How do you find all the real and complex roots of f(x) = 4x^4 + 4x^2 + 9x – 2?

Jan 23, 2017

Here's a sketch of how you could solve this algebraically...

Explanation:

$0 = 4 {x}^{4} + 4 {x}^{2} + 9 x - 2$

$\textcolor{w h i t e}{0} = \left(2 {x}^{2} - a x + b\right) \left(2 {x}^{2} + a x + c\right)$

$\textcolor{w h i t e}{0} = 4 {x}^{4} + \left(2 \left(b + c\right) - {a}^{2}\right) {x}^{2} + \left(b - c\right) a x + b c$

Equating coefficients and rearranging a little, we find:

$\left\{\begin{matrix}b + c = {a}^{2} / 2 + 2 \\ b - c = \frac{9}{a} \\ b c = - 2\end{matrix}\right.$

Then:

${\left({a}^{2} / 2 + 2\right)}^{2} = {\left(b + c\right)}^{2} = {\left(b - c\right)}^{2} + 4 b c = \frac{81}{a} ^ 2 - 8$

Multiplied out:

${\left({a}^{2}\right)}^{2} / 4 + 2 {a}^{2} + 4 = \frac{81}{a} ^ 2 - 8$

Hence (multiplying through by $4 {a}^{2}$) and rearranging:

${\left({a}^{2}\right)}^{3} + 8 {\left({a}^{2}\right)}^{2} + 48 \left({a}^{2}\right) - 324 = 0$

Use Cardano's method to find Real root of this cubic:

${a}^{2} = \frac{1}{3} \left(- 8 + \sqrt{5590 + 90 \sqrt{3921}} + \sqrt{5590 - 90 \sqrt{3921}}\right) \approx 3.6060$

We can choose the positive square root:

$a = \sqrt{\frac{1}{3} \left(- 8 + \sqrt{5590 + 90 \sqrt{3921}} + \sqrt{5590 - 90 \sqrt{3921}}\right)}$

Then:

$b = \frac{1}{2} \left({a}^{2} / 2 + 2 + \frac{9}{a}\right)$

$c = \frac{1}{2} \left({a}^{2} / 2 + 2 - \frac{9}{a}\right)$

Leaving two quadratics to solve:

$2 {x}^{2} - a x + \frac{1}{2} \left({a}^{2} / 2 + 2 + \frac{9}{a}\right) = 0$

$2 {x}^{2} + a x + \frac{1}{2} \left({a}^{2} / 2 + 2 - \frac{9}{a}\right) = 0$