# How do you solve the equation and identify any extraneous solutions for (x-4)/x + x/3=6?

Jul 17, 2015

Multiply through by $x$ to get a quadratic in $x$, then use quadratic formula to get:

$x = \frac{15 \pm \sqrt{273}}{2}$

#### Explanation:

Multiply all the terms by $3 x$ to get:

$3 \left(x - 4\right) + {x}^{2} = 18 x$

In theory this could introduce an extraneous $x = 0$ solution, but it does not...

Subtract $18 x$ from both sides to get:

${x}^{2} - 15 x - 12 = 0$

This quadratic is in the standard $a {x}^{2} + b x + c$ form, with $a = 1$, $b = - 15$ and $c = - 12$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$= \frac{15 \pm \sqrt{{15}^{2} - \left(4 \times 1 \times - 12\right)}}{2 \times 1}$
$= \frac{15 \pm \sqrt{225 + 48}}{2}$
$= \frac{15 \pm \sqrt{273}}{2}$