What are we losing or finding here?

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WHAT DO WE LOSE OR FIND HERE?

V. A. Petrov

Almost every written exam paper in mathematics includes some kind of equation. When solving it, the examinee must not only accurately check the calculations necessary to obtain the answer, but also be required to make sure that these calculations are correct. You need to be able to analyze each of the proposed transformations: does it lead to an equivalent equation, does it threaten the loss or acquisition of a solution.

Issues of tactics of “fighting” with built roots and with the loss of roots were considered in the article by A. G. Mordkovich “Someone loses, someone finds” («Kvant» No. 5, 1970) *). Now we invite you to check whether you are confident enough in this tactic. First, solutions to the equations are given. You need to analyze these decisions and answer the following question. Is the answer correct? If not, then what is here - are the roots lost or the structure found? Why? What is the correct answer? Compare your thoughts with ours on page 62.

1. $2\sin 4x = \sqrt{2(1 - \cos 4x)}$.

Solution. Using the formula
$$2\sin^2\alpha = 1 - \cos 2\alpha,$$
we get
$$2\sin 4x = \sqrt{4\sin^2 2x}$$
or
$$\sin 4x = \sin 2x,$$
which is already very easy to solve.

Answer: $x = n\pi$ and $x = \frac{\pi}{6} + n\frac{\pi}{3}$, where $n = 0, \pm 1, \pm 2, \dots$

*) This “tactics” is covered in more detail in the book by G.V. Dorofeev, M.K. Potanov, N.H. Rozov, very useful for applicants, “Mathematics Manual for Those Building Universities”, M., “Nauka”, 1970.

2. $8\cos x\cos 2x\cos 4x = 1$.

Solution. Multiplying both sides of the equation by sin $x$ and using the double angle sine formula three times, we get the simple equation
$$\sin 8x = \sin x.$$
Answer: $x = (2n + 1)\frac{\pi}{9}$ and $x = 2n\frac{\pi}{7}$, where $n$ is any integer.

3. $\sin \left(\frac{1}{5}\arccos x\right) = 1$.

Solution. From this equation we find
$$\arccos x = \frac{5\pi}{2} + 10n\pi$$
($n = 0, \pm 1, \pm 2, \dots$).

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From here, by the definition of arccos $x$, we obtain $x = \cos \left( \frac{5\pi}{2} + 10n\pi \right) = 0$.

Answer: $x = 0$.

4. $\log_{x-1} (x^2 - 5x + 7) = 1$.

Solution. Let’s find the ODZ:

$$x^2 - 5x + 7 > 0, \quad x - 1 > 0,$$

whence $x > 1$. According to the definition of a logarithmic function, we have $x^2 - 5x + 7 = x - 1; \quad x_1 = 2, \quad x_2 = 4$.

Since both values ​​are included in the ODZ, they are solutions to the original equation.

5. $\sqrt[3]{2x - 1} + \sqrt[3]{x - 1} = 1$.

Solution. Let’s cube both sides of the equation:

$$3x - 2 + 3\sqrt[3]{(2x - 1)(x - 1)} \times (\sqrt[3]{2x - 1} + \sqrt[3]{x - 1}) = 1.$$

The expression in parentheses represents the left side of the original equation - it can be replaced with the right side. After this, the equation is reduced to the form

$$(2x - 1) (x - 1) = (1 - x)^3.$$

Answer: $x_1 = 0, \quad x_2 = 1$.

6. Can the equation

$$\sqrt{x + a\sqrt{x + b}} + \sqrt{x} = c$$

have infinitely many solutions?

Solution. Move $\sqrt{x}$ to the right and square both sides:

$$x + a\sqrt{x + b} = c^2 - 2c\sqrt{x} + x.$$

After bringing similar terms we get: $(a + 2c)\sqrt{x} = c^2 - b$. For $a + 2c = 0$ and $c^2 - b = 0$ (and only in this case), the equation becomes an identity that is valid for all non-negative $x$.

Answer: The equation has infinitely many solutions at $a = -2c$ and $b = c^2$. The solutions in this case are $x \ge 0$.

7. $$\cos \frac{x}{2} \sqrt{x^2 + (7-\pi)x - 7\pi} = 1.$$

Solution. If, when raised to the power of some positive