Equation has exactly one real solution I am trying to show that the equation
$$
4x^2y^4+12x^2y^2+4x^2+4xy^2+4x+1=0
$$
has exactly one real solution $x,y$ and to determine it.
The first observation is: If $y=0$, we are left with
$$
4x^2+4x+1=0
$$
which is solved by $x=-1/2$.
Thus a real solution is given by
$$
x=-\frac{1}{2}, y=0
$$
and it remains to show that this is the only real solution.
This is where I am stuck...
 A: Setting $y^2=z,\;z≥0$ and rewriting the original polynomial as a quadratic  respect to $x$, then you have:
$$4(z^2+3z+1)x^2+4(z+1)x+1=0$$
If we work with real-valued pair of $(x,y)$, then note that $\Delta_x≥0$.
More explicitly, we have:
$$
\begin{align}\Delta_x&=4(z+1)^2-4(z^2+3z+1)\\
&=-4z≥0\end{align}
$$
$z≥0\wedge z≤0$ implies that, $z=y=0$. This means $\Delta_x=0$. This leads to:
$$\begin{align}x=x_1=x_2&=\frac {-2(z+1)}{4}\\
&=-\frac 12.\end{align}$$
Therefore, $(x,y)=\left(-\frac 12,0\right)$ is an only possible solution.

Observe that, you can also get the same result by rearranging the equation quadratic respect to $y^2=z$ and taking $\Delta_z≥0$.
A: Grouping terms, we have
$$4  \left(y^4+3 y^2+1\right)x^2+4 \left(y^2+1\right)x+1=0$$
$$\Delta=-16y^2$$ So, no real root except if $y=0$. In this case, what is left is what you wrote at the beginning.
A: I do not know if it helps, but you can rearrange the proposed equation as follows:
\begin{align*}
4x^{2}y^{4} + 12x^{2}y^{2} + 4x^{2} + 4xy^{2} + 4x + 1 = 0 & \Longleftrightarrow (4x^{2}y^{4} + 4xy^{2} + 1) + (12x^{2}y^{2} + 4x^{2} + 4x) = 0\\\\
& \Longleftrightarrow (2xy^{2} + 1)^{2} + 12x^{2}y^{2} + (4x^{2} + 4x + 1) = 1\\\\
& \Longleftrightarrow (2xy^{2} + 1)^{2} + 12x^{2}y^{2} + (2x + 1)^{2} = 1
\end{align*}
A: Let$$P(x,y)=
4x^2y^4+12x^2y^2+4x^2+4xy^2+4x+1
$$
and$$Q(y)=y^4+3y^2+1.$$
For all $(x,y)\in\Bbb R^2,$ since $Q(y)>0$ and
$$Q(y)P(x,y)=(2Q(y)x+y^2+1)^2+y^2,$$
we have $P(x,y)=0$ if and only if
$$y=0\text{ and }x=-\frac1{2Q(0)}=-\frac12.$$
