Re: having trouble factoring

Hello, deviantxane!

Where did this equation come from?

Are you sure it's correct?

Quote:

$\displaystyle \text{Solve: }\:4x - 2(x^2-1)^{-\frac{1}{2}}\:=\:0$

Note that $\displaystyle x \ne \pm1.$

We have: .$\displaystyle 4x - \frac{2}{\sqrt{x^2-1}}\:=\:0 \quad\Rightarrow\quad 4x \:=\:\frac{2}{\sqrt{x^2-1}}$

Multiply by $\displaystyle \sqrt{x^2-1}:\;\;4x\sqrt{x^2-1} \:=\: 2 \quad\Rightarrow\quad 2x\sqrt{x^2-1} \:=\: 1$

Square both sides: .$\displaystyle 4x^2(x^2-1) \:=\:1 \quad\Rightarrow\quad 4x^4 - 4x^2 - 1 \:=\:0$

Quadratic Formula: .$\displaystyle x^2 \;=\;\frac{4\pm\sqrt{(\text{-}4)^2 - 4(4)(\text{-}1)}}{2(4)} \;=\;\frac{4\pm4\sqrt{2}}{8} \;=\;\frac{1\pm\sqrt{2}}{8}$

For a real root: .$\displaystyle x \;=\;\pm\sqrt{\frac{1 + \sqrt{2}}{8}}$

Re: having trouble factoring

thanks, I looked back over my problem today and realized I had forgotten a calculus rule that lead me to the equation I was trying to factor thanks so much or your help.