simplest form.\[\sqrt{s ^{8}}+\sqrt{25s ^{8}}+2\sqrt{s^{8}}-\sqrt{s ^{4}}\]

Answer 4

\( \sqrt{s^8}=s^{8/2}=s^4\), so set y = \(s^4\), then \(y+5y+2y-{y ^{1/2}}\), which simplifies to \(8y-y^{1/2}=s^4-s^{4/2}=s^4-s^2\), after subbing back.

Answer 5

the options are \[8s ^{4}, 28s ^{4}\sqrt{s}, 8s ^{4} -s ^{2}, or 28s ^{4} \sqrt{s}-s ^{2}\]

Answer 6

Hmm... I wonder which one it is ? Did you see what we did above?

Answer 7

it was confusing

Answer 8

But now?

Answer 9

still

Answer 10

Ok, so we have made some substitutions to make the s^4 stuff easier to read. In the end we found the result \( s^4-s^2\)

Answer 11

is it the 3rd one

Answer 12

yep