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

The solution to (x^2-x-2)/(x^2-4x+3)>= 0 is x 3

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

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

x \leq - 1 or 1 < x \leq 2 or x > 3

Solution steps

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

Factor

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

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

Identify the intervals

x \leq - 1 or 1 < x \leq 2 or x > 3

x^{4} - 6 x^{3} + 9 x^{2} + 4 x - 12

5 y > 25

2.5 (10 x - 16) = 5

4 - \frac{5}{4 m + 8}

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

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