What is 8/(x^2+1)>= 4 ?

The solution to 8/(x^2+1)>= 4 is -1<= x<= 1

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8/(x^2+1)>= 4

\frac{8}{x ^{2} + 1} \geq 4

- 1 \leq x \leq 1

Interval Notation

[- 1, 1]

Solution steps

\frac{8}{x ^{2} + 1} \geq 4

Rewrite in standard form

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

Factor

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

\frac{- 4 ( x + 1 ) ( x - 1 )}{x ^{2} + 1} \geq 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - 4 ( x + 1 ) ( x - 1 ) ) ( - 1 )}{x ^{2} + 1} \leq 0 \cdot (- 1)

Simplify

\frac{4 ( x + 1 ) ( x - 1 )}{x ^{2} + 1} \leq 0

Divide both sides by

4

\frac{\frac{4 ( x + 1 ) ( x - 1 )}{x ^{2} + 1}}{4} \leq \frac{0}{4}

Simplify

\frac{( x + 1 ) ( x - 1 )}{x ^{2} + 1} \leq 0

Identify the intervals

- 1 \leq x \leq 1

6 (5 x - 3) \leq 36 x

a^{2} - 12 a + 35 = 0

3 x - 1 \leq \frac{7}{2} x < 12 x + 2

\frac{5 x ^{7} y ^{8}}{x ^{2} y ^{8}}

\frac{3 m ^{- 4}}{m ^{3}}