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