I am facing a strange problem in solving the following equation for x and y
$\frac{4x^2+(x^2+y^2-1)^2}{(x^2+(y-1)^2)^2}=1$ $4x^2+(x^2+y^2-1)^2=(x^2+(y-1)^2)^2$ On solving the terms in brackets, we get
$4y^3-8y^2+4y+4x^2y=0$ $4y(y^2-2y+1+x^2)=0$ $4y[(y-1)^2+x^2]=0$ $y=0 \space or \space (y-1)^2+x^2=0$ So one solution is $y=0$.
$\text{if}\space(y-1)^2+x^2=0\,\space \text{then} \space y=1\space \text{and}\space x=0$ But if I try to put $y=1$, and $x=0$ in original equation, I get $\frac{0}{0}$, which is not $1$.
So I want to ask is $(y=1,$ $x=0)$ solution or not ?
Do I need to check solution while solving an equation?
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algebra-precalculus
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1You could save yourself some manipulation bu using the difference of two squares for your more complicated terms. – 2012-06-28
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$(x,y)=(0,1)$ is not a solution. \begin{align*} &\frac{4x^2+(x^2+y^2-1)^2}{\left(x^2+(y-1)^2\right)^2}=1\\ \iff&\begin{cases}4x^2+(x^2+y^2-1)^2=\left(x^2+(y-1)^2\right)^2\\\left(x^2+(y-1)^2\right)^2\neq0\end{cases}\\ \iff&\begin{cases}y\left(x^2+(y-1)^2\right)=0\\\left(x^2+(y-1)^2\right)^2\neq0\end{cases}\\ \iff&\begin{cases}y=0\\x\neq0\hbox{ or }y\neq1\end{cases}\iff y=0 \end{align*}
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2@HappyMittal See Mark Bennet's comment. But the *boring* expansion is sometimes necessary. You can try to expand this: $4(y+z)^2(z+x)^2(x+y)^2((yz+zx+xy)(1/(y+z)^2+1/(z+x)^2+1/(x+y)^2)-9/4)$. It's from [Iran 96 TST](http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2089092). I spent many hours in expanding it! – 2012-06-28