# How do you find all zeros with multiplicities of f(x)=2x^4-7x^2+6?

May 28, 2017

See explanation.

#### Explanation:

The original equation can be transformed to a quadratic equation by a substitution: ${x}^{2} = t$

$2 {t}^{2} - 7 t + 6 = 0$

$\Delta = {\left(- 7\right)}^{2} - 4 \cdot 2 \cdot 6 = 49 - 48 = 1$

$\sqrt{\Delta} = 1$

${t}_{1} = \frac{7 - 1}{4} = \frac{3}{2}$ and ${t}_{2} = \frac{7 + 1}{4} = 2$

Now we have to find corresponding $x$ values:

${x}^{2} = \frac{3}{2}$

${x}_{1} = - \sqrt{\frac{3}{2}}$ and ${x}_{2} = \sqrt{\frac{3}{2}}$

If we rationalize the denominators we get:

${x}_{1} = - \frac{\sqrt{6}}{2}$ and ${x}_{2} = \frac{\sqrt{6}}{2}$

${x}^{2} = 2$

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

${x}_{1} = - \frac{\sqrt{6}}{2}$, ${x}_{2} = \frac{\sqrt{6}}{2}$, ${x}_{3} = - \sqrt{2}$, ${x}_{4} = - \sqrt{2}$