A) One and only one real number
B) Real with sum one
C) Real with sum zero
D) Real with product zero
Correct Answer: C
Solution :
When\[x<0\], \[|x|=-x\] \ Equation is \[{{x}^{2}}-x-6=0\Rightarrow x=-2,\,3\] \[\because \ x<0,\ \therefore \ x=-2\] is the solution. When\[x\ge 0\],\[|x|=x\] \[\therefore \] Equation is\[{{x}^{2}}+x-6=0\Rightarrow x=2,-3\] \[\because \] \[x\ge 0\], \ \[x=2\] is the solution. Hence \[x=2\], \[-2\] are the solutions and their sum is zero. Aliter: \[|{{x}^{2}}|+|x|-6=0\] Þ \[(|x|+3)(|x|-2)=0\] Þ \[|x|=-3\], which is not possible and \[|x|=2\] Þ \[x=\pm 2\].
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