If the resultant or discriminant of a polynomial is not zero, can we conclude critical points are distinct?
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Quadratic only has one critical point no matter what be its discriminant. – prog_SAHIL Sep 03 '18 at 05:24
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@prog_SAHIL, thanks for the nice reply, could you elaborate on your answer – Math123 Sep 03 '18 at 05:30
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No , this isn't the case.
Consider quadratic equations,
$$x^2-5x+6=0$$
Equation has $D\gt0,$
But the only critical point is $x=\frac{5}{2}$
$$x^2+x+1=0$$
Equation has $D\lt0,$
But the only critical point is $x=-\frac{1}{2}$
In fact, All the quadratic equation irrespective of their discriminant have only one critical point as,
$$p(x)=ax^2+bx+c$$
$p'(x)$ will be a linear equation in $x$ hence providing only one critical point.
$$p'(x)=2ax+b$$
$$x=-\frac{2a}{b}$$
prog_SAHIL
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No, consider for example $\,x^3+1\,$ whose discriminant is $\,\ne 0\,$, yet it has a double critical point at $\,0\,$.
All that can be concluded from a non-$0$ discriminant is that the polynomial itself has no zeros of multiplicity $\,\gt 1\,$.
dxiv
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