For what range of values of $a$ will the following fourth degree polynomial have all real roots:

$$x^4 - 2ax^2 + x + a^2 -a = 0$$

  • $\begingroup$ Is $a\in \Bbb R$? $\endgroup$ – xavierm02 Apr 27 '13 at 18:51
  • 3
    $\begingroup$ I'm not sure if it's particularly helpful, but the polynomial is the same as $$(x^2-a)^2+(x-a)=0.$$ $\endgroup$ – Clayton Apr 27 '13 at 18:51
  • $\begingroup$ @Clayton I was about to key in the same thing. From this, we can say that the root has to be less than $a$. $\endgroup$ – user17762 Apr 27 '13 at 18:52

First transform this equation into the equation with unknown $a$:

$a^{2}-(2x^{2} + 1)a + x^{4} + x = 0$

Solving the equation with unknown obtain two second degree equations:

$x^{2} + x - a = 0 $ and $ x^{2} - x + 1 - a = 0$

For $a < - \frac{1}{4}$ equation has no real roots

For $a = - \frac{1}{4}$ equation has two equal real roots

For $- \frac{1}{4} < a < \frac{3}{4}$ equation has two different real roots

For $a=\frac{3}{4}$ equation has real four roots of which three are equal

For $ a> \frac{3}{4}$ equation has four different real roots

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    $\begingroup$ Brilliant! I tried a lot to factorize it, but I never realized that it was a quadratic in a! :) $\endgroup$ – Quark Apr 28 '13 at 5:17

Let denote $f(x)=x^4 - 2ax^2 + x + a^2 -a = 0$. We have $$f'(x)=4x^3-4ax+1\quad;\quad f''(x)=12x^2-4a$$

If $f''\geq0$ then $f'$ is increasing and then $f$ has at most two real roots so a necessary condition to have 4 real roots is $$f'' \quad\text{change sign}\iff a>0$$ and the roots of $f''$ are $\alpha=-\sqrt{\frac{a}{3}}$ and $-\alpha$ and if we see the variations of $f'$ we find $$f \quad \text{has 4 real roots}\iff a>0,\quad f'(\alpha)>0,\quad f'(-\alpha)<0$$

Edit We have $$f'(\alpha)=8\left(\frac{a}{3}\right)^{3/2}+1>0\quad \forall a>0$$ and we can find easily (except miscalculation) that $$f'(-\alpha)<0\iff a>\frac{3}{4}$$ so we conclude $$f \quad \text{has 4 real roots}\iff a>\frac{3}{4} $$

  • $\begingroup$ @ShreevatsaR I edited my answer. $\endgroup$ – user63181 Apr 27 '13 at 19:15
  • $\begingroup$ Cool, thanks for this. :) $\endgroup$ – ShreevatsaR Apr 28 '13 at 4:30
  • $\begingroup$ @Sami Ben Romdhane But there is still one more condition right, since you have proved that f'(x) has three real roots (let them be l,m,n such that l>m>n, now we also have to take the condition f(l) < 0; f(m) >0 ; f(n) < 0 $\endgroup$ – Quark Apr 28 '13 at 5:01

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