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Question Number 88873 by jagoll last updated on 13/Apr/20
3+x2−5>∣x−1∣
Commented by john santu last updated on 13/Apr/20
⇒(3+x2−5)2>(x−1)26x2−5+4+x2>x2−2x+16x2−5>−2x−3(i)−2x−3<0∧x2−5⩾0⇒x>−32∧x⩽−5∨x⩾5wegetx⩾5(ii)−2x−3>0⇒36(x2−5)>4x2+12x+9⇒x<−94thesolution:x<−94∨x⩾5
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