Many problems with a parameter are reduced to the study of a square trinomial, so let's consider these problems in more detail.
I. When solving the simplest problems, a formula for the roots of a quadratic equation and Vieta's theorem is sufficient.
For what values of the parameter a a the set of solutions of the inequality $$x^2+ax-1
Since the coefficient of x 2 x^2 is positive, the solution to the inequality is the interval between the roots in the case $$D > 0$$ and the empty set if D ≤ 0 D \leq 0 .
Find the discriminant: D = a 2 + 4 D = a^2+4 ($$D>0$$ for all a a). Then the set of solutions is the interval
x ∈ (- a - a 2 + 4 2 ; - a + a 2 + 4 2) x \in (\dfrac(-a-\sqrt(a^2+4))(2); \dfrac(-a+ \sqrt(a^2+4))(2)) . It is required that the length of this interval be equal to 5, i.e.
A + a 2 + 4 2 = - a - a 2 + 4 2 + 5 ⇔ a 2 + 4 = 5 ⇔ a = ± 21 \dfrac(-a+\sqrt(a^2+4))(2) = \ dfrac(-a-\sqrt(a^2+4))(2) + 5 \Leftrightarrow \sqrt(a^2+4)=5 \Leftrightarrow a = \pm \sqrt(21) .
ANSWER
A = ± 21 a = \pm \sqrt(21)
For what values of the parameter p p does the equation x 2 + p 2 + 4 p x + p - 1 x^2+\sqrt(p^2+4p)\cdot x +p-1 have roots and the sum of the squares of the roots is minimal?
The sum of the squares of the roots of the equation is conveniently expressed using the Vieta theorem:
x 1 2 + x 2 2 = (x 1 + x 2) 2 - 2 x 1 x 2 = (- p 2 + 4 p) 2 - 2 (p - 1) = p 2 + 2 p + 2 x_1^2 +x_2^2 = (x_1+x_2)^2-2x_1x_2=(-\sqrt(p^2+4p))^2-2(p-1) = p^2 +2p + 2 .
But before applying the Vieta theorem, it is imperative to check that the equation has roots! To do this, we calculate the discriminant: D = p 2 + 4 p - 4 (p - 1) = p 2 + 4 D = p^2+4p-4(p-1) = p^2+4 . We see that the discriminant is positive for any allowed values p p , i.e. at
p ∈ (- ∞ ; - 4 ] ∪ [ 0 ; + ∞) (5) p \in (-\infty; -4]\bigcup, etc.), in which you need to draw a drawing yourself and draw the appropriate conclusions.
Remarks. 1. For equations and inequalities of the form
$$ax^2 + bx + c = 0,\: ax^2 + bx + c > 0, \: ax^2 + bx + c we need to consider the case a = 0 a =0 separately. Then it will turn out linear equation (inequality).
2. In most tasks, it is important to consider sign numbers a a - the direction of the branches of the parabola depends on this.
3. Note that the totality of two systems
$$\begin(cases) a > 0, \\ f(a) > 0 \end(cases) and \begin(cases) a
is equivalent to the inequality $$a f(a) > 0$$. Therefore, in the condition 1 ° 1^(\circ) one can write down one system $$\begin(cases) D>0, \\ a f(A) > 0, \\ x_(\text(c))
Other conditions can be simplified similarly:
$$2^(\circ) \Leftrightarrow \begin(cases) D>0, \\ a f(A) > 0, \\ x_(\text(c)) > A .\end(cases) \:\:\ : 3^(\circ) \Leftrightarrow a f(A) 0, \\ a f(A) > 0, \\ a f(B) > 0, \\ A
Let's move on to examples.
For which a a the equation (2 a - 2) x 2 + (a + 1) x + 1 = 0 (2a-2)x^2 + (a+1)x +1 = 0 has roots, and they all belong to the interval (-2; 0) (-2; 0) ?
1) If 2 a - 2 = 0 (a = 1) 2a-2=0\:(a=1) , then the equation becomes 2 x + 1 = 0 2x+1=0 . This equation has a single root x = - 0.5 x=-0.5 , which belongs to the interval (- 2 ; 0) (-2; 0) . Hence, a = 1 a =1 satisfies the condition of the problem.
2) If 2 a - 2 ≠ 0 2a-2 \neq 0 , then the equation is quadratic. Finding the discriminant:
D = (a + 1) 2 - 4 (2 a - 2) = a 2 - 6 a + 9 = (a - 3) 2 D=(a+1)^2-4(2a-2)=a^ 2-6a+9=(a-3)^2 .
