Question Number 49461 by Pk1167156@gmail.com last updated on 07/Dec/18
$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{solutions}\:\mathrm{for}\:{x}\:\mathrm{from} \\ $$$$\mathrm{the}\:\mathrm{equation}\:{x}^{\mathrm{2}} −\mid{x}\mid−\mathrm{2}=\mathrm{0}\:\:\mathrm{is} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 07/Dec/18
![∣x∣=x x>0 =−x x<0 x^2 −x−2=0 [considered x>0] x^2 −2x+x−2=0 x(x−2)+1(x−2)=0 (x−2)(x+1)=0 x=2 but we can not consider x=−1 because already cosidered x>0 now consider x<0 x^2 +x−2=0 x^2 +2x−x−2=0 x(x+2)−1(x+2)=0 (x+2)(x−1)=0 x=−2 but we can not consider x=1 because we already considerd x<0 solution is x=±2](https://www.tinkutara.com/question/Q49479.png)
$$\mid{x}\mid={x}\:\:{x}>\mathrm{0} \\ $$$$\:\:\:\:\:=−{x}\:{x}<\mathrm{0} \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{2}=\mathrm{0}\:\:\:\:\left[{considered}\:{x}>\mathrm{0}\right] \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{x}+{x}−\mathrm{2}=\mathrm{0} \\ $$$${x}\left({x}−\mathrm{2}\right)+\mathrm{1}\left({x}−\mathrm{2}\right)=\mathrm{0} \\ $$$$\left({x}−\mathrm{2}\right)\left({x}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}=\mathrm{2}\:\:{but}\:{we}\:{can}\:{not}\:{consider}\:{x}=−\mathrm{1} \\ $$$${because}\:{already}\:{cosidered}\:{x}>\mathrm{0} \\ $$$${now}\:{consider}\:{x}<\mathrm{0} \\ $$$${x}^{\mathrm{2}} +{x}−\mathrm{2}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} +\mathrm{2}{x}−{x}−\mathrm{2}=\mathrm{0} \\ $$$${x}\left({x}+\mathrm{2}\right)−\mathrm{1}\left({x}+\mathrm{2}\right)=\mathrm{0} \\ $$$$\left({x}+\mathrm{2}\right)\left({x}−\mathrm{1}\right)=\mathrm{0} \\ $$$${x}=−\mathrm{2}\:{but}\:{we}\:{can}\:{not}\:{consider}\:{x}=\mathrm{1}\:{because} \\ $$$${we}\:{already}\:{considerd}\:{x}<\mathrm{0} \\ $$$${solution}\:{is}\:{x}=\pm\mathrm{2} \\ $$
Answered by mr W last updated on 07/Dec/18
$${let}\:{t}=\mid{x}\mid\geqslant\mathrm{0} \\ $$$${x}^{\mathrm{2}} =\mid{x}\mid^{\mathrm{2}} \\ $$$$\Rightarrow{t}^{\mathrm{2}} −{t}−\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\left({t}+\mathrm{1}\right)\left({t}−\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow{t}=−\mathrm{1}\:\left(<\mathrm{0}\Rightarrow{not}\:{ok}\right)\:{or}\:{t}=\mathrm{2}\:\left(>\mathrm{0}\:\Rightarrow{ok}\right) \\ $$$$\Rightarrow{x}=\pm\mathrm{2} \\ $$$$\Rightarrow\mathrm{2}\:{solutions}! \\ $$