4
$\begingroup$

I would appreciate if somebody could help me with the following problem.I am not good at quartic equations,so could not attempt much.

Q:The number of integral values of $p$ for which the equation $x^4+4x^3-8x^2+p=0$ has all 4 real roots.

Let $\alpha,\beta,\gamma,\delta $ are four real roots.
According to Vieta's formula
$\alpha+\beta+\gamma+\delta=-4$
$\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta=-8$
$\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta=0$
$\alpha\beta\gamma\delta=p$

then i got stuck..what to do?

Thanks in advance.

$\endgroup$
  • 2
    $\begingroup$ You could also consider the equation $x^4 + 4x^3 - 8x^2 = 0$ and look at the behaviour of the roots of that (should be easily factorisable), and then consider p as a vertical translation $\endgroup$ – BadAtMaths Jul 25 '15 at 10:57
8
$\begingroup$

For a simple approach consider the function $y=x^4+4x^3-8x^2=x^2\cdot\left((x+2)^2-12\right)$ - the intersections with the line $y=-p$ will give the roots of the original. Since this is just a horizontal line in the normal $x,y$ plane, a quick sketch will show that the number of real roots is governed by the relationship of $p$ to the local minima/maxima of the quartic.

The form of the quartic makes this easy to sketch - and the double root at $x=0$ means the cubic you get on differentiating has an obvious root, leaving a quadratic to factor.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.