# Solving for $x$ in Quadratic Equations

So, I have finished study for linear equations for my methods course but now I have run into a problematic quadratic equation. I have tried researching for a method of tackling this question but I have come up with nothing. I have played around a bit with the question on the CAS Calculator, and have found the answer (Which is $x=-4$, $x=-2$) but this doesn't help me study for an exam. So here is the question.

Solve the following quadratic equations for $x$. $$x+6+\frac{8}{x}=0$$

The questions before this I had to either apply the Null Factor Law or use this equation:

$$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

Now I am just stuck in this question and three others with the same form. So how would I approach the question and what equation should I use along the way?

• Multiply through by $x$. You'll obtain $$x^2 + 6x + 8 = 0$$ which you can solve by any of the methods you've learned. Apr 11, 2017 at 9:35

The trick here is to multiply both sides by $x$.

This gives a quadratic equation: $$x^2+6x+8=0 \text{ where } x\neq 0$$ Can you now apply the quadratic formula?

Note that you can also factorize it then use the Null Factor Law.

The given equation is in fact not of the quadratic type, as it is not a polynomial. Anyway, you can rewrite it by reducing to the common denominator

$$x+6+\frac8x=\frac{x^2+6x+8}x=0,$$

giving you a strictly equivalent equation.

Now the solution of the latter is obtained

• when the numerator is zero (a plain quadratic equation),
• while the denominator is non-zero (this is trivial).

You should know what to do next.