# quadratic equation precalculus

from Stewart, Precalculus, 5th, p56, Q. 79

Find all real solutions of the equation

$$\dfrac{x+5}{x-2}=\dfrac{5}{x+2}+\dfrac{28}{x^2-4}$$

my solution

$$\dfrac{x+5}{x-2}=\dfrac{5}{x+2}+\dfrac{28}{(x+2)(x-2)}$$ $$(x+2)(x+5)=5(x-2)+28$$ $$x^2+2x-8=0$$

$$\dfrac{-2\pm\sqrt{4+32}}{2}$$ $$\dfrac{-2\pm6}{2}$$ $$x=-4\text{ or }2$$

official answer at the back of the book has only one real solution of $-4$

where did I go wrong?

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values of variable on which functions is define means function have value called as domain.If you put x=2 in your question two functions will be undefined that's why in the answer only 1 value is consider –  iostream007 May 18 '13 at 17:44
The very first thing I thought when I looked at the original equation is that $2$ and $-2$ cannot be solutions. That's something to look for. And I have no doubt that what Stewart had in mind is precisely that some students would do just what you did here. –  Michael Hardy May 18 '13 at 17:47

## 3 Answers

You multiplied both sides by $(x-2)(x+2)$. If this is zero, you may introduce extra solutions, hence you need to check your final answer to see if you have any extraneous solutions. In this case, you do! For $x=2$, two of the three fractions in the original equation are undefined.

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Why do we have to reject the solution $x=2$?

Hint: What happens when we put $x=2$ in the original equation?
Review your equations. Make sure that you didn't multiply by 0.

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$$\frac{x+5}{x-2}=\frac{5}{x+2}+\frac{28}{x^2-4}$$ multiply both sides with $(x-2)(x+2)\neq 0$ or $x\neq 2,x\neq -2$ we get $$(x+5)(x+2)=5(x-2)+28,x\neq 2,-2$$

$$x^2+2x-8=0,x\neq 2,-2$$

$$x^2-2x+4x-8=0,x\neq 2,-2$$ $$x(x-2)+4(x-2)=0,x\neq 2,-2$$ $$(x-2)(x+4)=0,x\neq 2,-2$$ so $x+4=0$ or $x=-4\neq 2,-2$ is unique solution because from $x-2=0$ follow solution $x=-2$ that is prohibited

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