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. – ÍgjøgnumMeg Apr 11 '17 at 9:35

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.

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.