What is (4x^2+4x-3)/(4x^2-1)>0 ?

The solution to (4x^2+4x-3)/(4x^2-1)>0 is x 1/2

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(4x^2+4x-3)/(4x^2-1)>0

\frac{4 x ^{2} + 4 x - 3}{4 x ^{2} - 1} > 0

x < - \frac{3}{2} or - \frac{1}{2} < x < \frac{1}{2} or x > \frac{1}{2}

Solution steps

\frac{4 x ^{2} + 4 x - 3}{4 x ^{2} - 1} > 0

Factor

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

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

Identify the intervals

x < - \frac{3}{2} or - \frac{1}{2} < x < \frac{1}{2} or x > \frac{1}{2}

t - 1 = - 5

\frac{x - 2}{x + 4} \leq 2

x^{2} + 6 x + 9 = 16

∣ 3 x ∣ = 3

x^{4} + 5 x^{3}

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