Equations with parameters: graphical solution method
8-9 grades
The article discusses a graphical method for solving some equations with parameters, which is very effective when you need to establish how many roots an equation has depending on the parameter a.
Problem 1. How many roots does the equation have? | | x | – 2 | = a depending on parameter a?
Solution. In the coordinate system (x; y) we will construct graphs of the functions y = | | x | – 2 | and y = a. Graph of the function y = | | x | – 2 | shown in the figure.
The graph of the function y = a is a straight line parallel to the Ox axis or coinciding with it (if a = 0).
From the drawing it can be seen that:
If a= 0, then straight line y = a coincides with the Ox axis and has the graph of the function y = | | x | – 2 | two common points; This means that the original equation has two roots (in in this case roots can be found: x 1.2 = d 2).
If 0< a < 2, то прямая y = a имеет с графиком функции y = | | x | – 2 | четыре общие точки и, следовательно, original equation has four roots.
If a= 2, then the line y = 2 has three common points with the graph of the function. Then the original equation has three roots.
If a> 2, then straight line y = a will have two points with the graph of the original function, that is, this equation will have two roots.
If a < 0, то корней нет;
If a = 0, a> 2, then there are two roots;
If a= 2, then three roots;
if 0< a < 2, то четыре корня.
Problem 2. How many roots does the equation have? | x 2 – 2| x | – 3 | = a depending on parameter a?
Solution. In the coordinate system (x; y) we will construct graphs of the functions y = | x 2 – 2| x | – 3 | and y = a.
Graph of the function y = | x 2 – 2| x | – 3 | shown in the figure. The graph of the function y = a is a straight line parallel to Ox or coinciding with it (when a = 0).
From the drawing you can see:
If a= 0, then straight line y = a coincides with the Ox axis and has the graph of the function y = | x2 – 2| x | – 3 | two common points, as well as the straight line y = a will have with the graph of the function y = | x 2 – 2| x | – 3 | two common points at a> 4. So, when a= 0 and a> 4 the original equation has two roots.
If 0< a < 3, то прямая y = a has with the graph of the function y = | x 2 – 2| x | – 3 | four common points, as well as the straight line y= a will have four common points with the graph of the constructed function at a= 4. So, at 0< a < 3, a= 4 the original equation has four roots.
If a= 3, then straight line y = a intersects the graph of a function at five points; therefore, the equation has five roots.
If 3< a < 4, прямая y = a пересекает график построенной функции в шести точках; значит, при этих значениях параметра исходное уравнение имеет шесть корней.
If a < 0, уравнение корней не имеет, так как прямая y = a не пересекает график функции y = | x 2 – 2| x | – 3 |.
If a < 0, то корней нет;
If a = 0, a> 4, then there are two roots;
if 0< a < 3, a= 4, then four roots;
If a= 3, then five roots;
if 3< a < 4, то шесть корней.
Problem 3. How many roots does the equation have?
depending on parameter a?
Solution. Let us construct a graph of the function in the coordinate system (x; y) 
but first let's present it in the form:
The lines x = 1, y = 1 are asymptotes of the graph of the function. Graph of the function y = | x | + a obtained from the graph of the function y = | x | displacement by a units along the Oy axis.
Function graphs 
intersect at one point at a> – 1; This means that equation (1) for these parameter values has one solution.
At a = – 1, a= – 2 graphs intersect at two points; This means that for these parameter values, equation (1) has two roots.
At – 2< a < – 1, a < – 2 графики пересекаются в трех точках; значит, уравнение (1) при этих значениях параметра имеет три решения.
If a> – 1, then one solution;
If a = – 1, a= – 2, then there are two solutions;
if – 2< a < – 1, a < – 1, то три решения.
Comment. When solving equation (1) of problem 3, special attention should be paid to the case when a= – 2, since the point (– 1; – 1) does not belong to the graph of the function 
but belongs to the graph of the function y = | x | + a.
Let's move on to solving another problem.
Problem 4. How many roots does the equation have?
x + 2 = a| x – 1 | (2)
depending on parameter a?
