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NDA Math Quadratic Equations and Inequations

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Question : 33 of 67

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:

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) \ge 0 \\ -(x-3)(x+2), & \text{if } (x-3)(x+2) < 0 \end{cases}.

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