Solve the following equation:

$$
x^{2}-4=\sqrt{x+4}
$$
I. Solution. On the appropriate domain of interpretation $(x \geq-4)$, the functions on both sides of the equation are inverses of each other.

$$
\text { 1987-09-252-1.eps }
$$

Figure 1

We can expect that the two curves will intersect on the line $y=x$ (Figure 1), where the abscissa satisfies $x^{2}-4=x$, that is,

$$
x^{2}-x-4=0
$$

By squaring both sides of the original equation and rearranging, we obtain the following equation:

$$
x^{4}-8 x^{2}-x+12=0
$$

It can be easily verified that the polynomial on the left side of (2) is a factor of the polynomial on the left side of (3), so it can be factored as:

$$
\left(x^{2}-x-4\right)\left(x^{2}+x-3\right)=0
$$

Thus, the roots of our equation are among the roots of $x^{2}-x-4=0$ and $x^{2}+x-3=0$:

$$
x=\frac{1 \pm \sqrt{17}}{2} \quad \text { or } \quad x=\frac{-1 \pm \sqrt{13}}{2}
$$

Due to the squaring, false roots have appeared, and we must exclude the roots with absolute values less than 2, since in that case $x^{2}-4$ is negative. On the set $x \geq-4,|x| \geq 2$, however, (1) and (3) are equivalent, so the two remaining roots, $x_{1}=\frac{1+\sqrt{17}}{2}$ and $x_{2}=\frac{-1-\sqrt{13}}{2}$, are solutions to the equation.

II. Solution. On the set $x \geq-4$, we seek the abscissas of the common points of the curves given by the equations $y=x^{2}-4$ and $y=\sqrt{x+4}$. After squaring the second equation, we get $y^{2}=x+4$. Subtract this from the first equation: $x^{2}-y^{2}=-(x-y)$, which gives $(x-y)(x+y+1)=0$. From the product form, for example, $y$ can be expressed as $y=x$ or $y=-x-1$. Substituting these into the first equation, the continuation involves solving two quadratic equations.

III. Solution. Write the given equation as follows:

$$
x^{2}-4+x+4+\frac{1}{4}=x+4+\sqrt{x+4}+\frac{1}{4}
$$

It is clear that both sides are perfect squares:

$$
\left(x+\frac{1}{2}\right)^{2}=\left(\sqrt{x+4}+\frac{1}{2}\right)^{2}
$$

From this, $\sqrt{x+4}=x$ or $\sqrt{x+4}=-x-1$, and the continuation is similar to that in the second solution (Figure 2).