What is (4x)/(x^2-9)<=-1 ?

The solution to (4x)/(x^2-9)<=-1 is -2-sqrt(13)<= x<-3\lor-2+sqrt(13)<= x<3

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(4x)/(x^2-9)<=-1

\frac{4 x}{x ^{2} - 9} \leq - 1

- 2 - 13 \leq x < - 3 or - 2 + 13 \leq x < 3

Interval Notation

[- 2 - 13, - 3) \cup [- 2 + 13, 3)

Decimal Notation

- 5.60555 \dots \leq x < - 3 or 1.60555 \dots \leq x < 3

Solution steps

\frac{4 x}{x ^{2} - 9} \leq - 1

Rewrite in standard form

\frac{4 x + x ^{2} - 9}{x ^{2} - 9} \leq 0

Factor

\frac{4 x + x ^{2} - 9}{x ^{2} - 9} : \frac{( x + 2 - 13 ) ( x + 2 + 13 )}{( x + 3 ) ( x - 3 )}

\frac{( x + 2 - 13 ) ( x + 2 + 13 )}{( x + 3 ) ( x - 3 )} \leq 0

Identify the intervals

- 2 - 13 \leq x < - 3 or - 2 + 13 \leq x < 3

5 u^{2} - 14 u + 8 = 0

9 k^{2} - (5 m - 1)^{2}

\frac{2 x - 3}{4} - \frac{3 + 2 x}{9} = 2 - \frac{x + 1}{6}

x + \frac{5}{3} \leq 2 x + \frac{1}{3} \leq 3 x + \frac{4}{3}

648 a + 1029 a^{2}

What is (4x)/(x^2-9)<=-1 ?
The solution to (4x)/(x^2-9)<=-1 is -2-sqrt(13)<= x<-3\lor-2+sqrt(13)<= x<3