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Question Number 200388 by mr W last updated on 18/Nov/23
find all values for k such that the eq.  x^3 −13x+k=0 has three integer roots.
$${find}\:{all}\:{values}\:{for}\:{k}\:{such}\:{that}\:{the}\:{eq}. \\ $$$${x}^{\mathrm{3}} −\mathrm{13}{x}+{k}=\mathrm{0}\:{has}\:{three}\:{integer}\:{roots}. \\ $$
Answered by Frix last updated on 18/Nov/23
(x−p)(x+(p/2)−q)(x+(p/2)+q)=0  x^3 −((3p^2 +4q^2 )/4)x−((p(p^2 −4q^2 ))/4)=0  ⇒  q=((√(52−3p^2 ))/2)∧k=−p(p^2 −13)  ⇒ p=±{1, 3, 4}∧k=±12
$$\left({x}−{p}\right)\left({x}+\frac{{p}}{\mathrm{2}}−{q}\right)\left({x}+\frac{{p}}{\mathrm{2}}+{q}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{3}} −\frac{\mathrm{3}{p}^{\mathrm{2}} +\mathrm{4}{q}^{\mathrm{2}} }{\mathrm{4}}{x}−\frac{{p}\left({p}^{\mathrm{2}} −\mathrm{4}{q}^{\mathrm{2}} \right)}{\mathrm{4}}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${q}=\frac{\sqrt{\mathrm{52}−\mathrm{3}{p}^{\mathrm{2}} }}{\mathrm{2}}\wedge{k}=−{p}\left({p}^{\mathrm{2}} −\mathrm{13}\right) \\ $$$$\Rightarrow\:{p}=\pm\left\{\mathrm{1},\:\mathrm{3},\:\mathrm{4}\right\}\wedge{k}=\pm\mathrm{12} \\ $$
Commented by mr W last updated on 18/Nov/23
thanks sir!
$${thanks}\:{sir}! \\ $$

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