# How do you solve x^2/(x - 4) (-) 7/x - 4 = 0?

Jun 2, 2015

Assuming $x \ne 0$ and $x \ne 4$ (otherwise terms of the equation are undefined)
and also assuming that the parentheses around the minus sign are extraneous.

We can write the equation with all terms on the left side having a common denominator of $\textcolor{red}{\left(x - 4\right)} \textcolor{b l u e}{\left(x\right)}$

$\textcolor{w h i t e}{\text{XXXXX}}$$\frac{\left[\left({x}^{2}\right) \cdot \textcolor{b l u e}{x}\right] - \left[7 \cdot \textcolor{red}{\left(x - 4\right)}\right] - \left[4 \cdot \left(\textcolor{red}{\left(x - 4\right)} \textcolor{b l u e}{\left(x\right)}\right)\right]}{\left(x - 4\right) \left(x\right)} = 0$

$\textcolor{w h i t e}{\text{XXXXX}}$$\frac{{x}^{3} - 7 x + 28 - 4 {x}^{2} + 16 x}{\left(x\right) \left(x - 4\right)} = 0$

$\textcolor{w h i t e}{\text{XXXXX}}$${x}^{3} - 4 {x}^{2} + 9 x + 28 = 0$

Then things really get ugly:

• Substitute $y + \frac{4}{3}$ for $x$ to get a depressed cubic equation (the equation won't be the only thing depressed when you're done with this) with the form y^3+Ay = B#
• Find $s$ and $t$ such that $3 \cdot s \cdot t = A$ and ${s}^{3} - {t}^{3} = B$
• Back-track to solve for $x$

You will notice that I have not done any of this (at least not here; formatting the steps is at least as horrifying as working out the solution).

If you really need this continued, first, please verify the question and that my assumptions were correct