What is x if #-8=1/(3x)+x#?

1 Answer
Nov 1, 2015

Answer:

You have two solutions:
#x=-4- sqrt(47/3)#, and
#x=-4+ sqrt(47/3)#

Explanation:

First of all, note that #x# can't be zero, otherwise #1/(3x)# would be a division by zero. So, provided #x\ne0#, we can rewrite the equation as

# (3x)/(3x)-8 = 1/(3x) + x (3x)/(3x)#

#iff#

#(-24x)/(3x) = 1/(3x) + (3x^2)/(3x)#

with the advantage that now all the terms have the same denominator, and we can sum the fractions:

#(-24x)/(3x) = (1+3x^2)/(3x)#

Since we assumed #x\ne 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#

which leads is to the quadratic equation

#3x^2+24x+1=0#.

To solve this, we can use the classic formula

#\frac{-b\pm sqrt(b^2-4ac)}{2a}#

where #a#, #b# and #c# play the role of #ax^2+bx+c=0#.

So, the solving formula becomes

#\frac{-24\pm sqrt(24^2-4*3*1)}{2*3}#

#=#

#\frac{-24\pm sqrt(576-12)}{6}#

#=#

#\frac{-24\pm sqrt(564)}{6}#

Since #564=36* 47/3#, we can simplfy it out the square root, obtaining

#\frac{-24\pm 6sqrt(47/3)}{6}#

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)#