How do you solve $\frac{9}{2} x + 2 = \frac{8}{x}$ $9/2x+2=8/x$ ?

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How do you solve $\frac{9}{2} x + 2 = \frac{8}{x}$ $9/2x+2=8/x$ ?

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Answer 1 · 2016-01-10T12:02:41

$x = \frac{- 2 \pm 2 \sqrt{37}}{9}$ $x=(-2+-2sqrt37)/9$

Explanation:

Multiply through by 2x to remove the fractions.

$9 x^{2} + 4 x = 16$ $9x^2+4x=16$
$9 x^{2} + 4 x - 16 = 0$ $9x^2+4x-16=0$
$x = \frac{- 4 \pm \sqrt{4^{2} - 4 \cdot 9 \cdot - 16}}{18}$ $x=(-4+-sqrt(4^2-4*9*-16))/18$
$x = \frac{- 4 \pm \sqrt{592}}{18}$ $x=(-4+-sqrt(592))/18$
$x = \frac{- 4 \pm 4 \sqrt{37}}{18}$ $x=(-4+-4sqrt37)/18$
$x = \frac{- 2 \pm 2 \sqrt{37}}{9}$ $x=(-2+-2sqrt37)/9$

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How do you solve $\frac{9}{2} x + 2 = \frac{8}{x}$ $9/2x+2=8/x$ ?

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