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Question Number 115695 by ZiYangLee last updated on 27/Sep/20

The product of all the solutions of the equation ∣(x−2)^2 ∣−3∣x−2∣+2=0 is

$$\mathrm{The}\:\mathrm{product}\:\mathrm{of}\:\mathrm{all}\:\mathrm{the}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\:\mid\left({x}−\mathrm{2}\right)^{\mathrm{2}} \mid−\mathrm{3}\mid{x}−\mathrm{2}\mid+\mathrm{2}=\mathrm{0}\:\:\mathrm{is} \\ $$

Answered by MJS_new last updated on 27/Sep/20

x=0 is a solution ⇒ answer is 0

$${x}=\mathrm{0}\:\mathrm{is}\:\mathrm{a}\:\mathrm{solution}\:\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\mathrm{0} \\ $$

Answered by mathmax by abdo last updated on 27/Sep/20

e⇒u^2 −3u+2 =0 with u=∣x−2∣(so u≥0) Δ=9−8=1 ⇒u_1 =((3+1)/2) =2 and u_2 =((3−1)/2) =1 ∣x−2∣=2 ⇒x−2=2 or x−2 =−2 ⇒x=4 or x=0 ∣x−2∣=1 ⇒x−2=1 or x−2 =−1 ⇒x=3 or x=1 ⇒Π(of roots) =0

$$\mathrm{e}\Rightarrow\mathrm{u}^{\mathrm{2}} −\mathrm{3u}+\mathrm{2}\:=\mathrm{0}\:\mathrm{with}\:\mathrm{u}=\mid\mathrm{x}−\mathrm{2}\mid\left(\mathrm{so}\:\mathrm{u}\geqslant\mathrm{0}\right) \\ $$$$\Delta=\mathrm{9}−\mathrm{8}=\mathrm{1}\:\Rightarrow\mathrm{u}_{\mathrm{1}} \:=\frac{\mathrm{3}+\mathrm{1}}{\mathrm{2}}\:=\mathrm{2}\:\:\mathrm{and}\:\mathrm{u}_{\mathrm{2}} =\frac{\mathrm{3}−\mathrm{1}}{\mathrm{2}}\:=\mathrm{1} \\ $$$$\mid\mathrm{x}−\mathrm{2}\mid=\mathrm{2}\:\Rightarrow\mathrm{x}−\mathrm{2}=\mathrm{2}\:\mathrm{or}\:\mathrm{x}−\mathrm{2}\:=−\mathrm{2}\:\Rightarrow\mathrm{x}=\mathrm{4}\:\mathrm{or}\:\mathrm{x}=\mathrm{0} \\ $$$$\mid\mathrm{x}−\mathrm{2}\mid=\mathrm{1}\:\Rightarrow\mathrm{x}−\mathrm{2}=\mathrm{1}\:\mathrm{or}\:\mathrm{x}−\mathrm{2}\:=−\mathrm{1}\:\Rightarrow\mathrm{x}=\mathrm{3}\:\mathrm{or}\:\mathrm{x}=\mathrm{1} \\ $$$$\Rightarrow\Pi\left(\mathrm{of}\:\mathrm{roots}\right)\:=\mathrm{0} \\ $$

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