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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 ∣
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{integers} \\ $$$$\mathrm{satisfying}\:\mid\mid\mathrm{x}−\mathrm{5}\mid−\mathrm{21}\mid\:=\:\mid\:\mathrm{x}^{\mathrm{2}} −\mathrm{10x}+\mathrm{4}\:\mid \\ $$
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
$$\mid\mid{x}−\mathrm{5}\mid−\mathrm{21}\mid=\mid\mid{x}−\mathrm{5}\mid^{\mathrm{2}} −\mathrm{21}\mid \\ $$$${let}\:{t}=\mid{x}−\mathrm{5}\mid\geqslant\mathrm{0} \\ $$$$\mid{t}−\mathrm{21}\mid=\mid{t}^{\mathrm{2}} −\mathrm{21}\mid \\ $$$${t}^{\mathrm{2}} −\mathrm{21}=\pm\left({t}−\mathrm{21}\right) \\ $$$${t}^{\mathrm{2}} −\mathrm{21}={t}−\mathrm{21}\:\Rightarrow{t}\left({t}−\mathrm{1}\right)=\mathrm{0}\:\Rightarrow{t}=\mathrm{0},\:\mathrm{1} \\ $$$$\Rightarrow\mid{x}−\mathrm{5}\mid=\mathrm{0},\:\mathrm{1}\:\Rightarrow{x}=\mathrm{4},\mathrm{5},\mathrm{6} \\ $$$${t}^{\mathrm{2}} −\mathrm{21}=−{t}+\mathrm{21}\:\Rightarrow{t}^{\mathrm{2}} +{t}−\mathrm{42}=\mathrm{0} \\ $$$$\Rightarrow\left({t}−\mathrm{6}\right)\left({t}+\mathrm{7}\right)=\mathrm{0}\:\Rightarrow{t}=\mathrm{6},−\mathrm{7}\left({rejected}\right) \\ $$$$\Rightarrow\mid{x}−\mathrm{5}\mid=\mathrm{6}\:\Rightarrow{x}=\mathrm{11},−\mathrm{1} \\ $$$${i}.{e}.\:{there}\:{are}\:\mathrm{5}\:{integers}: \\ $$$${x}=−\mathrm{1},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{11} \\ $$

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