Question

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

Answer 1

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)`