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I have the following equation to solve. I know that the answer is -5, I made several attempts at this, and arrive at a different answer. My first thought was to factor out the trinomial, but that didn't help. What are the correct steps to solve this equation for x?

$$\frac{x^2 + x -2}{x+3} ={-9}$$

James
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2 Answers2

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$$\frac{x^2 + x -2}{x+3} ={-9}$$ $$x^2+x-2=-9x-27$$ $$x^2+10x +25 = 0$$

now use the quadratic equation $x = \dfrac{-B \pm \sqrt{B^2-4AC}}{2A}$

and you get the result $x=-5$

Jan Hasenbichler
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  • Thanks, so simple it's embarrassing. I've used the quadratic equation many times before, I think the reason I didn't consider using it here is because it always yields two answers + -. In this case only -5 was given as the answer. – James Jul 24 '14 at 06:06
  • Yeah, there is a multiple answer this time because the square root equals 0 – Jan Hasenbichler Jul 24 '14 at 06:07
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I can't comment the other answer so just elaborating on each step:

$$\frac{x^2 + x -2}{x+3} ={-9}$$

Multiply both sides of the equation by $x + 3$ $$\frac{x^2 + x -2}{x+3} * \frac{x+3}{1} ={-9} * \frac{x + 3}{1}$$

$$x^2+x-2=-9x-27$$

Then subtract $-9x -27$ or alternatively add $9x + 27$ to both sides.

$$x^2+10x +25 = 0$$

Then, use quadratic formula or any other method you know.

chronocturnal
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  • Yeah you are right. Just use the quadratic equation or use the binomial theorem to see that it is actually $(x+5)^2$, so $x=-5$ must be a multiple zero – Jan Hasenbichler Jul 24 '14 at 06:02