Find the values of x for which $x^{2} - x - 2 > 0$ is true

(- \infty, - 1) \cup (2, \infty)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(- \infty, - 2) \cup (5, \infty)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(- \infty, - 10) \cup (, - 6 \infty)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

none of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $(- \infty, - 1) \cup (2, \infty)$
$x^{2} - x - 2 > 0$
$(x - 2) (x + 1) > 0$
Therefore $x < - 1$ and $x > 2$
The solution set is $(- \infty, - 1) \cup (2, \infty)$

Suggest Corrections

Similar questions

x

{log}_{0.2} (\frac{x + 2}{x}) \leq 1

Find the interval of $x$ such that $∣ ∣ \frac{x^{2} - 3 x - 1}{x^{2} + x + 1} ∣ ∣ < 3$ is satisfied is

$6 x^{2} + 17 x + 5$ =?

(a) $(2 x + 1)$

(b) $(2 x + 5) (3 x + 1)$

(d) None of these

The value of the integral $\int (x^{2} + x) (x^{- 8} + 2 x^{- 9})^{1 / 10} d x$ is

x

x^{2} - | x | - 2 \geq 0

Join BYJU'S Learning Program