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How do you solve #9/2x+2=8/x#?

1 Answer

Jan 10, 2016

#x=(-2+-2sqrt37)/9#

Explanation:

Multiply through by 2x to remove the fractions.

#9x^2+4x=16#
#9x^2+4x-16=0#
#x=(-4+-sqrt(4^2-4*9*-16))/18#
#x=(-4+-sqrt(592))/18 #
#x=(-4+-4sqrt37)/18#
#x=(-2+-2sqrt37)/9#

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