For what values of $m$, will the polynomial $P(x)=x^4+(m-3)x+m^2$ have four distinct real roots?

Question

I'm not quite too sure how to approach this question, so any explanation using any technique would be greatly appreciated. Thank you!

Answer 1

Assume it has 4 real roots $a,b,c,d$. By vieta's formulas,

$$abcd=m^2$$ $$a+b+c+d=0$$ $$ab+bc+cd+da+ac+bd=0$$ $$abc+bcd+cda+dab=m-3$$

But since $(a+b+c+d)^2=a^2+b^2+c^2+d^2+2(ab+bc+cd+da+ac+bd)$,

$$a^2+b^2+c^2+d^2=0\implies a=b=c=d=0$$

So this means if $P(x)$ has all real roots they must be all $0$, contradicting distinctness.

Therefore no value of $m$ makes $P(x)$ have 4 distinct real roots.

Answer 2

Hint:  it cannot have $\,4\,$ distinct real roots. Either use Descartes' rule of signs, or look at $\,P''(x)\,$.