Answer to Question #223089 in Algebra for Sanamtha

Question

Solve the inequality

3x-5/x+2 ≥ x-3

Answer 1

$3x-\frac{5}x+2 ≥ x-3$

Determine the defined range

$3x-\frac{5}x+2 ≥ x-3, x≠0$

Move the expression to the left hand side and change its sign

$3x-\frac{5}x+2 ≥ x-3$

Calculate

$3x-\frac{5}x+5-x ≥0$

Write all numerators above the common denominator

$\frac{3x^2-{5}+5x-x^2 }{x}≥ 0$

$\frac{2x^2-{5}+5x }{x}≥ 0$

Separate into possible cases

$2x^2-5+5x≥0$

$x>0$

$2x^2-5+5x≤0$

$x<0$

Solve the inequality for x

$x\in(-∞, \frac{-5-\sqrt{65}}{4}) ∪ (\frac{-5+\sqrt{65}}{4},+∞)$

$x>0$

$x\in(\frac{-5-\sqrt{65}}{4} ,\frac{-5+\sqrt{65}}{4})$

$x<0$

$x\in(\frac{-5+\sqrt{65}}{4}, +∞)$

$x\in(\frac{-5+\sqrt{65}}{4}, 0)$

$x\in(\frac{-5-\sqrt{65}}{4}, 0) ∪ (\frac{-5+\sqrt{65}}{4}, +∞),x≠0$

Find the intersection of the solution and the defined range

$x\in(\frac{-5-\sqrt{65}}{4}, 0) ∪ (\frac{-5+\sqrt{65}}{4}, +∞)$