<PRE><B>Question 982809</B>
1 + {{{3/x = 2/x^2}}}
multiply each term by x^2, cancel the denominators and you have
x^2 + 3x = 2
subtract 2 from both sides, to form a quadratic equation
x^2 + 3x - 2 = 0
use the quadratic formula; {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
a = 1; b = 3; c =-2
{{{x = (-3 +- sqrt( 3^2-4*1*-2 ))/(2*1) }}}
:
{{{x = (-3 +- sqrt(9-(-8 )))/2 }}}
:
{{{x = (-3 +- sqrt(9+8))/2 }}}
To solutions
{{{x = (-3 + sqrt(17))/2 }}}
x = .56
and
{{{x = (-3 - sqrt(17))/2 }}}
x = -3.56
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