# How do you find the zeroes for f(x)=2x^4-6x^2+1?

Jul 8, 2015

$x = \pm \sqrt{\frac{6 - 2 \sqrt{7}}{4}}$ or $x = \pm \sqrt{\frac{6 + 2 \sqrt{7}}{4}}$

#### Explanation:

The first step is to substitute $t = {x}^{2}$ which changes the degree of the equation to 2.
Now you get: $2 {t}^{2} - 6 t + 1 = 0$ and you solve it as any other quadratic equation.

$\Delta = {6}^{2} - 4 \cdot 2 \cdot 1 = 36 - 8 = 28$
$\sqrt{\Delta} = 2 \sqrt{7}$

${t}_{1} = \frac{6 - 2 \sqrt{7}}{4}$
${t}_{2} = \frac{6 + 2 \sqrt{7}}{4}$

Before we go further we must remember, that $t = {x}^{2}$, so $t$ cannot be negative. Both values fulfill this condition, so we can ca;lculate $x$ as:

${x}_{1} = \sqrt{{t}_{1}}$
${x}_{2} = - \sqrt{{t}_{1}}$
${x}_{3} = \sqrt{{t}_{2}}$
${x}_{4} = - \sqrt{{t}_{2}}$