Since the discriminant is a perfect square, we find the roots (as a rule, the above methods with the location of the roots are convenient to use if the formulas for the roots are cumbersome. If the discriminant is a perfect square and the roots are “good”, then it is easier to solve the problem directly):
To fulfill the conditions of the problem, it is required that the inequality $$-2 \dfrac(3)(2)$$ be satisfied.
ANSWER
A ∈ ( 1 ) ∪ (3 2 ; + ∞) a \in \(1\)\bigcup (\dfrac(3)(2); +\infty) .
For what values of a a does the inequality $$4^(\textrm(sin)\:x)-2\cdot (a-3) \cdot 2^(\textrm(sin)\:x) + a+3 > 0$$ for all x x ?
Denote 2 sin x = y 2^(\textrm(sin)\:x)=y . Since - 1 ≤ sin x ≤ 1 -1 \leq \textrm(sin)\:x \leq 1 , we get that 1 2 ≤ 2 sin x ≤ 2 \dfrac(1)(2) \leq 2^(\textrm (sin)\:x) \leq 2 . The original inequality takes the form
$$y^2-2(a-3)y+(a+3) > 0$$
This problem is equivalent to the following one: “for which a a the inequality $$y^2-2(a-3)y+(a+3) > 0$$ is satisfied for all y ∈ [ 1 2 ; 2 ] y \in [\dfrac(1)(2);2] ?
The plot of the left side of this inequality is a parabola with upward branches. The requirements of the task will be fulfilled in two cases. 1) $$D
a) This location of the parabola (the roots are to the left of the segment [ 1 2 ; 2 ] [\dfrac(1)(2);2]) is given by the conditions (we write down and solve the system):
$$\begin(cases) D \geq 0,\\ x_(\text(c)) 0 \end(cases) \Leftrightarrow \begin(cases) (a-3)^3-(a+3) \geq 0,\\ a-3 0 \end(cases) \Leftrightarrow \begin(cases) a \in (-\infty;1]\bigcup]6;+\infty),\\ a 0 \end(cases) \ Leftrightarrow a \leq 1 $$.
b) This case is given by $$D
c) Similarly to case a), we obtain the system:
$$\!\!\!\! \begin(cases) D \geq 0,\\ x_(\text(c)) > 2,\\ f(2) > 0 \end(cases) \Leftrightarrow \begin(cases) (a-3)^3 -(a+3) \geq 0,\\ a-3 > 2,\\ 4 - 4(a-3) +a+3 > 0 \end(cases) \Leftrightarrow \begin(cases) a\in ( -\infty;1]\bigcup ?
1) Consider the case a = 0 a = 0 (then the equation is not quadratic). The equation becomes - 5 x - 6 = 0 -5x-6=0 . Roots on the interval [ 0 ; 2 ] is not, so a = 0 a = 0 is not suitable.
2) The equation is quadratic. Denote the left side of the equation by f (x) f(x) . The equation has on the segment [ 0 ; 2 ] exactly one root in two cases.
A) The equation has a single root, and it belongs to the segment [ 0 ; 2]. This is possible when D = 0 D = 0 . We calculate the discriminant:
D = (2 a - 5) 2 - 4 a (a - 6) = 4 a + 25 D = (2a-5)^2-4a(a-6) = 4a+25 .
The discriminant vanishes at a = - 25 4 a=-\dfrac(25)(4) . In this case, the original equation takes the form - 25 4 x 2 - 35 2 x - 49 4 = 0 -\dfrac(25)(4)x^2-\dfrac(35)(2)x - \dfrac(49)(4 ) = 0 , whence x = - 7 5 x = -\dfrac(7)(5) . Roots on the interval [ 0 ; 2 ] no, it means that this case is not realized for any values of the parameter a a .
B) The equation has two roots ($$D>0 \Leftrightarrow a>-\dfrac(25)(4)$$), one of which belongs to the segment [ 0 ; 2] , while the other does not. To fulfill this condition, it is necessary and sufficient that either (a) the function f (x) f(x) takes at the ends of the segment [ 0 ; 2] values of different signs - then the root lies in the interval (0; 2) (0; 2) (as an example (you can consider other possible locations of the parabola yourself) see Fig. 7), or (b) at one of the ends segment vanishes - then the root lies on one of the ends of the segment.
(a) The condition “the numbers f (0) f(0) and f (2) f(2) have different signs” is equivalent to the inequality $$f(0)\cdot f(2)
$$\left(a-6\right)\left(4a+2\left(2a-5\right)+\left(a-6\right)\right)
(b) If f (0) = 0 f(0) = 0 , then a = 6 a=6 . Then the equation becomes 6 x 2 + 7 x = 0 6x^2+7x=0 . Its roots are the numbers x = 0 x=0 and x = - 7 6 x=-\dfrac(7)(6) , i.e. on the segment [ 0 ; 2 ] it has exactly one root.
