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Question Number 134169 by liberty last updated on 28/Feb/21
Find the number of integers satisfying ∣∣x−5∣−21∣ = ∣ x^2 −10x+4 ∣
Findthenumberofintegerssatisfying∣∣x−5∣−21∣=∣x2−10x+4∣
Answered by mr W last updated on 28/Feb/21
∣∣x−5∣−21∣=∣∣x−5∣^2 −21∣ let t=∣x−5∣≥0 ∣t−21∣=∣t^2 −21∣ t^2 −21=±(t−21) t^2 −21=t−21 ⇒t(t−1)=0 ⇒t=0, 1 ⇒∣x−5∣=0, 1 ⇒x=4,5,6 t^2 −21=−t+21 ⇒t^2 +t−42=0 ⇒(t−6)(t+7)=0 ⇒t=6,−7(rejected) ⇒∣x−5∣=6 ⇒x=11,−1 i.e. there are 5 integers: x=−1,4,5,6,11
∣∣x−5∣−21∣=∣∣x−5∣2−21∣lett=∣x−5∣⩾0∣t−21∣=∣t2−21∣t2−21=±(t−21)t2−21=t−21⇒t(t−1)=0⇒t=0,1⇒∣x−5∣=0,1⇒x=4,5,6t2−21=−t+21⇒t2+t−42=0⇒(t−6)(t+7)=0⇒t=6,−7(rejected)⇒∣x−5∣=6⇒x=11,−1i.e.thereare5integers:x=−1,4,5,6,11

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