Q162 / 270 · NDA 2017 (Phase 2) · Mathematics

The sum of all real roots of the equation $|\mathrm{x}-3|^2+|\mathrm{x}-3|-2=0$ is

  • A
    $2$
  • B
    $3$
  • C
    $4$
  • Correct answer
    D
    $6$

Solution

$\begin{aligned} & |\mathrm{x}-3|^2+|\mathrm{x}-3|-2=0 \\ & \text {Let }|\mathrm{x}-3|=\mathrm{t} \\ & \therefore \mathrm{t}^2+\mathrm{t}-2=0 \Rightarrow \mathrm{t}^2+2 \mathrm{t}-\mathrm{t}-2=0\end{aligned}$ $\begin{aligned} & \Rightarrow \mathrm{t}(\mathrm{t}+2)-1(\mathrm{t}+2)=0 \\ & \Rightarrow(\mathrm{t}+2)(\mathrm{t}-1)=0 \\ & \Rightarrow \mathrm{t}=-2 \text { or } \mathrm{t}=1\end{aligned}$ Since $t$ is modulus of a number, it cannot be negative. $\begin{aligned}& \therefore \mathrm{t}=1 \Rightarrow|\mathrm{x}-3|=1 \Rightarrow \mathrm{x}-3=1 \text { or } \mathrm{x}-3=-1 \\& \Rightarrow \mathrm{x}=4 \text { or } 2 \end{aligned}$ Sum of roots $=4+2=6$.

Hence, the correct answer is option: 'D'.

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