If f (2) = 0 f(2) = 0 , then a = 16 9 a=\dfrac(16)(9) . Then we get 16 9 x 2 - 13 9 x - 38 9 = 0 \dfrac(16)(9)x^2 - \dfrac(13)(9)x - \dfrac(38)(9) = 0 , whence x = 2 x=2 or x = - 19 16 x=-\dfrac(19)(16) , i.e. again only one of the two roots belongs to the segment [ 0 ; 2].
Hence, both values a = 6 a=6 and a = 16 9 a=\dfrac(16)(9) and satisfy the condition of the problem (for f (2) = 0 f(2) = 0 or f (0) = 0 f(0) = 0 be sure to find the second root and see if it is on the segment [ 0 ; 2 ] ).
Combining the results, we obtain a ∈ [ 16 9 ; 6 ] a\in [\dfrac(16)(9); 6].
ANSWER
16 9 ≤ a ≤ 6 \dfrac(16)(9) \leq a \leq 6
For what values of the parameter a a the equation | x 2 - 4 | x | + 3 | = a |x^2-4|x|+3| = a has exactly 8 solutions?
Let's plot the graphs of the left and right parts on the xOy plane.
To plot the left side, first draw a parabola y = x 2 - 4 x + 3 y = x^2-4x+3 . Then we reflect all the points of this parabola, which lie below the abscissa axis, relative to this axis and get the graph of the function y = | x 2 - 4 x + 3 | y=|x^2-4x+3| (Fig. 8a). Next, we discard all points lying to the left of the abscissa axis, and reflect the remaining points relative to this axis - we get the graph of the function y = | x 2 - 4 | x | + 3 | y=|x^2-4|x|+3| .
The plot of the right side is a horizontal line y = a y=a . The equation has 8 solutions when this line intersects the graph y = | x 2 - 4 | x | + 3 | y=|x^2-4|x|+3| at eight points. It is easy to see that this happens at $$0 ANSWER
A ∈ (0 ; 1) a\in (0;1)
Find all values of the parameter p p for which the equation 4 x + 2 x + 2 + 7 = p - 4 - x - 2 2 1 - x 4^x+2^(x+2)+7=p-4^( -x)-2\cdot 2^(1-x) has at least one solution.
Rewrite the equation as (4 x + 4 - x) + 4 (2 x + 2 - x) = p - 7 (4^x+4^(-x))+4\cdot (2^x+2^ (-x))=p-7 and make the substitution 2 x + 2 - x = t 2^x+2^(-x)=t . Squaring both sides of the last equality, we get that t 2 = (2 x + 2 - x) 2 = 4 x + 2 + 4 - x t^2=(2^x+2^(-x))^2= 4^x+2+4^(-x) , whence 4 x + 4 - x = t 2 - 2 4^x+4^(-x) = t^2-2 . The equation becomes t 2 - 2 + 4 t = p - 7 ⇔ (t + 2) 2 = p - 1 t^2-2+4t = p-7 \Leftrightarrow (t+2)^2 = p-1 .
Find the set of values of the left side of the equation. Since (we use the fact that the sum of two reciprocal positive numbers is at least two: a + 1 a ≥ 2 a+\dfrac(1)(a) \geq 2 for $$a>0$$ 0 (equality is possible only for a = 1 a = 1) This can be proved, for example, using the Cauchy inequality: for positive numbers, the arithmetic mean is not less than the geometric mean (a 1 + a 2 + . . . + a k k ≥ a 1 a 2 . . a k k) ( \dfrac(a_1+a_2+...+a_k)(k) \geq \sqrt[k](a_1\cdot a_2\cdot .. \cdot a_k)) , and equality is achieved only in the case a 1 = a 2 = . . . = a k a_1=a_2=...=a_k For two positive numbers, this inequality becomes a + b 2 ≥ a b \dfrac(a+b)(2) \geq \sqrt(ab) If b is substituted here = 1 a b = \dfrac(1)(a) , then we get the required inequality.) t ≥ 2 t \geq 2 , we find that the left side of the equation takes values from the interval [ 16 ; +∞) .
Decision. We transform the left side of this inequality as follows:
(2-x) a 2 + (x 2 -2x + 3) a-3x \u003d ax 2 - a 2 x - 2ax + 2a 2 + 3a - 3x \u003d
Ax (x - a) -2a (x - a) - 3 (x-a) \u003d (x - a) (ax - 2a - 3).
