CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Critical Point

Solve the fol...

Question

Solve the following inequality:
$\frac{2 x}{x^{2} - 9} ⩽ \frac{1}{x + 2}$

Open in App

Solution

$\frac{2 x}{x^{2} - 9} \leq \frac{1}{x + 2} \frac{2 x}{x^{2} - 9} - \frac{1}{x + 2} \leq 0 \frac{2 x^{2} + 4 x - x^{2} + 9}{(x + 2) (x^{2} - 9)} \leq 0 \frac{2 x^{2} + 4 x - x^{2} + 9}{(x + 2) (x - 3) (x + 3)} \leq 0$
As discriminant of $x^{2} + 4 x + 9 < 0$ for every $x \in R, x^{2} + 4 x + 9 > 0 \forall x \in R$
$\frac{1}{(x + 2) (x - 3) (x + 3)} \leq 0 x \in (- \infty, - 3) \cup (- 2, 3)$

Suggest Corrections

thumbs-up

0

Similar questions

Q. Solve the following inequality:

\frac{2 x + 3}{x^{2} + x - 12} ⩽ \frac{1}{2}

Q. Solve the following inequality:

(x^{2} - 2 x) (2 x - 2) - 9 \frac{2 x - 2}{x^{2} - 2 x} ⩽ 0

Q. Solve the following inequality:

\frac{(x + 2) (x^{2} - 2 x + 1)}{4 + 3 x - x^{2}} ⩾ 0

Q. Solve the following inequality:

\sqrt{- x^{2} + x + 2} + 2 x + 1 > 0.

Q. Solve the following inequality:

1 - x < \sqrt{x^{2} - 2 x} .

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Extrema

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon