# How to find the roots of this polynomial

The polynomial: $8x^4-8x^2+1=\frac{\sqrt{3}}{2}$

I can simplify with $u=x^{2}$ to $8u^2-8u+{\frac{\sqrt{3}}{2}}=0$ Mistake $\left(1-\frac{\sqrt{3}}{2}\right)$

apply the quadratic formula: $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ $$\frac{-(-8)\pm\sqrt{(-8)^2-4(8)\frac{\sqrt{3}}{2}} } {2a}$$

reduces to: $u=1\pm \frac{\sqrt{64-16\sqrt{3}}}{2}$

That is what I have done. then I just take the square root of each value of u, and that will give me all 4 values?

I checked my result with wolfalpha. I don't understand why when it say completing the square did it add $1/4$ and not $\left(\frac{-b}{2}\right)$?

• There is a 1 in your first equation that is missing in your second equation. – Gerry Myerson Sep 15 '12 at 13:10
• Also, your third line doesn't follow from either the second or the first. Looks like you have applied the quadratic formula incorrectly. – Gerry Myerson Sep 15 '12 at 13:12
• @GerryMyerson where did I make the mistake? – yiyi Sep 15 '12 at 13:22
• @GerryMyerson I get it $1-\frac{\sqrt{3}}{2}$ doesn't equal $\frac{\sqrt{3}}{2}$ – yiyi Sep 15 '12 at 13:23
• You could post and accept your answer, then, so we have one and the question doesn't keep coming up. – Ross Millikan Sep 15 '12 at 13:36

$8u^2-8u+\left(1-\frac{\sqrt{3}}{2}\right)=0$