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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Basics of Mathematics

NDA 2017 · Mathematics · Basics of Mathematics

The sum of all real roots of the equation |x-3|^2+|x-3|-2=0 is

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

Options

  1. A22
  2. B33
  3. C44
  4. D66

Correct answer

D. 66

Step-by-step solution

∣x−3∣2+∣x−3∣−2=0Let ∣x−3∣=t∴t2+t−2=0⇒t2+2t−t−2=0\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}⇒t(t+2)−1(t+2)=0⇒(t+2)(t−1)=0⇒t=−2 or t=1\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}SinceSincetismodulusofanumber,itcannotbenegative.is modulus of a number, it cannot be negative.∴t=1⇒∣x−3∣=1⇒x−3=1 or x−3=−1&⇒x=4 or 2\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}SumofrootsSum of roots=4+2=6$.
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