Solution. Note that x = 1 is not a root given equation, since equality 3 = a· 0 cannot be true for any parameter value a. Let's divide both sides of the equation by | x – 1 |(| x – 1 | No. 0), then equation (2) will take the form 
In the coordinate system xOy we will plot the function 
The graph of this function is shown in the figure. Graph of the function y = a is a straight line parallel to the Ox axis or coinciding with it (if a = 0).
If aЈ – 1, then there are no roots;
if – 1< aЈ 1, then one root;
If a> 1, then there are two roots.
Let's consider the most complex equation.
Problem 5. At what values of the parameter a equation
a x 2 + | x – 1 | = 0 (3)
has three solutions?
Solution. 1. The control value of the parameter for this equation will be the number a= 0, at which equation (3) takes the form 0 + | x – 1 | = 0, whence x = 1. Therefore, when a= 0, equation (3) has one root, which does not satisfy the conditions of the problem.
2. Consider the case when a № 0.
Let us rewrite equation (3) in the following form: a x 2 = – | x – 1 |. Note that the equation will have solutions only when a < 0.
In the coordinate system xOy we will construct graphs of the functions y = | x – 1 | and y = a x 2 . Graph of the function y = | x – 1 | shown in the figure. Graph of the function y = a x 2 is a parabola whose branches are directed downward, since a < 0. Вершина параболы - точка (0; 0).
Equation (3) will have three solutions only when the straight line y = – x + 1 is tangent to the graph of the function y= a x 2 .
Let x 0 be the abscissa of the point of tangency of the straight line y = – x + 1 with the parabola y = a x 2 . The tangent equation has the form
y = y(x 0) + y "(x 0)(x – x 0).
Let's write down the tangency conditions:
This equation can be solved without using the concept of derivative.
Let's consider another method. Let us use the fact that if the straight line y = kx + b has a single common point with the parabola y = a x 2 + px + q, then the equation a x 2 + px + q = kx + b must have a unique solution, that is, its discriminant is zero. In our case we have the equation a x 2 = – x + 1 ( a No. 0). Discriminant equation
Problems to solve independently
6. How many roots does the equation have depending on the parameter a?
1)| | x | – 3 | = a;
2)| x + 1 | + | x + 2 | = a;
3)| x 2 – 4| x | + 3 | = a;
4)| x 2 – 6| x | + 5 | = a.
1) if a<0, то корней нет; если a=0, a>3, then two roots; If a=3, then three roots; if 0<a<3, то четыре корня;
2) if a<1, то корней нет; если a=1, then there is an infinite set of solutions from the interval [– 2; – 1]; If a> 1, then there are two solutions;
3) if a<0, то корней нет; если a=0, a<3, то четыре корня; если 0<a<1, то восемь корней; если a=1, then six roots; If a=3, then there are three solutions; If a>3, then there are two solutions;
4) if a<0, то корней нет; если a=0, 4<a<5, то четыре корня; если 0<a< 4, то восемь корней; если a=4, then six roots; If a=5, then three roots; If a>5, then there are two roots.
7. How many roots does the equation have | x + 1 | = a(x – 1) depending on parameter a?
Note. Since x = 1 is not a root of the equation, this equation can be reduced to the form 
.
Answer: if a J –1, a > 1, a=0, then one root; if – 1<a<0, то два корня; если 0<aЈ 1, then there are no roots.
8. How many roots does the equation x + 1 = have? a| x – 1 |depending on parameter a?
Draw a graph (see figure).
Answer: if aЈ –1, then there are no roots; if – 1<aЈ 1, then one root; If a>1, then there are two roots.
9. How many roots does the equation have?
2| x | – 1 = a(x – 1)
depending on parameter a?
Note. Reduce the equation to form 
Answer: if a J –2, a>2, a=1, then one root; if –2<a<1, то два корня; если 1<aЈ 2, then there are no roots.
10. How many roots does the equation have?
depending on parameter a?
Answer: if aЈ 0, a i 2, then one root; if 0<a<2, то два корня.
11. At what values of the parameter a equation
x 2 + a| x – 2 | = 0
has three solutions?
