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If the resultant or discriminant of a polynomial is not zero, can we conclude critical points are distinct?

Bernard
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Math123
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2 Answers2

2

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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2

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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