Mathematics 26 Online OpenStudy (anonymous):

(x^2-1)^1/2 - (x^2-1)^-1/2(2-x^2) all divided by x^2-1 OpenStudy (anonymous):

$\frac{(x^2-1)^{1/2}- (x^2-1)^{-1/2}(2-x^2)}{x^2-1}$ like that? OpenStudy (anonymous):

yes and i have solve for x OpenStudy (anonymous):

and it =0 of course OpenStudy (anonymous):

you might multiply numerator and denominator by $(x^2-1)^{1/2}$ to clear the fraction but it is not an equation so you cannot solve for x OpenStudy (anonymous):

oh = 0. ok. still clear the fraction get $\frac{(x^2-1)^{1/2}- (x^2-1)^{-1/2}(2-x^2)}{x^2-1}\times \frac{(x^2-1)^{1/2}}{(x^2-1)^{1/2}}$ $=\frac{x^2-1-(2-x^2)}{(x^2-1)^{3/2}}$ $=\frac{2x^2-3}{(x^2-1)^{3/2}}$ OpenStudy (anonymous):

if this equals zero it means the numerator is equal zero so you can solve via $2x^2-3=0$ $x^2=\frac{3}{2}$ $x=\pm\sqrt{\frac{3}{2}}$ OpenStudy (anonymous):

I know that $(x ^{2}-1)^{1\div2}$ cancels out but where does the -1/2 go because wouldn't that be -1/4 exponent

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