question_answer

The roots of the equation \[|{{x}^{2}}-x-6|=x+2\]are

A) \[-2,1,4\]            

B)        \[0,2,4\]

C)  \[0,1,4\]             

D)         \[-2,2,4\]

Correct Answer: D

Solution :

We have \[|{{x}^{2}}-x-6|=x+2\]                 ?(i) or            \[|(x-3)(x+2)|=x+2\] Case I: When \[-2\le x\]or            \[x\ge 3\] From Eq. (i) \[{{x}^{2}}-x-6=x+2\] \[\Rightarrow \]\[{{x}^{2}}-2x-8=0\] \[\Rightarrow \]\[(x-4)(x+2)=0\Rightarrow x=-2,4\] Case II: When \[-2<x<3\] From Eq. (i) \[-({{x}^{2}}-x-6)=x+2\] \[\Rightarrow \]               \[-{{x}^{2}}+x+6=x+2\] \[\Rightarrow \]               \[-{{x}^{2}}=-4\Rightarrow x=\pm \,2\] \[\Rightarrow \]\[x=2\](\[\because \,x=-2\]not lies in the interval) Hence, the roots are \[\{-2,\text{ }2,\text{ }4\}\] Alternate Solution:         Case I: When\[-2\le x\]or \[x\ge 3\] Let          \[y={{x}^{2}}-x-6=x+2\] Now,     \[y={{x}^{2}}-x-6\] \[\left( {{x}^{2}}-x+\frac{1}{4} \right)=y+6+\frac{1}{4}\]                                 \[{{\left( x-\frac{1}{2} \right)}^{2}}=y+\frac{25}{4}\]                 and        \[y=x+2\]                 \[\Rightarrow \]               \[\frac{y}{2}-\frac{x}{2}=1\] Case II: When \[-2<x<3\] Let   \[y=-({{x}^{2}}-x-6)=x+2\] Now,       \[y=-({{x}^{2}}-x-6)\]                 \[-\left( {{x}^{2}}-x+\frac{1}{4} \right)=y-6-\frac{1}{4}\]                 \[{{\left( x-\frac{1}{2} \right)}^{2}}=-\left( y-\frac{25}{4} \right)\]                 and        \[y=x+2\] It is clear from die above two figures that the curve will intersect the line^ in three different points i.e., \[\{-2,\text{ }2,\text{ }4\}\].