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

then i got stuck..what to do?

Thanks in advance.

  • 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

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.


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