Question

(x^2-1)^1/2 - (x^2-1)^-1/2(2-x^2) all divided by x^2-1

Answer 1

It's all you Jim

Answer 2

Is the expression \[\Large \frac{(x^2-1)^{\frac{1}{2}} - \frac{(x^2-1)^{-\frac{1}{2}}}{(2-x^2)}}{x^2-1}\]

or is it


\[\Large \frac{\frac{(x^2-1)^{\frac{1}{2}} - (x^2-1)^{-\frac{1}{2}}}{(2-x^2)}}{x^2-1}\]


???

Answer 3

or is it \[\large \frac{(x^2-1)^{\frac{1}{2}} - (x^2-1)^{-\frac{1}{2}}(2-x^2)}{x^2-1}\]

Answer 4

second equation and i need to solve for x

Answer 5

You can't solve for x here because this isn't an equation. You can only simplify. Make sure you've copied the entire problem correctly.

Answer 6

it says the answer is \[\pm \sqrt{3/2}\] and it says solve for x and and in parentheses it in says put in simplest form.

Answer 7

Ok,so that means that this is an equation, but I don't see any equal signs. Can you repost the actual equation?

Answer 8

\[(x ^{2}-1)^{1/2}-(x ^{2}-1)^{-1/2}(2-x ^{2}) \div (x ^{2}-1)= 0\]

Answer 9

ah ok, thx

Answer 10

\[\large (x^{2}-1)^{1/2}-(x^{2}-1)^{-1/2}(2-x^{2}) \div (x^{2}-1)= 0\]


\[\large (x^{2}-1)^{1/2}-(x^{2}-1)^{-1/2}(2-x^{2})= 0*(x^{2}-1)\]


\[\large (x^{2}-1)^{1/2}-\frac{1}{(x^{2}-1)^{1/2}}(2-x^{2})= 0\]


\[\large \sqrt{x^{2}-1}-\frac{1}{\sqrt{x^{2}-1}}(2-x^{2})= 0\]


\[\large x^{2}-1-(2-x^{2})= 0\]


\[\large x^{2}-1-2+x^{2}= 0\]


\[\large 2x^{2}-3= 0\]


\[\large 2x^{2}= 3\]


\[\large x^{2}= \frac{3}{2}\]


\[\large x= \pm\sqrt{\frac{3}{2}}\]

Answer 11

did the two radicals cancel out?

Answer 12

I multiplied EVERY term by \[\sqrt{x^2-1}\] to clear out the fractions.

In doing that, the first term became \[\sqrt{x^2-1}*\sqrt{x^2-1}=(\sqrt{x^2-1})^2=x^2-1\]


For the second term, they simply canceled.

Answer 13

oh ok thank you so much