$x^4-14x^2+24x-k=0$ has $4$ real and unequal roots, what is the range of $k$?

Question

$x^4-14x^2+24x-k=0$ has $4$ real and unequal roots, then $k$ lies between $11$ and $117$ or $-11$ and $-8$ or $8$ and $11$ or $0$ and $8$

How can you find the range of $k$. It's an entrance exam question so the solution should not be too long

Thanks.

Answer 1

Let $f(x)=x^4-14x^2+24x-k$. Then $f'(x)=4(x-1)(x-2)(x+3)$ so to have four unequal real roots of $f$ you want $k$ such that $$ f(-3)<0, f(1)>0 \text{ and } f(2)<0. $$ Since $f(-3)=-117-k$, $f(1)=11-k$, and $f(2)=8-k$, you get the answer $8<k<11$.

Answer 2

A sketch:

I would put it in a different way: rewrite the equation as $\;x^4-14x^2+24x=k$, and set $$g(x)=x^4-14x^2+24x.$$ Now $g'(x)=4(x^3-7x+6)$ has three roots: $-3, 1,2$, so that the quartic $g(x)$ has one local maximum at $1$ and two local minima, at $-3$ and $2$. Furthermore, it tends to $+\infty$ when $x\to\pm\infty$.

We conclude that the equation $g(x)=k$ has four distinct roots if and only if $k\in \bigl[\max\bigl(g(-3),g(2)\bigr), g(1)\bigr]$.