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If-x-2a-2-5-x-2a-2-7-0-x-a-2-8a-10-a-2-9a-11-Find-a-

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Question Number 226536 by hardmath last updated on 02/Dec/25
If (x+(2a^2 +5))(x−(2a^2 +7)) ≤ 0 x∈[−(a^2 +8a−10) ; (a^2 +9a−11)] Find: a = ?
$$\mathrm{If}\:\:\:\left(\mathrm{x}+\left(\mathrm{2a}^{\mathrm{2}} +\mathrm{5}\right)\right)\left(\mathrm{x}−\left(\mathrm{2a}^{\mathrm{2}} +\mathrm{7}\right)\right)\:\leqslant\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\mathrm{x}\in\left[−\left(\mathrm{a}^{\mathrm{2}} +\mathrm{8a}−\mathrm{10}\right)\:;\:\left(\mathrm{a}^{\mathrm{2}} +\mathrm{9a}−\mathrm{11}\right)\right] \\ $$$$\mathrm{Find}:\:\boldsymbol{\mathrm{a}}\:=\:? \\ $$
Answered by gregori last updated on 02/Dec/25
(1) −2a^2 −5 = −a^2 −8a+10 a^2 −18a+15 = 0 (a−3)(a−5)= 0 (2) 2a^2 +7 = a^2 +9a−11 a^2 −9a +18 = 0 (a−3)(a−6)=0 ∴ a = 3
$$\:\:\left(\mathrm{1}\right)\:−\mathrm{2}{a}^{\mathrm{2}} −\mathrm{5}\:=\:−{a}^{\mathrm{2}} −\mathrm{8}{a}+\mathrm{10} \\ $$$$\:\:\:\:\:\:\:\:\:\:{a}^{\mathrm{2}} −\mathrm{18}{a}+\mathrm{15}\:=\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\left({a}−\mathrm{3}\right)\left({a}−\mathrm{5}\right)=\:\mathrm{0} \\ $$$$\:\left(\mathrm{2}\right)\:\mathrm{2}{a}^{\mathrm{2}} +\mathrm{7}\:=\:{a}^{\mathrm{2}} +\mathrm{9}{a}−\mathrm{11}\: \\ $$$$\:\:\:\:\:\:\:\:{a}^{\mathrm{2}} −\mathrm{9}{a}\:+\mathrm{18}\:=\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\left({a}−\mathrm{3}\right)\left({a}−\mathrm{6}\right)=\mathrm{0} \\ $$$$\:\:\therefore\:{a}\:=\:\mathrm{3} \\ $$$$ \\ $$
Answered by mr W last updated on 03/Dec/25
−(a^2 +8a−10)≤a^2 +9a−11 2a^2 +17a−21≥0 ⇒a≤((−17−(√(457)))/4) or a≥((−17+(√(457)))/4) ...(i) 2a^2 +5≥a^2 +8a−10 a^2 −8a+15≥0 (a−3)(a−5)≥0 ⇒a≤3, a≥5 ...(ii) 2a^2 +7≥a^2 +9a−11 a^2 −9a+18≥0 (a−3)(a−6)≥0 ⇒a≤3, a≥6 ...(iii) summary: (((√(457))−17)/4)≤a≤3 or a≥6 ✓
$$−\left({a}^{\mathrm{2}} +\mathrm{8}{a}−\mathrm{10}\right)\leqslant{a}^{\mathrm{2}} +\mathrm{9}{a}−\mathrm{11} \\ $$$$\mathrm{2}{a}^{\mathrm{2}} +\mathrm{17}{a}−\mathrm{21}\geqslant\mathrm{0} \\ $$$$\Rightarrow{a}\leqslant\frac{−\mathrm{17}−\sqrt{\mathrm{457}}}{\mathrm{4}}\:{or}\:{a}\geqslant\frac{−\mathrm{17}+\sqrt{\mathrm{457}}}{\mathrm{4}}\:\:\:…\left({i}\right) \\ $$$$ \\ $$$$\mathrm{2}{a}^{\mathrm{2}} +\mathrm{5}\geqslant{a}^{\mathrm{2}} +\mathrm{8}{a}−\mathrm{10} \\ $$$${a}^{\mathrm{2}} −\mathrm{8}{a}+\mathrm{15}\geqslant\mathrm{0} \\ $$$$\left({a}−\mathrm{3}\right)\left({a}−\mathrm{5}\right)\geqslant\mathrm{0} \\ $$$$\Rightarrow{a}\leqslant\mathrm{3},\:{a}\geqslant\mathrm{5}\:\:\:\:…\left({ii}\right) \\ $$$$ \\ $$$$\mathrm{2}{a}^{\mathrm{2}} +\mathrm{7}\geqslant{a}^{\mathrm{2}} +\mathrm{9}{a}−\mathrm{11} \\ $$$${a}^{\mathrm{2}} −\mathrm{9}{a}+\mathrm{18}\geqslant\mathrm{0} \\ $$$$\left({a}−\mathrm{3}\right)\left({a}−\mathrm{6}\right)\geqslant\mathrm{0} \\ $$$$\Rightarrow{a}\leqslant\mathrm{3},\:{a}\geqslant\mathrm{6}\:\:\:\:…\left({iii}\right) \\ $$$$ \\ $$$${summary}: \\ $$$$\frac{\sqrt{\mathrm{457}}−\mathrm{17}}{\mathrm{4}}\leqslant{a}\leqslant\mathrm{3}\:{or}\:{a}\geqslant\mathrm{6}\:\:\checkmark \\ $$
Commented by hardmath last updated on 03/Dec/25
thankyou cool dear professor, asnwer:3
$$\mathrm{thankyou}\:\mathrm{cool}\:\mathrm{dear}\:\mathrm{professor},\:\mathrm{asnwer}:\mathrm{3} \\ $$
Commented by mr W last updated on 03/Dec/25
answer is: (((√(457))−17)/4)≤a≤3 or a≥6 a=3 is only one of infinite many possbilities.
$${answer}\:{is}: \\ $$$$\frac{\sqrt{\mathrm{457}}−\mathrm{17}}{\mathrm{4}}\leqslant{a}\leqslant\mathrm{3}\:{or}\:{a}\geqslant\mathrm{6} \\ $$$${a}=\mathrm{3}\:{is}\:{only}\:{one}\:{of}\:{infinite}\:{many}\: \\ $$$${possbilities}. \\ $$
Commented by hardmath last updated on 03/Dec/25
thank you dear professor
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$

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