Determine the exact value of $k$ if the cubic equation $x^3 + kx +4 = 0$ has 2 distinct real roots
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1What have you tried? Please show your attempts. – Lion Heart Jun 28 '22 at 12:54
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First off, $k$ has to be negative. Can you see why? – Oscar Lanzi Jun 28 '22 at 12:54
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Note: I assume you meant to say that it should have exactly two distinct real roots. $x^3-10x+4$ has three distinct real roots (so, in particular, it has two distinct real roots). Your case ought to have one real root of multiplicity $1$ and a second, distinct, real root with multiplicity $2$. Phrasing it that way should guide the computation. – lulu Jun 28 '22 at 13:05
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Read wikipedia for the cubic $x^3+px+q=0$ and the discriminant. – Dietrich Burde Jun 28 '22 at 13:07
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I think that if it is a double root it is also a root of the derivative => k must be < 0 ? – Callie12 Jun 28 '22 at 13:07
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Using Vietas theorem makes solution more easy. Let double root is $x_1$ and other root is $x_2$. Then $2x_1+x_2=0$, $x_1^2x_2=-4$, $x_1^2+2x_1x_2=k$. First two equations give $x_1$, $x_2$ and third equation gives $k$. – Ivan Kaznacheyeu Jun 29 '22 at 07:10
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Let $f(x)=x^3+kx+4$. Being a real cubic equation, $f(x)=0$ will have at least one real root. It will have other roots real if $f_{max}>0$ and $f_{min}<0$. It will have one root as double root if $f_{max}f_{min}=0.$ Here $f'(x)=0 \implies x=\pm \sqrt{-k/3}$. Then $f_{max}=f(-\sqrt{-k/3})=4-\frac{2k}{3}\sqrt{-k/3}, f_{min}=f(-\sqrt{-k/3})=4+\frac{2k}{3}\sqrt{-k/3}.$ Next, $f_{min}f_{max}=0 \implies k=-3(4)^{1/3}.$
So when $k=-3(4)^{1/3}=-4.7622$, apart from one essential real root there will be one double root. Consequently for this special value of $k$, $f(x)=0$ will have two distinct real roots. S below see the plot of $f(x)$
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Very interesting rules regarding $f_{max/min}$. Would you please provide a link where I could learn more about those rules? Thank you. – NoChance May 12 '23 at 01:51