What is (400t^2+40t+1)/(25t^2-1)>= 0 ?

The solution to (400t^2+40t+1)/(25t^2-1)>= 0 is t 1/5

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(400t^2+40t+1)/(25t^2-1)>= 0

\frac{400 t ^{2} + 40 t + 1}{25 t ^{2} - 1} \geq 0

t < - \frac{1}{5} or t = - \frac{1}{20} or t > \frac{1}{5}

Solution steps

\frac{400 t ^{2} + 40 t + 1}{25 t ^{2} - 1} \geq 0

Factor

\frac{400 t ^{2} + 40 t + 1}{25 t ^{2} - 1} : \frac{( 20 t + 1 ) ^{2}}{( 5 t + 1 ) ( 5 t - 1 )}

\frac{( 20 t + 1 ) ^{2}}{( 5 t + 1 ) ( 5 t - 1 )} \geq 0

Identify the intervals

t < - \frac{1}{5} or t = - \frac{1}{20} or t > \frac{1}{5}

\frac{2 x}{x ^{2} - y ^{2}} - \frac{1}{x + y} - \frac{y}{x - x y}

x^{3} - 3 x^{2} y - x + 3 y

60 + \frac{3}{5} (178 - 60)

a^{2} = 5^{2} + 1 3^{2}

What is (400t^2+40t+1)/(25t^2-1)>= 0 ?
The solution to (400t^2+40t+1)/(25t^2-1)>= 0 is t<-1/5 \lor t=-1/20 \lor t> 1/5