Note. Reduce the equation to the form x 2 = – a| x – 2 |.
Answer: when a J –8.
12. At what values of the parameter a equation
a x 2 + | x + 1 | = 0
has three solutions?
Note. Use problem 5. This equation has three solutions only if the equation a x 2 + x + 1 = 0 has one solution, and the case a= 0 does not satisfy the conditions of the problem, that is, the case remains when 
13. How many roots does the equation have?
x | x – 2 | = 1 – a
depending on parameter a?
Note. Reduce the equation to the form –x |x – 2| + 1 = a
depending on parameter a?
Note. Construct graphs of the left and right sides of this equation.
Answer: if a<0, a>2, then there are two roots; if 0Ј aЈ 2, then one root.
16. How many roots does the equation have?
depending on parameter a?
Note. Construct graphs of the left and right sides of this equation. To graph a function 
Let's find the intervals of constant sign of the expressions x + 2 and x:
Answer: if a>– 1, then one solution; If a= – 1, then there are two solutions; if – 3<a<–1, то четыре решения; если aЈ –3, then there are three solutions.
For each value of the parameter a a solve the inequality | 2 x + a | ≤ x + 2 |2x+a| \leq x+2 .
First, let's solve an auxiliary problem. Let's consider this inequality as an inequality with two variables x x and a a and draw on the coordinate plane x O a xOa all points whose coordinates satisfy the inequality.
If 2 x + a ≥ 0 2x+a \geq 0 (i.e. on the straight line a = - 2 x a=-2x and higher), then we get 2 x + a ≤ x + 2 ⇔ a ≤ 2 - x 2x+ a \leq x+2 \Leftrightarrow a \leq 2-x .
The set is shown in Fig. 11.
Now let's solve the original problem using this drawing. If we fix a a , then we get a horizontal line a = const a = \textrm(const) . To determine the values of x x, you need to find the abscissa of the points of intersection of this line with the set of solutions to the inequality. For example, if a = 8 a=8, then the inequality has no solutions (the straight line does not intersect the set); if a = 1 a=1 , then the solutions are all x x from the interval [ - 1 ; 1 ] [-1;1], etc. So, three options are possible.
1) If $$a>4$$, then there are no solutions.
2) If a = 4 a=4, then x = - 2 x=-2.
ANSWER
at $$a
for a = 4 a=4 - x = - 2 x=-2 ;
for $$a>4$$ - there are no solutions.
Find all values of the parameter a a for which the inequality $$3-|x-a| > x^2$$ a) has at least one solution; b) has at least one positive solution.
Let's rewrite the inequality in the form $$3-x^2 > |x-a)$$. Let's construct graphs of the left and right parts on the x O y xOy plane. The graph of the left side is a parabola with branches downwards with the vertex at the point (0; 3) (0;3) . The graph intersects the x-axis at points (± 3 ; 0) (\pm \sqrt(3);0) . The graph of the right side is an angle with the vertex on the x-axis, the sides of which are directed upward at an angle of 45 ° 45^(\circ) to the coordinate axes. The abscissa of the vertex is the point x = a x=a .
a) In order for an inequality to have at least one solution, it is necessary and sufficient that at least at one point the parabola is above the graph y = | x - a | y=|x-a| . This is accomplished if the vertex of the angle lies between points A A and B B of the abscissa axis (see Fig. 12 - points A A and B B are not included). Thus, it is necessary to determine at what position of the vertex one of the branches of the angle touches the parabola.
Let's consider the case when the vertex of the corner is at point A A . Then the right branch of the angle touches the parabola. Its slope is equal to one. This means that the derivative of the function y = 3 - x 2 y = 3-x^2 at the point of tangency is equal to 1 1, i.e. - 2 x = 1 -2x=1, whence x = - 1 2 x = -\frac( 1)(2) . Then the ordinate of the tangent point is y = 3 - (1 2) 2 = 11 4 y = 3 - (\frac(1)(2))^2 = \frac(11)(4) . The equation of a straight line having an angular coefficient k = 1 k=1 and passing through a point with coordinates (- 1 2 ; 11 4) (-\frac(1)(2); \frac(11)(4)) is the following * ( \^* : y - 11 4 = 1 · (x + 1 2) y - \frac{11}{4} = 1 \cdot (x+ \frac{1}{2}) , откуда y = x + 13 4 y = x + \frac{13}{4} .!}
This is the equation of the right branch of the corner. The abscissa of the point of intersection with the x axis is equal to - 13 4 -\frac(13)(4), i.e. point A A has coordinates A (- 13 4 ; 0) A(-\frac(13)(4); 0) . For reasons of symmetry, point B B has coordinates: B (13 4 ; 0) B(\frac(13)(4); 0) .
