Examples of irrational inequalities and methods for solving them. Irrational inequalities

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Any inequality that includes a function under the root is called irrational. There are two types of such inequalities:

In the first case, the root less function g (x), in the second - more. If g(x) - constant, the inequality is greatly simplified. Please note: outwardly these inequalities are very similar, but their solution schemes are fundamentally different.

Today we will learn how to solve irrational inequalities of the first type - they are the simplest and most understandable. The inequality sign can be strict or non-strict. The following statement is true for them:

Theorem. Any irrational inequality of the form

Equivalent to the system of inequalities:

Not weak? Let's look at where this system comes from:

  1. f (x) ≤ g 2 (x) - everything is clear here. This is the original inequality squared;
  2. f (x) ≥ 0 is the ODZ of the root. Let me remind you: the arithmetic square root exists only from non-negative numbers;
  3. g(x) ≥ 0 is the range of the root. By squaring inequality, we burn away the negatives. As a result, extra roots may appear. The inequality g(x) ≥ 0 cuts them off.

Many students “get hung up” on the first inequality of the system: f (x) ≤ g 2 (x) - and completely forget the other two. The result is predictable: wrong decision, lost points.

Since irrational inequalities are enough complex topic, let's look at 4 examples at once. From basic to really complex. All problems are taken from entrance exams Moscow State University named after M. V. Lomonosov.

Examples of problem solving

Task. Solve the inequality:

Before us is a classic irrational inequality: f(x) = 2x + 3; g(x) = 2 is a constant. We have:

Of the three inequalities, only two remained at the end of the solution. Because the inequality 2 ≥ 0 always holds. Let's cross the remaining inequalities:

So, x ∈ [−1.5; 0.5]. All points are shaded because the inequalities are not strict.

Task. Solve the inequality:

We apply the theorem:

Let's solve the first inequality. To do this, we will reveal the square of the difference. We have:

2x 2 − 18x + 16< (x − 4) 2 ;
2x 2 − 18x + 16< x 2 − 8x + 16:
x 2 − 10x< 0;
x (x − 10)< 0;
x ∈ (0; 10).

Now let's solve the second inequality. There too quadratic trinomial:

2x 2 − 18x + 16 ≥ 0;
x 2 − 9x + 8 ≥ 0;
(x − 8)(x − 1) ≥ 0;
x ∈ (−∞; 1]∪∪∪∪)

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