Thursday, October 24, 2019

polynomials - Question about quartic equation having all 4 real roots


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.


Answer



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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