What is (x^2+1)/(x^2-3x+2)<= 0 ?

The solution to (x^2+1)/(x^2-3x+2)<= 0 is 1<x<2

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(x^2+1)/(x^2-3x+2)<= 0

\frac{x ^{2} + 1}{x ^{2} - 3 x + 2} \leq 0

1 < x < 2

Interval Notation

(1, 2)

Solution steps

\frac{x ^{2} + 1}{x ^{2} - 3 x + 2} \leq 0

Factor

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

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

Identify the intervals

1 < x < 2

\frac{3 ( - 2 + 1 )}{2} - \frac{- 6 ( 1 - 3 )}{3 - ( - 5 )}

12 x^{2} + 36 = 0

4 \times 6

7 \cdot 7

∣ 2 x - 5 ∣ > 13

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