Determine the values of k for which the equation will have zero, two or four real roots

Question

Determine the values of k for which the equation $\frac{x^4+1}{x^2}-k=0$ will have zero, two or four real roots.

So far I have found the derivative of the function which is $y'=\frac{2x^4-2}{x^3}$ and then I set $y'=0$, which gives us that the critical points are $x=±1$. Now plugging in $x=1$ or $x=-1$ into the equation gives us that $k=2$, and so the function will have two real roots when $k=2$.

I'm not sure how to prove the rest, but I know that the answer is when $k>2$ the function has four real roots and when $k<2$ the function doesn't have any real roots.

Answer 1

The equation is equivalent to $\;x^4-kx^2+1=0$, and if you set $\;u=x^2$, solving it (in $\mathbf R$) amounts to solve the equation $$u^2-tu+1=0 \tag{1}$$ in positive numbers. Note that the latter equation, if it has any real roots, ($\Delta=k^2-4\ge 0$) it has roots with the same sign since their product is equal to $1$. So

• The original equation has no root if $$\begin{array}{lll} \text{– either }& \Delta<0,&\text{i.e. }\; -2<k<2,\\ \text{– or }& \Delta\ge0,&\text{i.e. }\;\, |k|\ge 2\enspace\text{ and the roots are negative, i.e. their sum }\; k<0\\ &&\text{so in the end }\;k\le -2. \end{array}$$ Summing this up, we conclude there's no root if and only if $\;k<2$.
• It has two roots if and only if eq. $(1)$ has a positive double root, which happens when $k=2$.
• It has four roots if and only if eq. $(1)$ has two positive roots, which happens if and only if $|k|>2$ and $k>0$, i.e. $\;k>2$.

Answer 2

Use that $$x^4-kx^2+1=0$$ and Substitute $$x^2=t$$

Answer 3

Multiplying through by $x^2$, the original equation becomes $x^4 - kx^2 + 1 = 0$ ($x \neq 0$). You can find the roots by using the quadratic formula and decide if they are real or complex by looking at the discriminant.