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4-2.Quadratic Equations and Inequations
hard
If $|{x^2} - x - 6| = x + 2$, then the values of $x$ are
A
$-2, 2, -4$
B
$-2, 2, 4$
C
$3, 2, -2$
D
$4, 4, 3$
Solution
(b) $|{x^2} – x – 6| = x + 2$, then
Case $I$ : ${x^2} – x – 6 < 0$
==> $(x – 3)(x + 2) < 0$
==> $ – 2 < x < 3$
In this case, the equation becomes
${x^2} – x – 6 = – x – 2$or ${x^2} – 4 = 0\,\,\, \Rightarrow x = \pm 2$
Clearly $x = 2$ satisfies the domain of the equation in this case. So $x = 2$ is a solution.
Case $II$ : ${x^2} – x – 6 \ge 0.$So $x \le – 2$or $x \ge 3$
Then equation reduces to
${x^2} – x – 6 = 0 = x + 2$
i.e. ${x^2} – 2x – 8 = 0$or $x = – 2,\,4$
Both these values lies in the domain of the equation in this case, so $x = – 2,\,4$ are the roots.
Hence roots are $x = – 2,\,2,\,4$.
Standard 11
Mathematics