This inequality will take the form: (x - a) (ax - 2a - 3) ≥ 0.
If a = 0, we get - Зх ≥ 0 x ≤ 0.
If a ≠ 0, then -3 a
As a 0, then the solution to this inequality will be the interval of the numerical axis located between the roots of the equation corresponding to the inequality.
Let's find out the mutual arrangement of numbers a and , taking into account the condition - 3 ≤ a
3 ≤a
A = -1.
In all considered cases, we present the solutions of this inequality depending on the values of the parameter:
We get that only x = -1 is the solution to this inequality for any value of the parameter a .
Answer: -1
- Conclusion.
Why I chose the project on the topic “Development guidelines solutions of quadratic equations and inequalities with parameters”? Since when solving any trigonometric, exponential, logarithmic equations, inequalities, systems, we most often come to consider sometimes linear, and most often quadratic equations and inequalities. When solving the most complex problems with parameters, most tasks are reduced with the help of equivalent transformations to the choice of solutions of the type: a (x - a) (x - c) > 0 (
We have reviewed theoretical basis for solving quadratic equations and inequalities with parameters. We remembered the necessary formulas and transformations, considered various arrangements of graphs of a quadratic function depending on the value of the discriminant, on the sign at the highest coefficient, on the location of the roots, the top of the parabola. We identified a scheme for solving and selecting results, compiled a table.
The project shows analytical and graphical methods for solving quadratic equations and inequalities. Students in a vocational school need visual perception of the material for better assimilation of the material. It is shown how you can change the variable x and accept the parameter as an equal value.
For a visual assimilation of this topic, the solution of 8 tasks with parameters, 1 - 2 for each section, is considered. Example 1 considers the number of solutions for different values parameter, in example No. 3, the solution of the quadratic equation is analyzed under a variety of initial conditions. For solutions square inequalities made a graphic illustration. In example No. 5, the method of replacing the parameter as an equal value is used. The project includes consideration of example No. 8 from the tasks included in section C, for intensive preparation for passing the exam.
For high-quality training of students in solving problems with parameters, it is recommended to use multimedia technologies in full, namely: use presentations for lectures, electronic textbooks and books, own developments from the media library. Binary lessons of mathematics + computer science are very effective. The Internet is an irreplaceable assistant to the teacher and the student. The presentation requires imported objects from existing educational resources. The most convenient and acceptable in work is the DER "Using Microsoft Office at school".
The development of methodological recommendations on this topic will facilitate the work of young teachers who came to work at the school, replenish the teacher's portfolio, serve as a model for special subjects, sample solutions will help students cope with complex tasks.
- Literature.
1. Gorshtein P.I., Polonsky V.B., Yakir M.S. Tasks with parameters. "Ileksa", "Gymnasium", Moscow - Kharkov, 2002.
2. Balayan E.N. Collection of tasks in mathematics to prepare for the exam and olympiads. 9-11 grades. "Phoenix", Rostov-on-Don, 2010.
3. Yastrebinetsky G.A. Tasks with parameters. M., "Enlightenment", 1986.
4. Kolesnikova S.I. Mathematics. Solving complex problems of the Unified State Exam. M. "IRIS - press", 2005.
5. Rodionov E.M., Sinyakova S.L. Mathematics. Allowance for entering universities. Training center "Landmark" MSTU. N.E. Bauman, M., 2004.
6. Scanavi M.I. Collection of tasks in mathematics for applicants to universities: In 2 books. Book 1, M., 2009.
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Solving inequalities with a parameter.
Inequalities that have the form ax > b, ax< b, ax ≥ b, ax ≤ b, где a и b – действительные числа или выражения, зависящие от параметров, а x – неизвестная величина, называются linear inequalities.
Solution principles linear inequalities with a parameter are very similar to the principles of the solution linear equations with a parameter.
Example 1
Solve the inequality 5x - a > ax + 3.
Decision.
First, let's transform the original inequality:
5x - ax > a + 3, we take x out of brackets on the left side of the inequality:
(5 - a) x > a + 3. Now consider the possible cases for the parameter a:
If a > 5 then x< (а + 3) / (5 – а).
If a = 5, then there are no solutions.
If a< 5, то x >(a + 3) / (5 - a).
This solution will be the answer to the inequality.
Example 2
Solve the inequality x(a - 2) / (a - 1) - 2a / 3 ≤ 2x - a for a ≠ 1.
Decision.
