What is ((x^2+2x-2))/(3+2x-x^2)<x ?

The solution to ((x^2+2x-2))/(3+2x-x^2) 3

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((x^2+2x-2))/(3+2x-x^2)<x

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

- 1 < x < 2 or x > 3

Interval Notation

(- 1, 2) \cup (3, \infty)

Solution steps

\frac{x ^{2} + 2 x - 2}{3 + 2 x - x ^{2}} < x

Rewrite in standard form

\frac{x ^{3} - x ^{2} - x - 2}{3 + 2 x - x ^{2}} < 0

Factor

\frac{x ^{3} - x ^{2} - x - 2}{3 + 2 x - x ^{2}} : \frac{( x - 2 ) ( x ^{2} + x + 1 )}{- ( x + 1 ) ( x - 3 )}

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

Multiply both sides by

- 1

(reverse the inequality)

\frac{( x - 2 ) ( x ^{2} + x + 1 ) ( - 1 )}{- ( x + 1 ) ( x - 3 )} > 0 \cdot (- 1)

Simplify

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

Identify the intervals

- 1 < x < 2 or x > 3

(3^{7})^{4}

∣ 4 x + 3 ∣ = 8

- 32 = w - 12

\frac{( \frac{\frac{2 x t}{3}}{\frac{x}{4 t}} )}{\frac{2 t}{3}}

\frac{1}{6} - \frac{1}{36}

What is ((x^2+2x-2))/(3+2x-x^2)<x ?
The solution to ((x^2+2x-2))/(3+2x-x^2)<x is -1<x<2\lor x>3