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Question Number 79085 by key of knowledge last updated on 22/Jan/20

α and β (or more) are root of: (x^2 +x)^2 −2(x^2 +x)−k=0 and αβ<0 ⇒k∈?

$$\alpha\:\mathrm{and}\:\beta\:\left(\mathrm{or}\:\mathrm{more}\right)\:\:\mathrm{are}\:\mathrm{root}\:\mathrm{of}: \\ $$ $$\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}\right)−\mathrm{k}=\mathrm{0} \\ $$ $$\mathrm{and}\:\:\alpha\beta<\mathrm{0}\:\:\:\:\:\:\:\:\Rightarrow\mathrm{k}\in? \\ $$

Commented bymr W last updated on 22/Jan/20

k∈[−1,0) right?

$${k}\in\left[−\mathrm{1},\mathrm{0}\right) \\ $$ $${right}? \\ $$

Answered by MJS last updated on 23/Jan/20

(x^2 +x)^2 −2(x^2 +x)−k=0 ⇒ x^2 +x=1±(√(k+1)) ⇒ x=−(1/2)±((√(5+4(√(k+1))))/2)∨x=−(1/2)±((√(5−4(√(k+1))))/2) to get 4 real solutions −1≤k≤(9/(16)) for k≥−1 there′s always a pair of roots α, β with αβ<0

$$\left({x}^{\mathrm{2}} +{x}\right)^{\mathrm{2}} −\mathrm{2}\left({x}^{\mathrm{2}} +{x}\right)−{k}=\mathrm{0} \\ $$ $$\Rightarrow \\ $$ $${x}^{\mathrm{2}} +{x}=\mathrm{1}\pm\sqrt{{k}+\mathrm{1}} \\ $$ $$\Rightarrow \\ $$ $${x}=−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{5}+\mathrm{4}\sqrt{{k}+\mathrm{1}}}}{\mathrm{2}}\vee{x}=−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{5}−\mathrm{4}\sqrt{{k}+\mathrm{1}}}}{\mathrm{2}} \\ $$ $$\mathrm{to}\:\mathrm{get}\:\mathrm{4}\:\mathrm{real}\:\mathrm{solutions} \\ $$ $$−\mathrm{1}\leqslant{k}\leqslant\frac{\mathrm{9}}{\mathrm{16}} \\ $$ $$\mathrm{for}\:{k}\geqslant−\mathrm{1}\:\mathrm{there}'\mathrm{s}\:\mathrm{always}\:\mathrm{a}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{roots}\:\alpha,\:\beta \\ $$ $$\mathrm{with}\:\alpha\beta<\mathrm{0} \\ $$

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