## Task A-2.3.

Solve the equation in the set of real numbers

$$
\frac{1}{x^{2}}+\frac{1}{(1-x)^{2}}=24
$$
## First Solution.

Notice that the given equation is equivalent to the system

$$
\left\{\begin{array}{c}
x+y=1 \\
\frac{1}{x^{2}}+\frac{1}{y^{2}}=24
\end{array}\right.
$$

From the first equation, by squaring, we get $x^{2}+y^{2}=1-2 x y$,

and from the second equation, we have $x^{2}+y^{2}=24 x^{2} y^{2}$.

Therefore, $24(x y)^{2}+2(x y)-1=0$, and solving this quadratic equation for $x y$ gives

$$
x y=\frac{-2 \pm \sqrt{4+4 \cdot 24}}{2 \cdot 24}=\frac{-2 \pm 10}{48}
$$

i.e., $x y=-\frac{1}{4}$ or $x y=\frac{1}{6}$.

If $x y=-\frac{1}{4}$, then from $x+y=1$ we conclude that $x$ and $y$ are the solutions of the equation

$x^{2}-x-\frac{1}{4}=0$, hence $x_{1}=\frac{1+\sqrt{2}}{2}, x_{2}=\frac{1-\sqrt{2}}{2}$.

Similarly, if $x y=\frac{1}{6}$, then $x$ and $y$ are the solutions of the equation $x^{2}-x+\frac{1}{6}=0$, hence $x_{3}=\frac{3+\sqrt{3}}{6}, x_{4}=\frac{3-\sqrt{3}}{6}$.

Each of these numbers is a solution to the given equation.