# Solve an equation for x where $y = \frac{ x^2 - x + 1}{ x^2 + x + 1 }$

Solve an equation for x where

$$y = \frac{ x^2 - x + 1}{ x^2 + x + 1 }$$

Also, is there an easier way to find the range of the equation, rather than reversing it and finding it's domain?

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My suggestion would be to first clear the denominator:

$\displaystyle y(x^2+x+1)=x^2-x+1$

Distributing gives

$yx^2+yx+y=x^2-x+1$

Now if we move everything over to the left hand side

$yx^2 -x^2 +yx+x+y-1=0$

If you carefully combine the terms, you can deal with

$(y-1)x^2+(y+1)x+(y-1)=0$

This is really a quadratic equation in $x$ where $a=y-1$, $b=y+1$, and $c=y-1$. If you run the quadratic formula, you will be able to solve for $x$.

Regarding the second part of your question, the only other way I can see how to determine the range of a function is to use some techniques from calculus. Do you have any knowledge about that subject?

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thanks, I knew it would be simple like that, but for the life of me I kept getting it wrong. As for calculus, yes, and naturally that would work but I was looking for an algebraic method for not-so-solvable equations. – ericosg May 29 '12 at 8:20
@ericosg: in my answer I tried to give a insight which methods could you use to find a range without solving the equation. – Ilya May 29 '12 at 8:27

You can do the following: what is the range? This is the set of points $y$ for which there exists $x$ s.t. $$y = \frac{x^2-x+1}{x^2+x+1}.$$ Since the denominator has no real zeroes, we can multiply both sides by it and obtain $$(y-1)x^2+(y+1)x+(y-1) =0.\tag{1}$$ So the range is those $y$ for which the solution of $(1)$ exists. Clearly, these are all $y$ for which the discriminant of the equation is non-negative: $$(y+1)^2 - 4(y-1)^2\geq 0. \tag{2}$$ Soving the latter inequality gives you the range. Note that you should consider the case $y=1$ separately since in that case $(1)$ is a quadratic equation and not linear - but $1$ solves $(2)$ and hence will be included in the answer in any case.

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You have already got answers for the first part of your question, for the second part, to get an answer without using calculus, notice that $y = 1 - \frac{2}{x + \frac{1}{x} + 1}$. Now consider $x + \frac{1}{x}$. Its range is the whole number line except for (-2,2). To see why consider $x+\frac{1}{x}-2$ in the case where x is positive and $x+\frac{1}{x}+2$ in the case where x is negative. You should be easily able to show that both are non-negative by getting a square. Now that you know the range for $x+\frac{1}{x}$, you can easily translate it into a range for $y = 1 - \frac{2}{x + \frac{1}{x} + 1}$

EDIT: I see you already have good answer to the second part too, still this is a different way to look at it.

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Nice use of $x+\frac{1}{x}$. – André Nicolas May 29 '12 at 10:28