MyQuestionIcon

1

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

Inequalities Involving Modulus Function

Solve the fol...

Question

Solve the following inequality:
$\sqrt{x^{2} - 3 x + 2} > 2 x - 5.$

Open in App

Solution

$x^{2} - 3 x + 2 \geq 0$
$(x - 2) (x - 1) \geq 0$
$x ϵ (- \infty, 1] \cup [2, \infty)$
$\sqrt{x^{2} - 3 x + 2} > 2 x - 5$
Squarring both sides
$x^{2} - 3 x + 2 > 4 x^{2} - 20 x + 25$
$3 x^{2} - 17 x + 23 < 0$
As discriminant less than $0$ So, $3 x^{2} - 17 x + 23 > 0 \forall x ϵ R$
$x ϵ (- \infty, - 1] \cup [2, \infty)$

Suggest Corrections

thumbs-up

0

Similar questions

Q. Solve the following inequality:

\frac{3 x - 5}{x^{2} + 4 x - 5} > \frac{1}{2}

Q. Solve the following inequality:

\frac{5 - 2 x}{3 x^{2} - 2 x - 16} < 1

Q. Solve the following inequality:

\sqrt{- x^{2} + 6 x - 5} > 8 - 2 x .

Q. Solve the following inequality:

\frac{3 x^{2} - 4 x - 6}{2 x - 5} < 0

Q. Solve the following inequality:

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

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Modulus

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Inequalities Involving Modulus Function

Standard IX Mathematics

Join BYJU'S Learning Program

CrossIcon