1. ## Equations Reducible To Quadratics

For x^4 - 8x^2-4= 0 ,

Does x work out to be = +/-√(4+/-2√5)

N.B. +/- means + or -

I've subbed in my answer but it doesnt seem to work. I've tried finding x using quadratic formula and completing the square.

2. Hello,

I think you are doing the right thing. Just one word: since $\displaystyle 4-2\sqrt{5}<0$, $\displaystyle \pm\sqrt{4-2\sqrt{5}}$ are imaginary numbers.

Bye.

3. Originally Posted by xwrathbringerx
For x^4 - 8x^2-4= 0 ,

Does x work out to be = +/-√(4+/-2√5)

N.B. +/- means + or -

I've subbed in my answer but it doesnt seem to work. I've tried finding x using quadratic formula and completing the square.

Here's how it's solved...

$\displaystyle x^4 - 8x^2 - 4 = 0$.

Let $\displaystyle X = x^2$.

So we have $\displaystyle X^2 - 8X - 4 = 0$.

Using the Quadratic formula $\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ we find

$\displaystyle X = \frac{8 \pm \sqrt{64 + 16}}{2} = \frac{8 \pm \sqrt{80}}{2}$

$\displaystyle = \frac{8 \pm 4\sqrt{5}}{2} = 4 + 2\sqrt{5}$ or $\displaystyle 4 - 2\sqrt{5}$.

Since $\displaystyle x^2 = X$ we have

$\displaystyle x^2 = 4 + 2\sqrt{5}$ or $\displaystyle x^2 = 4 - 2\sqrt{5}$.

So $\displaystyle x = \pm \sqrt{4 \pm 2\sqrt{5}}$.

So yes, you're right.