What is (16x^2-81)/(x^2-36)<= 0 ?

The solution to (16x^2-81)/(x^2-36)<= 0 is -6<x<=-9/4 \lor 9/4 <= x<6

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(16x^2-81)/(x^2-36)<= 0

\frac{16 x ^{2} - 81}{x ^{2} - 36} \leq 0

- 6 < x \leq - \frac{9}{4} or \frac{9}{4} \leq x < 6

Interval Notation

(- 6, - \frac{9}{4}] \cup [\frac{9}{4}, 6)

Decimal Notation

- 6 < x \leq - 2.25 or 2.25 \leq x < 6

Solution steps

\frac{16 x ^{2} - 81}{x ^{2} - 36} \leq 0

Factor

\frac{16 x ^{2} - 81}{x ^{2} - 36} : \frac{( 4 x + 9 ) ( 4 x - 9 )}{( x + 6 ) ( x - 6 )}

\frac{( 4 x + 9 ) ( 4 x - 9 )}{( x + 6 ) ( x - 6 )} \leq 0

Identify the intervals

- 6 < x \leq - \frac{9}{4} or \frac{9}{4} \leq x < 6

16 r^{2} + 10 = 0

lo g_{5} (\frac{1}{3125})

c = 180 (4) + 250

49 a^{2} - 169

48 a^{2} - 27 b^{2}

What is (16x^2-81)/(x^2-36)<= 0 ?
The solution to (16x^2-81)/(x^2-36)<= 0 is -6<x<=-9/4 \lor 9/4 <= x<6