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3x-2-12x-5-x-2-4x-1-5-0-




Question Number 206251 by TonyCWX08 last updated on 10/Apr/24
3x^2 −12x−5(√(x^2 −4x−1))−5=0
$$\mathrm{3}{x}^{\mathrm{2}} −\mathrm{12}{x}−\mathrm{5}\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{1}}−\mathrm{5}=\mathrm{0} \\ $$
Answered by A5T last updated on 10/Apr/24
3(x^2 −4x−1)−2−5(√(x^2 −4x−1))=0  Let (√(x^2 −4x−1))=p; 3p^2 −5p−2=0⇒p=2 or ((−1)/3)  ⇒x^2 −4x−5=0 ⇒x=5 or −1
$$\mathrm{3}\left({x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{1}\right)−\mathrm{2}−\mathrm{5}\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{1}}=\mathrm{0} \\ $$$${Let}\:\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{1}}={p};\:\mathrm{3}{p}^{\mathrm{2}} −\mathrm{5}{p}−\mathrm{2}=\mathrm{0}\Rightarrow{p}=\mathrm{2}\:{or}\:\frac{−\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{5}=\mathrm{0}\:\Rightarrow{x}=\mathrm{5}\:{or}\:−\mathrm{1} \\ $$
Commented by Frix last updated on 10/Apr/24
(√(x^2 −4x−1))=p ⇒ p≥0  I see you didn′t use p=−(1/3) anyway but I  think this should be noticed.
$$\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{1}}={p}\:\Rightarrow\:{p}\geqslant\mathrm{0} \\ $$$$\mathrm{I}\:\mathrm{see}\:\mathrm{you}\:\mathrm{didn}'\mathrm{t}\:\mathrm{use}\:{p}=−\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{anyway}\:\mathrm{but}\:\mathrm{I} \\ $$$$\mathrm{think}\:\mathrm{this}\:\mathrm{should}\:\mathrm{be}\:\mathrm{noticed}. \\ $$
Commented by A5T last updated on 10/Apr/24
Yea, that was the reason.
$${Yea},\:{that}\:{was}\:{the}\:{reason}. \\ $$

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