Example 8. Solve the equation

$$
\frac{1}{x}+\frac{1}{\sqrt{1-x^{2}}}=\frac{35}{12}
$$
Solution. By squaring both sides of the equation and combining like terms, we obtain the equation

$$
\frac{1}{x^{2}\left(1-x^{2}\right)}+\frac{2}{x \sqrt{1-x^{2}}}-\left(\frac{35}{12}\right)^{2}=0
$$

which is a consequence of the original equation.

$$
\begin{aligned}
\text { Let } y=\frac{1}{x \sqrt{1-x^{2}}} ; \text { then } \\
y^{2}+2 y-\left(\frac{35}{12}\right)^{2}=0
\end{aligned}
$$

This equation has two roots: $y_{1}=25 / 12, y_{2}=-49 / 12$.

Thus, a consequence of the original equation is the combination of systems

$$
\left\{\begin{array} { r l } 
{ \frac { 1 } { x } + \frac { 1 } { \sqrt { 1 - x ^ { 2 } } } } & { = \frac { 3 5 } { 1 2 } , } \\
{ \frac { 1 } { x } \frac { 1 } { \sqrt { 1 - x ^ { 2 } } } } & { = \frac { 2 5 } { 1 2 } , }
\end{array} \left\{\begin{array}{rl}
\frac{1}{x}+\frac{1}{\sqrt{1-x^{2}}} & =\frac{35}{12} \\
\frac{1}{x} \frac{1}{\sqrt{1-x^{2}}} & =\frac{-49}{12}
\end{array}\right.\right.
$$

Let's solve the first system of the combination. According to Vieta's theorem, $1 / x$ and $1 / \sqrt{1-x^{2}}$ are the roots of the quadratic equation

$$
z^{2}-\frac{35}{12} z+\frac{25}{12}=0
$$

solving it, we find its roots: $5 / 3$ and $5 / 4$. Therefore, the first system of the combination is equivalent to the combination of systems:

![](https://cdn.mathpix.com/cropped/2024_05_21_aca6fef227914c05e0acg-056.jpg?height=251&width=794&top_left_y=691&top_left_x=631)

The solution of this combination consists of two numbers: $x_{1}=3 / 5$ and $x_{2}=4 / 5$.

Let's solve the second system of the combination. The roots of the quadratic equation $z^{2}-\frac{35}{12} z-\frac{49}{12}=0$ are $\frac{7(5-\sqrt{73})}{24}$ and $\frac{7(5+\sqrt{73})}{24}$. According to Vieta's theorem, the second system of the combination (2) is equivalent to the combination of systems:

$$
\left\{\begin{array} { l } 
{ \frac { 1 } { x } = \frac { 7 ( 5 + \sqrt { 7 3 } ) } { 2 4 } , } \\
{ \frac { 1 } { \sqrt { 1 - x ^ { 2 } } } = \frac { 7 ( 5 - \sqrt { 7 3 } ) } { 2 4 } , }
\end{array} \left\{\begin{array}{l}
\frac{1}{x}=\frac{7(5-\sqrt{73})}{24} \\
\frac{1}{\sqrt{1-x^{2}}}=\frac{7(5+\sqrt{73})}{24}
\end{array}\right.\right.
$$

The expression $\frac{1}{\sqrt{1-x^{2}}}$ is positive, while the number $\frac{7(5-\sqrt{73})}{24}$ is negative, so the first system of this combination has no solutions.

The solution of the second system of this combination is the number $-(5+\sqrt{73}) / 14$.

Thus, the set of all roots of the original equation is contained in the set $\{-(5+\sqrt{73}) / 14,3 / 5,4 / 5\}$. Direct substitution shows that the three numbers: $-(5+\sqrt{73}) / 14$, $3 / 5,4 / 5$-constitute the set of all solutions of the original equation.

When both sides of an irrational equation are raised to a power that allows the elimination of radicals, the appearance of extraneous roots of the original equation typically occurs for the following reasons:

a) due to the possible expansion of the domain of the original equation (i.e., the domain of the obtained equation is wider than the domain of the original equation);

b) due to raising the left and right sides of the equation to an even power, which are equal in absolute value but one is positive and the other is negative.

The solution of irrational equations by replacing the equation with its consequence (with subsequent verification of the roots) can be carried out as follows:

1. Find the domain of the original equation.
2. Transition from the equation to its consequence.
3. Find the roots of the obtained equation.
4. Check whether the found roots are roots of the original equation.

The verification consists of the following:

a) the membership of each found root in the domain of the original equation is checked. Those roots that do not belong to the domain are extraneous for the original equation;

b) for each root that belongs to the domain of the original equation, it is checked whether the left and right sides of each of the equations arising in the process of solving the original equation and raised to an even power have the same sign. Those roots for which the parts of any of the equations raised to an even power have different signs are extraneous for the original equation;

c) only those roots that belong to the domain of the original equation and for which both sides of each of the equations arising in the process of solving the original equation and raised to an even power have the same sign are checked by direct substitution into the original equation.

This method of solving with the specified verification method allows avoiding cumbersome calculations in the case of direct substitution of each of the found roots of the last equation into the original equation.