What is 1/(2-x^2)>= (16x)/(x^2-2)+8x ?

The solution to 1/(2-x^2)>= (16x)/(x^2-2)+8x is x<-sqrt(2)\lor-1/2 <= x<sqrt(2)

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1/(2-x^2)>= (16x)/(x^2-2)+8x

\frac{1}{2 - x ^{2}} \geq \frac{16 x}{x ^{2} - 2} + 8 x

x < - 2 or - \frac{1}{2} \leq x < 2

Interval Notation

(- \infty, - 2) \cup [- \frac{1}{2}, 2)

Decimal Notation

x < - 1.41421 \dots or - 0.5 \leq x < 1.41421 \dots

Solution steps

\frac{1}{2 - x ^{2}} \geq \frac{16 x}{x ^{2} - 2} + 8 x

Rewrite in standard form

\frac{- 1 - 8 x ^{3}}{( x + 2 ) ( x - 2 )} \geq 0

Factor

\frac{- 1 - 8 x ^{3}}{( x + 2 ) ( x - 2 )} : \frac{- ( 2 x + 1 ) ( 4 x ^{2} - 2 x + 1 )}{( x + 2 ) ( x - 2 )}

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

Multiply both sides by

- 1

(reverse the inequality)

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

Simplify

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

Identify the intervals

x < - 2 or - \frac{1}{2} \leq x < 2

x^{2} + 10 x + 42 = - 3 x + 2

\frac{- 5 - 2 i}{- 3 + 5 i}

1 = 7 x

\frac{π}{2} - \frac{4 π}{11}

What is 1/(2-x^2)>= (16x)/(x^2-2)+8x ?
The solution to 1/(2-x^2)>= (16x)/(x^2-2)+8x is x<-sqrt(2)\lor-1/2 <= x<sqrt(2)