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Question Number 206251 by TonyCWX08 last updated on 10/Apr/24

3x^2 −12x−5(√(x^2 −4x−1))−5=0

3x2−12x−5x2−4x−1−5=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

3(x2−4x−1)−2−5x2−4x−1=0Letx2−4x−1=p;3p2−5p−2=0⇒p=2or−13⇒x2−4x−5=0⇒x=5or−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.

x2−4x−1=p⇒p⩾0Iseeyoudidn′tusep=−13anywaybutIthinkthisshouldbenoticed.

Commented by A5T last updated on 10/Apr/24

Yea, that was the reason.

Yea,thatwasthereason.

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