# How do you find all the zeros of f(x) = 5x^4 - x^2 + 2?

##### 2 Answers
Jul 9, 2016

$x = \pm \left(\sqrt{\frac{2 \sqrt{10} + 1}{20}}\right) \pm \left(\sqrt{\frac{2 \sqrt{10} - 1}{20}}\right) i$

#### Explanation:

Since $f \left(x\right)$ has no odd terms, we could treat it as a quadratic in ${x}^{2}$, but the resulting quadratic has negative discriminant:

$\Delta = {\left(- 1\right)}^{2} - 4 \left(5\right) \left(2\right) = 1 - 40 = - 39$

As a result it has only Complex zeros, of which we would then want to find square roots. This is a possible approach, but there is an alternative one...

Note that:

$\left({a}^{2} - k a b + {b}^{2}\right) \left({a}^{2} + k a b + {b}^{2}\right) = {a}^{4} + \left(2 - {k}^{2}\right) {a}^{2} {b}^{2} + {b}^{4}$

Let:

$a = \sqrt[4]{5} x$

$b = \sqrt[4]{2}$

Then:

${a}^{4} + \left(2 - {k}^{2}\right) {a}^{2} {b}^{2} + {b}^{4} = 5 {x}^{4} + \left(2 - {k}^{2}\right) \sqrt{10} {x}^{2} + 2$

If we let:

$k = \sqrt{\frac{20 + \sqrt{10}}{10}}$

Then:

$\left(2 - {k}^{2}\right) \sqrt{10} = \left(2 - \frac{20 + \sqrt{10}}{10}\right) \sqrt{10} = - 1$

And:

$k a b = \sqrt{\frac{20 + \sqrt{10}}{10}} \sqrt[4]{10} x = \sqrt{\frac{20 + \sqrt{10}}{10} \cdot \sqrt{10}} = \sqrt{2 \sqrt{10} + 1}$

So:

$f \left(x\right) = \left({a}^{2} - k a b + {b}^{2}\right) \left({a}^{2} + k a b + {b}^{2}\right)$

$= \left(\sqrt{5} {x}^{2} - \left(\sqrt{2 \sqrt{10} + 1}\right) x + \sqrt{2}\right) \left(\sqrt{5} {x}^{2} + \left(\sqrt{2 \sqrt{10} + 1}\right) x + \sqrt{2}\right)$

Then the zeros of these quadratic factors are given by the quadratic formula as:

$x = \frac{\pm \sqrt{2 \sqrt{10} + 1} \pm \sqrt{\left(2 \sqrt{10} + 1\right) - 4 \sqrt{10}}}{2 \sqrt{5}}$

$= \pm \left(\sqrt{\frac{2 \sqrt{10} + 1}{20}}\right) \pm \left(\sqrt{\frac{2 \sqrt{10} - 1}{20}}\right) i$

Jul 9, 2016

$x = \pm \left(\sqrt{\frac{2 \sqrt{10} + 1}{20}}\right) \pm \left(\sqrt{\frac{2 \sqrt{10} - 1}{20}}\right) i$

#### Explanation:

$f \left(x\right) = 5 {x}^{4} - {x}^{2} + 2$

I will use the result I derived for another question (see https://socratic.org/s/aw38evei), that the square roots of $a \pm b i$ are:

$\pm \left(\sqrt{\frac{\sqrt{{a}^{2} + {b}^{2}} + a}{2}}\right) \pm \left(\sqrt{\frac{\sqrt{{a}^{2} + {b}^{2}} - a}{2}}\right) i$

First treat $f \left(x\right)$ as a quadratic in ${x}^{2}$ and use the quadratic formula to find:

${x}^{2} = \frac{1 \pm \sqrt{{\left(- 1\right)}^{2} - 4 \left(5\right) \left(2\right)}}{2 \cdot 5} = \frac{1 \pm \sqrt{- 39}}{10} = \frac{1}{10} \pm \frac{\sqrt{39}}{10} i$

Let $a = \frac{1}{10}$ and $b = \frac{\sqrt{39}}{10}$

Then:

$\sqrt{{a}^{2} + {b}^{2}} = \sqrt{\frac{1}{100} + \frac{39}{100}} = \sqrt{\frac{40}{100}} = \frac{2 \sqrt{10}}{10}$

So the square roots of $a + b i$ are:

$\pm \left(\sqrt{\frac{\sqrt{{a}^{2} + {b}^{2}} + a}{2}}\right) \pm \left(\sqrt{\frac{\sqrt{{a}^{2} + {b}^{2}} - a}{2}}\right) i$

$= \pm \left(\sqrt{\frac{\frac{2 \sqrt{10}}{10} + \frac{1}{10}}{2}}\right) \pm \left(\sqrt{\frac{\frac{2 \sqrt{10}}{10} - \frac{1}{10}}{2}}\right) i$

$= \pm \left(\sqrt{\frac{2 \sqrt{10} + 1}{20}}\right) \pm \left(\sqrt{\frac{2 \sqrt{10} - 1}{20}}\right) i$

These are the zeros of our quartic.