1/square root of x^2 -4
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Originally Posted by yess 1/square root of x^2 -4 Do you mean $\displaystyle \frac{1}{\sqrt{x^2}}-4$ or $\displaystyle \frac{1}{\sqrt{x^2-4}}$ ? I'll assume the first. Note that $\displaystyle \sqrt{x^2}=|x|$.
sorry! its the second one you wrote =]
Multiply it by $\displaystyle \frac{\sqrt{x^2 - 4}}{\sqrt{x^2 - 4}}$.
Originally Posted by yess sorry! its the second one you wrote =] Well the radicand is not a perfect square.. so perhaps they want you to rationalize the denominator.
so i worked it all out to 1/(x^2 - 4) is that right?
No. What is $\displaystyle \frac{1}{\sqrt{x^2 - 4}}\cdot \frac{\sqrt{x^2 - 4}}{\sqrt{x^2 - 4}}$?
Originally Posted by Prove It No. What is $\displaystyle \frac{1}{\sqrt{x^2 - 4}}\cdot \frac{\sqrt{x^2 - 4}}{\sqrt{x^2 - 4}}$? (√(x^2 - 4))/(x^2 - 4) ?
Correct.
so is that as far as it simplifies then??
Originally Posted by yess so is that as far as it simplifies then?? You can either write it as $\displaystyle \frac{\sqrt{x^2 - 4}}{x^2 - 4}$ or $\displaystyle \frac{\sqrt{(x-2)(x+2)}}{(x-2)(x+2)}$, whichever you think is simpler...
Originally Posted by Prove It You can either write it as $\displaystyle \frac{\sqrt{x^2 - 4}}{x^2 - 4}$ or $\displaystyle \frac{\sqrt{(x-2)(x+2)}}{(x-2)(x+2)}$, whichever you think is simpler... ook so i just leave it the way it is like that? i squared top and bottom to get [(x-2)(x+2)/(x-2)(x+2)(x-2)(x+2)] so 1/(x^2 -4) but thats too far?
$\displaystyle \frac{\sqrt{x^2 - 4}}{x^2 - 4} \neq \left(\frac{\sqrt{x^2 - 4}}{x^2 - 4}\right)^2$.
Originally Posted by yess Simplify the following: 1/square root of x^2 -4 Where has the question come from? What is expected by "simplify"? Please give the broader context that the question comes from. There is insufficient information to know what sort of answer is expected.
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