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UPSC NDA I 2019 Mathematics Solved Paper

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Question : 38 of 120

Marks: +1, -0

What are the roots of the equation

|x^{2}-x-6|=x+2 ?

−2,1,4
0,2,4
0,1,4
−2,2,4

Solution: 👈: Video Solution

Given:

\;|x^{2}-x-6|=x+2

\; \text{ We can factorize }\; x^{2}-x-6 \;\text{ as }\;(x-3)(x+2)

\Rightarrow|(x-3)(x+2)|=\begin{cases} (x-3)(x+2), & \text{ if } (x-3)(x+2) \geq 0 \\ -(x-3)(x+2), & \text{ if } (x-3)(x+2) < 0 \end{cases}

\; \text{ Case } -1: \; \text{ If } \;(x-3)(x+2) \geq 0

\Rightarrow x \in (-\infty,-2] \cup [3,\infty)

\Rightarrow x^{2}-x-6=x+2

\Rightarrow x^{2}-2x-8=0

\Rightarrow (x-4)(x+2)=0

\Rightarrow x=4 \; \text{ or } \; -2 \in (-\infty,-2] \cup [3,\infty)

\; \text{ So, the roots of the given quadratic equation are }\; 4 \;\text{ and }\; -2 \;\text{. }\;

\; \text{ Case } -2 \; \text{ : If } \;(x-3)(x+2) < 0

\Rightarrow x \in [-2,3]

\Rightarrow -(x^{2}-x-6)=x+2

\Rightarrow x^{2}-4=0

\Rightarrow x=2 \; \text{ or } \; -2 \in [-2,3]

\; \text{ So, the roots of the given quadratic equation are }\; 2 \;\text{ and }\; -2

\; \text{ Hence, the roots of the given quadratic equation are }\; -2,2 \;\text{ and }\; 4 .

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