What is (x-2)/(x^2-3x-10)<= 0 ?

The solution to (x-2)/(x^2-3x-10)<= 0 is x<-2\lor 2<= x<5

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(x-2)/(x^2-3x-10)<= 0

\frac{x - 2}{x ^{2} - 3 x - 10} \leq 0

x < - 2 or 2 \leq x < 5

Interval Notation

(- \infty, - 2) \cup [2, 5)

Solution steps

\frac{x - 2}{x ^{2} - 3 x - 10} \leq 0

Factor

\frac{x - 2}{x ^{2} - 3 x - 10} : \frac{x - 2}{( x + 2 ) ( x - 5 )}

\frac{x - 2}{( x + 2 ) ( x - 5 )} \leq 0

Identify the intervals

x < - 2 or 2 \leq x < 5

8^{2} + 2^{2} = x^{2}

\frac{\frac{z ^{2} + 20 z + 96}{2 z + 3}}{( z + 8 ) ^{2}}

125 + (x - y)^{3}

13 \geq 2 x - 3 \geq 5

y^{3} - 5 y^{2} + 6 y - 30

What is (x-2)/(x^2-3x-10)<= 0 ?
The solution to (x-2)/(x^2-3x-10)<= 0 is x<-2\lor 2<= x<5