What is x^2-7-(81)/(10(x+5))+(81)/(10(x-5))>= 0 ?

The solution to x^2-7-(81)/(10(x+5))+(81)/(10(x-5))>= 0 is x 5

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x^2-7-(81)/(10(x+5))+(81)/(10(x-5))>= 0

x^{2} - 7 - \frac{81}{10 ( x + 5 )} + \frac{81}{10 ( x - 5 )} \geq 0

x < - 5 or x = - 4 or x = 4 or x > 5

Interval Notation

(- \infty, - 5) \cup x = - 4 \cup x = 4 \cup (5, \infty)

Solution steps

x^{2} - 7 - \frac{81}{10 ( x + 5 )} + \frac{81}{10 ( x - 5 )} \geq 0

Rewrite in standard form

\frac{10 x ^{4} - 320 x ^{2} + 2560}{( x + 5 ) ( x - 5 )} \geq 0

Factor

\frac{10 x ^{4} - 320 x ^{2} + 2560}{( x + 5 ) ( x - 5 )} : \frac{10 ( x + 4 ) ^{2} ( x - 4 ) ^{2}}{( x + 5 ) ( x - 5 )}

\frac{10 ( x + 4 ) ^{2} ( x - 4 ) ^{2}}{( x + 5 ) ( x - 5 )} \geq 0

Divide both sides by

10

\frac{\frac{10 ( x + 4 ) ^{2} ( x - 4 ) ^{2}}{( x + 5 ) ( x - 5 )}}{10} \geq \frac{0}{10}

Simplify

\frac{( x + 4 ) ^{2} ( x - 4 ) ^{2}}{( x + 5 ) ( x - 5 )} \geq 0

Identify the intervals

x < - 5 or x = - 4 or x = 4 or x > 5

∣ x - 4 ∣ = ∣ 3 x + 6 ∣

- 16 v^{3} w - 16 v^{2} w - 56 v w

\frac{x}{- 10} - 3 = 3

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

2 x^{2} + 14 x \geq 0

What is x^2-7-(81)/(10(x+5))+(81)/(10(x-5))>= 0 ?
The solution to x^2-7-(81)/(10(x+5))+(81)/(10(x-5))>= 0 is x<-5\lor x=-4\lor x=4\lor x>5