|
x + 1 +
3
√
x + 2 +
3
√
x + 3 = 0.
• Solve the equation 4(
√
x
2
− 1)
3
− 3x
2
√
x
2
− 1 = x
3
.
• Solve the inequality
3
2x − x
√
x − 1 +
√
x +
3
√
1
− 2x ≤ 0.
(Dorofeev, Sedova, and Troitskaya, 2010, pp. 44, 47, 51)
The first of these problems allows for a mental solution based on
the properties of the monotonic function — the left-hand side of the
equation is an increasing function, and therefore it assumes the value 0
at no more than one point. By trial and error, it is possible to determine
that x = −2 is a root of this equation. Thus, this is the only solution
to this equation.
The solution to the second of these problems becomes noticeably
more simple if one uses the trigonometric substitution x =
1
cos t
. As for
the third problem, a “trained eye” will see that substituting
3
√
1
− 2x = a,
√
x = b,
leads to a simpler inequality, a + b ≤
3
√
a
3
+ b
3
. After raising the
inequality to the third power and making the appropriate identity
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transformations, we obtain an expression of the form f(a) ≤ 0, where
f(a) is a second-degree polynomial whose roots relative to the variable
a are obvious: a = 0 and a = − b. After this, we can return to the
original variable and, for example, use the interval method.
4.2.4
The final attestation in algebra for 11th graders
Upon completing their mathematics education, all graduates of sec-
ondary general education schools in Russia must take the Uniform
State Exam (USE). Without going into a detailed discussion on the
structure and aims of the USE here,
1
we will confine ourselves to
describing several problems from the USE just on the two topics
examined in this chapter.
The transformation of algebraic expressions. The “difficult” part (C)
of the exam contains no problems specifically on this topic, although
the solutions to problems in other topics require sufficient proficiency
in carrying out transformations of algebraic expressions. The following
problem is typical of an easier section:
Find the value of the expression
2
√
7
2
14
.
As can be seen, the students’ ability to operate with roots of degree
n is tested on quite primitive examples. The problems have a purely
technical character.
Equations and inequalities. The problems in this “easy” section on
a different topic also presuppose command of the standard algorithm.
The knowledge provided by the basic course is sufficient to solve them.
Consider the following example:
Find the root of the equation
√
−72 − 17x = −x. If the equation has
more than one root, indicate the lesser of them.
1
Editorial note: For a more detailed treatment of the USE, see Chapter 8 of this
volume.
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On Algebra Education in Russian Schools
187
The following problem may serve as an example from a more
difficult section of the exam (in this section, this problem is about
average in difficulty):
Solve the inequality
√
7
− x <
√
x
3
−6x
2
+14x−7
√
x−1
.
Formally, this problem is only for advanced-course graduates,
since the basic course does not address solving irrational inequalities.
Although the probability that this problem will be solved by basic-level
graduates is by no means zero — if it does not scare them off
immediately, this inequality can be solved by graduates who have
completed the basic program (it is not so difficult to see that one must
solve the inequality 8x − x
2
− 7 < x
3
− 6x
2
+ 14x − 7 on the interval
(0, 7]) — the general tendency to learn standard algorithms by rote
and to apply them “head-on” can do the student a disservice. Thus, the
student who, after studying the basic course decides to do the “college”
part of the exam as well, can turn out to be psychologically unprepared
for such work.
The “difficult part” of the exam contains many problems pertaining
to material that is shared by the basic and advanced courses. Their
formulations are thus understandable to all students. However, without
a developed capacity for mathematical thought, without the skills
associated with advanced mathematical activity, it is unrealistic for
students to hope to solve them. To some extent, it may be said
that students who have completed the basic course, but who by the
time they graduate from high school have decided for one or another
reason to go on to colleges that require applicants to take a USE in
mathematics, are thus given a chance to display their giftedness.
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