What is (16x^2-81)/(x^2-64)<= 0 ?

The solution to (16x^2-81)/(x^2-64)<= 0 is -8<x<=-9/4 \lor 9/4 <= x<8

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(16x^2-81)/(x^2-64)<= 0

\frac{16 x ^{2} - 81}{x ^{2} - 64} \leq 0

- 8 < x \leq - \frac{9}{4} or \frac{9}{4} \leq x < 8

Interval Notation

(- 8, - \frac{9}{4}] \cup [\frac{9}{4}, 8)

Decimal Notation

- 8 < x \leq - 2.25 or 2.25 \leq x < 8

Solution steps

\frac{16 x ^{2} - 81}{x ^{2} - 64} \leq 0

Factor

\frac{16 x ^{2} - 81}{x ^{2} - 64} : \frac{( 4 x + 9 ) ( 4 x - 9 )}{( x + 8 ) ( x - 8 )}

\frac{( 4 x + 9 ) ( 4 x - 9 )}{( x + 8 ) ( x - 8 )} \leq 0

Identify the intervals

- 8 < x \leq - \frac{9}{4} or \frac{9}{4} \leq x < 8

y (3 x - 2) - 3 x + 2

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3 x^{2} + 15 x - 18 = 0

\frac{4}{3} + 5 y = \frac{5}{2}

- y^{3} - 3 y^{2} + 9 y + 27

What is (16x^2-81)/(x^2-64)<= 0 ?
The solution to (16x^2-81)/(x^2-64)<= 0 is -8<x<=-9/4 \lor 9/4 <= x<8