First slide

Theory of equations

Question

The roots of the equation $|x^{2} - x - 6| = x + 2$ are

Moderate

A

$- 2, 1, 4$
B

$0, 2, 4$
C

$0, 1, 4$
D

$- 2, 2, 4$

Solution

We have,
$|x^{2} - x - 6| = \{\begin{aligned} x^{2} - x - 6, if x \leq - 2 or x \geq 3 \\ - (x^{2} - x - 6), if - 2 < x < 3 \end{aligned}$
CASE I When $x \leq - 2$ or , $x > 3$
In this case, we have $|x^{2} - x - 6| = x^{2} - x - 6$
$\begin{array}{l} ∴ |x^{2} - x - 6| = x + 2 \\ \Rightarrow x^{2} - x - 6 = x + 2 \\ \Rightarrow x^{2} - 2 x - 8 = 0 \\ \Rightarrow (x - 4) (x + 2) = 0 . \\ \Rightarrow x = - 2, 4 \end{array}$
CASE II When $- 2 < x < 3$
In this case, we have $|x^{2} - x - 6| = - (x^{2} - x - 6)$
$|x^{2} - x - 6| = x + 2$
$\Rightarrow - (x^{2} - x - 6) = x + 2$
$\begin{array}{l} \Rightarrow x^{2} - 4 = 0 \\ \Rightarrow x = \pm 2 \\ \Rightarrow x = 2 \end{array}$
Hence, the roots are -2, 2, 4.

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App