What is (x^2-3x+2)/(4-x^2)<= 0 ?

The solution to (x^2-3x+2)/(4-x^2) = 1

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(x^2-3x+2)/(4-x^2)<= 0

\frac{x ^{2} - 3 x + 2}{4 - x ^{2}} \leq 0

x < - 2 or x \geq 1

Interval Notation

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

Solution steps

\frac{x ^{2} - 3 x + 2}{4 - x ^{2}} \leq 0

Factor

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

\frac{( x - 1 ) ( x - 2 )}{- ( x + 2 ) ( x - 2 )} \leq 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( x - 1 ) ( x - 2 ) ( - 1 )}{- ( x + 2 ) ( x - 2 )} \geq 0 \cdot (- 1)

Simplify

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

\frac{x - 1}{x + 2} \geq 0

Identify the intervals

x < - 2 or x \geq 1

\frac{x + 3}{8} - \frac{x + 5}{5} \geq \frac{3}{10}

\frac{9}{i}

\frac{1 - ( 1 + 0.055 ) ^{- 20}}{0.055}

16 = x^{2}

2 x + 3 = - 1

What is (x^2-3x+2)/(4-x^2)<= 0 ?
The solution to (x^2-3x+2)/(4-x^2)<= 0 is x<-2\lor x>= 1