MyQuestionIcon

1

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

Surds

solve: X2 +...

Question

solve: $(X^{2} + 3 X) (2 X + 3) - 16 \frac{2 X + 3}{X^{2} + 3 X} \geq 0$

Open in App

Solution

$(X^{2} + 3 X) (2 X + 3) - 16 \frac{(2 X + 3)}{(X^{2} + 3 X)} \geq 0$

$(X^{2} + 3 X) (2 X + 3) \geq 16 \frac{(2 X + 3)}{(X^{2} + 3 X)}$

$(X^{2} + 3 X) \geq 16 \frac{1}{(X^{2} + 3 X)}$

$(X^{2} + 3 X)^{2} \geq 16$

$(X^{2} + 3 X) \geq \sqrt{16}, (X^{2} + 3 X) \leq \sqrt{16}$

$(X^{2} + 3 X) \geq 4, (X^{2} + 3 X) \leq 4$

$X \geq 1, X \leq - 4$

Suggest Corrections

thumbs-up

0

Similar questions

(x^{2} + 3 x) (2 x + 3) - 16 \frac{2 x + 3}{x^{2} + 3 x} \geq 0

(x^{2} + 3 x) (2 x + 3) - 16 \frac{2 x + 3}{x^{2} + 3 x}

Q. Solve the following inequality:

(x^{2} + 3 x) (2 x + 3) - 16 \frac{2 x + 3}{x^{2} + 3 x} ⩾ 0

Q. Solve the following inequality:

\frac{1 - 2 x - 3 x^{2}}{3 x - x^{2} - 5} > 0

Q. Solve following equation:

\sqrt{3} x^{2} - 2 \sqrt{3} x + 3 \sqrt{3} = 0

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Surds and Rationalization of Surds

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Standard IX Mathematics

Join BYJU'S Learning Program

CrossIcon