Equations Reducible To Quadratics

**xwrathbringerx** · October 1st 2008, 03:49 PM

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.

Can anyone please help?

**wisterville** · October 1st 2008, 11:14 PM

Hello,

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

Bye.

**Prove It** · October 2nd 2008, 12:42 AM

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.

Can anyone please help?

Here's how it's solved...

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

Let $X = x^2$ .

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

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

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

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

Since $x^2 = X$ we have

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

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

So yes, you're right.

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