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

The solution to (x+1)/(x^2+2)> 1/4 is 2-sqrt(6)<x<2+sqrt(6)

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

\frac{x + 1}{x ^{2} + 2} > \frac{1}{4}

2 - 6 < x < 2 + 6

Interval Notation

(2 - 6, 2 + 6)

Decimal Notation

- 0.44948 \dots < x < 4.44948 \dots

Solution steps

\frac{x + 1}{x ^{2} + 2} > \frac{1}{4}

Rewrite in standard form

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

Factor

- x^{2} + 4 x + 2 : - (x - 2 - 6) (x - 2 + 6)

\frac{- ( x - 2 - 6 ) ( x - 2 + 6 )}{x ^{2} + 2} > 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - ( x - 2 - 6 ) ( x - 2 + 6 ) ) ( - 1 )}{x ^{2} + 2} < 0 \cdot (- 1)

Simplify

\frac{( x - 2 - 6 ) ( x - 2 + 6 )}{x ^{2} + 2} < 0

Identify the intervals

2 - 6 < x < 2 + 6

\frac{( - x ^{2} + 4 x - 3 )}{( 2 x - 6 )} < 0

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

x^{4} + 6 x^{3} + 6 x^{2} + 6 x + 5 > 0

x \cdot (x - 1) \geq 0

(x^{2} - 9)^{2} \cdot (x + 1) (x^{2} - 2 x - 3) (x - 1) \geq 0

What is (x+1)/(x^2+2)> 1/4 ?
The solution to (x+1)/(x^2+2)> 1/4 is 2-sqrt(6)<x<2+sqrt(6)