From here we get that a ∈ (- 13 4 ; 13 4) a\in (-\frac(13)(4); \frac(13)(4)) .
b) The inequality has positive solutions if the vertex of the corner is located between the points F F and B B (see Fig. 13). Finding the position of the point F F is not difficult: if the vertex of the corner is at the point F F, then its right branch (the straight line given by the equation y = x - a y = x-a passes through the point (0; 3) (0;3). From here we find that a = - 3 a=-3 and point F F has coordinates (- 3 ; 0) (-3;0) . Therefore, a ∈ (- 3 ; 13 4) a \in (-3; \frac(13)(4) ) .
ANSWER
a) a ∈ (- 13 4 ; 13 4) , a\in (-\frac(13)(4); \frac(13)(4)),\:\:\: b) a ∈ (- 3 ; 13 4) a \in (-3; \frac(13)(4)) .
* {\^* Полезные формулы: !}
- \-- a straight line passing through the point (x 0 ; y 0) (x_0;y_0) and having an angular coefficient k k is given by the equation y - y 0 = k (x - x 0) y-y_0=k(x-x_0 ) ;
- \-- the angular coefficient of the straight line passing through the points (x 0 ; y 0) (x_0;y_0) and (x 1 ; y 1) (x_1;y_1), where x 0 ≠ x 1 x_0 \neq x_1, is calculated by formula k = y 1 - y 0 x 1 - x 0 k = \dfrac(y_1-y_0)(x_1-x_0) .
Comment. If you need to find the value of the parameter at which the straight line y = k x + l y=kx+l and the parabola y = a x 2 + b x + c y = ax^2+bx+c touch, then you can write the condition that the equation k x + l = a x 2 + b x + c kx+l = ax^2+bx+c has exactly one solution. Then another way to find the values of the parameter a a for which the vertex of the angle is at point A A is the following: equation x - a = 3 - x 2 x-a = 3-x^2 has exactly one solution ⇔ D = 1 + 4 (a + 3) = 0 ⇔ a = - 13 4 \Leftrightarrow D = 1 + 4(a+3) = 0 \Leftrightarrow a = -\ dfrac(13)(4) .
Please note that in this way it is impossible to write down the condition for a line to touch an arbitrary graph. For example, the line y = 3 x - 2 y = 3x - 2 touches the cubic parabola y = x 3 y=x^3 at the point (1 ; 1) (1;1) and intersects it at the point (- 2 ; - 8) (-2;-8), i.e. the equation x 3 = 3 x + 2 x^3 = 3x+2 has two solutions.
Find all values of the parameter a a , for each of which the equation (a + 1 - | x + 2 |) (x 2 + 4 x + 1 - a) = 0 (a+1-|x+2|)(x^2 +4x+1-a) = 0 has a) exactly two distinct roots; b) exactly three different roots.
Let's do the same as in example 25. Let's depict the set of solutions to this equation on the plane x O a xOa . It is equivalent to the combination of two equations:
1) a = | x + 2 | - 1 a = |x+2| -1 is an angle with branches up and the vertex at the point (- 2 ; - 1) (-2;-1) .
2) a = x 2 + 4 x + 1 a = x^2 + 4x + 1 - this is a parabola with branches up and the vertex at the point (- 2 ; - 3) (-2;-3) . See fig. 14.
