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Solving $$x^2+x^2t^2+2xt^2+t^2=1$$ for $x$ yields $$x=\frac{1-t^2}{t^2+1}.$$ I had to use software because I could not manipulate it algebraically, and sadly the program did not tell me how it reached that solution.

What is the method?

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up vote 9 down vote accepted

This is a quadratic equation in $x$. Use the Quadratic Formula.

Recall that the solutions of $ax^2+bx+c=0$ are $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.

When you do the details, you will find that $x=-1$ is also a solution.

Another way: In this case, we can also use factorization. The polynomial $$x^2(1+t^2)+2xt^2+t^2-1$$ factors as $$(x+1)\left((1+t^2)x+(t^2-1)\right),$$ and then the roots are easy to pick up.

Still another way: By inspection, $x=-1$ is a root. But the product of the roots of $ax^2+bx+c=0$ is $\frac{c}{a}$. In our case, the product of the roots is $\frac{t^2-1}{1+t^2}$. Since one of the roots is $-1$, the other one must be $\frac{1-t^2}{1+t^2}$.

Remark: My guess is that you are solving the equation because you are trying to find where the line through $(-1,0)$ with slope $t$ meets the unit circle. If that is the case, then the solution $x=-1$ is obvious even without looking at the equation. Thus the "Still another way" is the most natural way to proceed: If you have a quadratic, and you know one root, the other root requires almost no work. Just use the fact that the product of the roots is $\frac{c}{a}$ or the fact that the sum of the roots is $-\frac{b}{a}$.

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Collect the coefficients in $x$ to get a polynomial in $x$: $$ (t^2+1) x^2 +(2t^2)x+(t^2-1) = 0 $$ Now solve using quadratic formula: $$ \frac{-2t^2 \pm \sqrt{(2t^2)^2 - 4(t^2+1)(t^2-1)}}{2(t^2+1)} = \\ \frac{-2t^2 \pm 2}{2(t^2+1)} = \frac{-t^2 \pm 1}{t^2 + 1}$$ So either $x = -1$ or $x = \frac{-t^2 + 1}{t^2 + 1}$.

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You got a great hint earlier, but you might want to also look at the implicit plot and see another hidden fact.

See, for example:

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Nice link! How did we ever do without good ole Wolfram Alpha? – amWhy May 20 '13 at 0:41
I'm just thinking back to the time when simply having a graphing calculator seemed to be the most innovative tech available! – amWhy May 20 '13 at 0:47

This equation can be written as $x^2+t^2(x+1)^2=1$ which $\implies t^2=\frac{(1+x)(1-x)}{(1+x)^2}\implies t^2=\frac{(1-x)}{(1+x)}$ (assuming $x\neq -1$) $ \implies t^2(1+x)=1-x\implies t^2+t^2x=1-x\implies x(t^2+1)=1-t^2\implies x=\frac{1-t^2}{1+t^2}$

Here, we have to check whether $x=-1$ satisfies the equation separately. Putting in the equation $x=-1$, it satisfies the equation. Thus $x=-1, \frac{1-t^2}{1+t^2}$

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