• # question_answer 2) For the equation $|{{x}^{2}}|+|x|-6=0$, the roots are [EAMCET 1988, 93] A) One and only one real number B) Real with sum one C) Real with sum zero D) Real with product zero

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 $x=\pm 2$.