zero of a function

Question

I would like to find zeros of this function $f(x) = 6x^4 + x^2 - 1$

I've tried Descartes's rules but I couldn't find the zero from any of my possibilities

Answer 1

Start with factoring and then solve each factor

$$f(x) = 6x^4 + x^2 - 1=(3x^2-1)(2x^2+1)=0 \implies $$

$$ x=\pm \frac {\sqrt 3}{3}, x=\pm i \frac {\sqrt 2}{2}$$

Answer 2

Hint:

Start with finding the zeros of $$g(y)=6y^2+y-1$$

Then consider the relationship between $f(x)$ and $g(y)$

Answer 3

Hint:

What if we set $u=x^2$? Then our equation becomes $$f(u)=6u^2+u-1$$ Can you solve that?

Answer 4

Hint:

This is a biquadratic equation. Set $u=x^2$ and find the non-negative root of the quadratic equation $$6u^2+u-1=0.$$ Note that, as the leading coefficient and the constant term do not have the same sign, the quadratic equation will have one positive and one negative root. So the given biquadratic equation has two opposite real roots and two opposite imaginary roots.

Answer 5

Substituate $y=x^2$, then

$$6x^4+x^2-1=6y^2+y-1$$

to find the roots of the original function, start to solve the below one $$6y^2+y-1=0$$

$$y_{1,2}=\frac{-1 \pm \sqrt{1^2-4\cdot 6\cdot(-1)}}{2\cdot 6}=\frac{-1 \pm \sqrt{25}}{12}=\frac{-1 \pm 5}{12}$$

$$y_1=\frac{-1 + 5}{12}=\frac{4}{12}=\frac{1}{3}$$ $$y_2=\frac{-1 - 5}{12}=-\frac{6}{12}=-\frac{1}{2}$$

Now you can express $y$ by $x$

$$y_1=x^2=\frac{1}{3} \implies x_1=\frac{\sqrt{3}}{3}, \ \ \ x_2=-\frac{\sqrt{3}}{3}$$

$$y_2=x^2=-\frac{1}{2} \implies x_3=\frac{\sqrt{2}}{2}i, \ \ \ x_4=-\frac{\sqrt{2}}{2}i.$$

Obviously thre are four solutions including only two real solutions.