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a=-1.89;b=1;
ContourPlot [y^2  ==x^3+a x+b, {x,-2,2},{y,-2,2},GridLines->Automatic]
Clear[a];FindRoot [0==x^3+a x+b,{a,-1.9}]
NSolve[ 0==x^3+a x+b,a]

Trying to find constant $a\approx 1.89 $ more accurately for cross-over point. Please help.

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Narasimham
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  • It is $$y^2=x^3+ax+b$$? – Dr. Sonnhard Graubner Jun 19 '20 at 14:05
  • See for example Cardano's formula for the roots of the cubic. In your case you don't have a crossover point, since there is a tiny gap between the second and third solution. In that region $y^2$ is negative. Try for example $x=0.79$ or $x=0.795$ – Andrei Jun 19 '20 at 14:12

1 Answers1

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It's not entirely clear from your question, but I believe you seek $a$ such that the cubic curve $$y^2=x^3-a x+b \tag{1}$$ has a cross-over point. (I'm changing the sign of $a$ on purpose, and I'm leaving $b$ as a free parameter for generality.) This will happen at the threshold that $$x^3-ax+b=0 \tag{2}$$ has a double-root.

Such a double-root is necessarily also a root of the derivative: $3x^2-a=0$. This gives $x=\pm\sqrt{a/3}$. Substituting this back into $(2)$ gives $$0 = \pm\frac{a^{3/2}}{3\sqrt{3}}\mp\frac{a^{3/2}}{\sqrt{3}}+b = \mp\frac{2a^{3/2}}{3\sqrt{3}}+b \quad\to\quad a^{3/2}=\pm \frac32 b\sqrt{3} \quad\to\quad a=3\left(\frac{b}{2}\right)^{2/3} \tag{3}$$ When $b=1$, this gives $a=\dfrac{3}{2^{2/3}}=1.889881\ldots$ which (with the sign change) corresponds to OP's approximate $-1.89$.

Blue
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