What is (x^2-3x-4)/(x^2-4x-5)<0 ?

The solution to (x^2-3x-4)/(x^2-4x-5)<0 is 4<x<5

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(x^2-3x-4)/(x^2-4x-5)<0

\frac{x ^{2} - 3 x - 4}{x ^{2} - 4 x - 5} < 0

4 < x < 5

Interval Notation

(4, 5)

Solution steps

\frac{x ^{2} - 3 x - 4}{x ^{2} - 4 x - 5} < 0

Factor

\frac{x ^{2} - 3 x - 4}{x ^{2} - 4 x - 5} : \frac{( x + 1 ) ( x - 4 )}{( x + 1 ) ( x - 5 )}

\frac{( x + 1 ) ( x - 4 )}{( x + 1 ) ( x - 5 )} < 0

Identify the intervals

4 < x < 5

6 x^{2} - 27 x - 54

15 + 10 x = 10 x - 11

(1.2 m)^{4}

- 50 x^{4} + 32 y^{2}

What is (x^2-3x-4)/(x^2-4x-5)<0 ?
The solution to (x^2-3x-4)/(x^2-4x-5)<0 is 4<x<5