First of all, note that xx can't be zero, otherwise 1/(3x)13x would be a division by zero. So, provided x\ne0x≠0, we can rewrite the equation as
(3x)/(3x)-8 = 1/(3x) + x (3x)/(3x)3x3x−8=13x+x3x3x
iff⇔
(-24x)/(3x) = 1/(3x) + (3x^2)/(3x)−24x3x=13x+3x23x
with the advantage that now all the terms have the same denominator, and we can sum the fractions:
(-24x)/(3x) = (1+3x^2)/(3x)−24x3x=1+3x23x
Since we assumed x\ne 0x≠0, we can claim that the two fractions are equal if and only if the numerators are equal: so the equation is equivalent to
-24x = 1+3x^2−24x=1+3x2
which leads is to the quadratic equation
3x^2+24x+1=03x2+24x+1=0.
To solve this, we can use the classic formula
\frac{-b\pm sqrt(b^2-4ac)}{2a}−b±√b2−4ac2a
where aa, bb and cc play the role of ax^2+bx+c=0ax2+bx+c=0.
So, the solving formula becomes
\frac{-24\pm sqrt(24^2-4*3*1)}{2*3}−24±√242−4⋅3⋅12⋅3
==
\frac{-24\pm sqrt(576-12)}{6}−24±√576−126
==
\frac{-24\pm sqrt(564)}{6}−24±√5646
Since 564=36* 47/3564=36⋅473, we can simplfy it out the square root, obtaining
\frac{-24\pm 6sqrt(47/3)}{6}−24±6√4736
and finally we can simplify the whole expression:
\frac{-cancel(6)*4\pm cancel(6)sqrt(47/3)}{cancel(6)}
into
-4\pm sqrt(47/3)
