What is (8x-1)/x <= x+1 ?

The solution to (8x-1)/x = (7+3sqrt(5))/2

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(8x-1)/x <= x+1

\frac{8 x - 1}{x} \leq x + 1

0 < x \leq \frac{7 - 3 5}{2} or x \geq \frac{7 + 3 5}{2}

Solution steps

\frac{8 x - 1}{x} \leq x + 1

Rewrite in standard form

\frac{7 x - x ^{2} - 1}{x} \leq 0

Factor

7 x - x^{2} - 1 : - (x - \frac{7 + 3 5}{2}) (x - \frac{7 - 3 5}{2})

\frac{- ( x - \frac{7 + 3 5}{2} ) ( x - \frac{7 - 3 5}{2} )}{x} \leq 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - ( x - \frac{7 + 3 5}{2} ) ( x - \frac{7 - 3 5}{2} ) ) ( - 1 )}{x} \geq 0 \cdot (- 1)

Simplify

\frac{( x - \frac{7 + 3 5}{2} ) ( x - \frac{7 - 3 5}{2} )}{x} \geq 0

Identify the intervals

0 < x \leq \frac{7 - 3 5}{2} or x \geq \frac{7 + 3 5}{2}

216 x^{2} - 512 = 0

3 n^{4} - 147

2 = 2 (x - 3)

∣ t - 3 ∣ = 4

e + 9 \geq 15

What is (8x-1)/x <= x+1 ?
The solution to (8x-1)/x <= x+1 is 0<x<= (7-3sqrt(5))/2 \lor x>= (7+3sqrt(5))/2