Let's transform the original inequality:
x(a - 2) / (a - 1) - 2x ≤ 2a/3 - a;
Ah/(a – 1) ≤ -a/3. Multiply by (-1) both parts of the inequality, we get:
ax/(a – 1) ≥ a/3. Let's explore the possible cases for the parameter a:
1 case. Let a/(a – 1) > 0 or a € (-∞; 0)ᴗ(1; +∞). Then x ≥ (а – 1)/3.
2nd case. Let a/(а – 1) = 0, i.e. a = 0. Then x is any real number.
3rd case. Let a/(а – 1)< 0 или а € (0; 1). Тогда x ≤ (а – 1)/3.
Answer: x € [(a - 1) / 3; +∞) for a € (-∞; 0)ᴗ(1; +∞);
x € [-∞; (a – 1)/3] for a € (0; 1);
x € R for a = 0.
Example 3
Solve the inequality |1 + x| ≤ ax with respect to x.
Decision.
It follows from the condition that the right side of the inequality ax must be non-negative, i.e. ax ≥ 0. By the rule of expansion of the module from the inequality |1 + x| ≤ ax we have a double inequality
Ax ≤ 1 + x ≤ ax. We rewrite the result in the form of a system:
(ax ≥ 1 + x;
(-ax ≤ 1 + x.
Let's transform to the form:
((а – 1)x ≥ 1;
((a + 1)x ≥ -1.
We investigate the resulting system on intervals and at points (Fig. 1):
For a ≤ -1 x € (-∞; 1/(a - 1)].
At -1< а < 0 x € [-1/(а – 1); 1/(а – 1)].
When a \u003d 0 x \u003d -1.
At 0< а ≤ 1 решений нет.
Graphical method for solving inequalities
Plotting greatly simplifies the solution of equations containing a parameter. The use of the graphical method in solving inequalities with a parameter is even clearer and more expedient.
Graphical solution of inequalities of the form f(x) ≥ g(x) means finding the values of the variable x for which the graph of the function f(x) lies above the graph of the function g(x). To do this, it is always necessary to find the intersection points of the graphs (if they exist).
Example 1
Solve the inequality |x + 5|< bx.
Decision.
We build graphs of functions y = |x + 5| and y = bx (Fig. 2). The solution of the inequality will be those values of the variable x for which the graph of the function y = |x + 5| will be below the graph of the function y = bx. 
The figure shows:
1) For b > 1, the lines intersect. The abscissa of the intersection point of the graphs of these functions is the solution of the equation x + 5 = bx, whence x = 5/(b - 1). The graph y \u003d bx is higher for x from the interval (5 / (b - 1); +∞), which means that this set is the solution to the inequality.
2) Similarly, we find that at -1< b < 0 решением является х из интервала (-5/(b + 1); 5/(b – 1)).
3) For b ≤ -1 x € (-∞; 5/(b - 1)).
4) For 0 ≤ b ≤ 1, the graphs do not intersect, which means that the inequality has no solutions.
Answer: x € (-∞; 5/(b - 1)) for b ≤ -1;
x € (-5/(b + 1); 5/(b – 1)) at -1< b < 0;
there are no solutions for 0 ≤ b ≤ 1; x € (5/(b – 1); +∞) for b > 1.
Example 2
Solve the inequality a(a + 1)x > (a + 1)(a + 4).
Decision.
1) Let's find the "control" values for the parameter a: a 1 = 0, a 2 = -1.
2) Let's solve this inequality on each subset of real numbers: (-∞; -1); (-one); (-ten); (0); (0; +∞).
a) a< -1, из данного неравенства следует, что х >(a + 4)/a;
b) a \u003d -1, then this inequality will take the form 0 x > 0 - there are no solutions;
c)-1< a < 0, из данного неравенства следует, что х < (a + 4)/a;
d) a = 0, then this inequality has the form 0 x > 4 – there are no solutions;
e) a > 0, this inequality implies that x > (a + 4)/a.
Example 3
Solve the inequality |2 – |x||< a – x.
Decision.
We plot the function y = |2 – |x|| (Fig. 3) and consider all possible cases of the location of the line y \u003d -x + a. 
Answer: the inequality has no solutions for a ≤ -2;
x € (-∞; (a - 2)/2) with a € (-2; 2];
x € (-∞; (a + 2)/2) for a > 2.
When solving various problems, equations and inequalities with parameters, a significant number of heuristic techniques open up, which can then be successfully applied in any other branches of mathematics.
Tasks with parameters play an important role in shaping logical thinking and mathematical culture. That is why, having mastered the methods of solving problems with parameters, you will successfully cope with other problems.
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