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Question Number 88873 by jagoll last updated on 13/Apr/20

3+(√(x^2 −5)) > ∣x−1∣

3+x2−5>∣x−1∣

Commented byjohn santu last updated on 13/Apr/20

⇒(3+(√(x^2 −5)))^2 > (x−1)^2 6(√(x^2 −5)) +4+x^2 > x^2 −2x+1 6(√(x^2 −5)) > −2x−3 (i) −2x−3 < 0 ∧ x^2 −5 ≥ 0 ⇒x > −(3/2) ∧ x≤−(√5) ∨ x≥ (√5) we get x ≥ (√5) (ii)−2x−3 > 0 ⇒36(x^2 −5) > 4x^2 +12x+9 ⇒x < −(9/4) the solution : x <−(9/4) ∨ x ≥(√5)

⇒(3+x2−5)2>(x−1)2 6x2−5+4+x2>x2−2x+1 6x2−5>−2x−3 (i)−2x−3<0∧x2−5⩾0 ⇒x>−32∧x⩽−5∨x⩾5 wegetx⩾5 (ii)−2x−3>0⇒36(x2−5)> 4x2+12x+9⇒x<−94 thesolution:x<−94∨x⩾5

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