We find the intersection points of two graphs. The right branch of the angle is given by the equation y = x + 1 y=x+1 . Solving the equation
x + 1 = x 2 + 4 x + 1 x+1 = x^2+4x+1
we find that x = 0 x=0 or x = - 3 x=-3 . Only the value x = 0 x=0 is suitable (since for the right branch x + 2 ≥ 0 x+2 \geq 0). Then a = 1 a=1 . Similarly, we find the coordinates of the second intersection point - (- 4 ; 1) (-4; 1) .
Let's return to the original problem. The equation has exactly two solutions for those a a for which the horizontal line a = const a=\textrm(const) intersects the set of solutions to the equation at two points. From the graph we see that this is true for a ∈ (- 3 ; - 1) ∪ ( 1 ) a\in (-3;-1)\bigcup\(1\) . There will be exactly three solutions in the case of three intersection points, which is only possible when a = - 1 a=-1 .
ANSWER
a) a ∈ (- 3 ; - 1) ∪ ( 1 ) ; a\in (-3;-1)\bigcup\(1\);\:\:\: b) a = - 1 a=-1 .
$$\begin(cases) x^2-x-a \leq 0,\\ x^2+2x-6a \leq 0 \end(cases) $$
has exactly one solution.
Let us depict the solutions of the system of inequalities on the plane x O a xOa . Let's rewrite the system in the form $$ \begin(cases) a \leq -x^2+x,\\ a \geq \dfrac(x^2+6x)(6) .\end(cases) $$
The first inequality is satisfied by points lying on the parabola a = - x 2 + x a = -x^2+x and below it, and the second is satisfied by points lying on the parabola a = x 2 + 6 x 6 a = \dfrac(x^2 +6x)(6) and above. We find the coordinates of the vertices of the parabolas and their intersection points, and then build a graph. The top of the first parabola is (1 2 ; 1 4) (\dfrac(1)(2);\dfrac(1)(4)), the top of the second parabola is (- 1 ; - 1 6) (-1; -\dfrac( 1)(6)), the intersection points are (0; 0) (0;0) and (4 7; 12 49) (\dfrac(4)(7); \dfrac(12)(49)). The set of points satisfying the system is shown in Fig. 15. It can be seen that the horizontal line a = const a=\textrm(const) has exactly one common point with this set (which means the system has exactly one solution) in the cases a = 0 a=0 and a = 1 4 a= \dfrac(1)(4) .
ANSWER
A = 0 , a = 1 4 a=0,\: a=\dfrac(1)(4)
Find the smallest value of the parameter a a , for each of which the system
$$\begin(cases) x^2+y^2 + 3a^2 = 2y + 2\sqrt(3)ax,\\ \sqrt(3)|x|-y=4 \end(cases) $$
has a unique solution.
Let's transform the first equation, highlighting complete squares:
(x 2 - 2 3 a x + 3 a 2) + (y 2 - 2 y + 1) = 1 ⇔ (x - a 3) 2 + (y - 1) 2 = 1. 18 (x^2- 2\sqrt(3)ax+3a^2)+(y^2-2y+1)=1 \Leftrightarrow (x-a\sqrt(3))^2+(y-1)^2 =1. \:\:\:\left(18\right)
Unlike previous problems, here it is better to depict a drawing on the x O y xOy plane (a drawing in the “variable - parameter” plane is usually used for problems with one variable and one parameter - the result is a set on the plane. In this problem we are dealing with two variables and a parameter. Drawing a set of points (x; y; a) (x;y;a) in three-dimensional space is a difficult task; moreover, such a drawing is unlikely to be visual). Equation (18) specifies a circle with center (a 3 ; 1) (a\sqrt(3);1) of radius 1. The center of this circle, depending on the value of a a, can be located at any point on the line y = 1 y=1.
The second equation of the system is y = 3 | x | - 4 y = \sqrt(3)|x|-4 sets the angle with the sides up at an angle of 60 ° 60^(\circ) to the abscissa axis (the angular coefficient of the straight line is the tangent of the angle of inclination tg 60 ° = 3 \textrm(tg )(60^(\circ)) = \sqrt(3)), with the vertex at the point (0; - 4) (0;-4) .
This system of equations has exactly one solution if the circle touches one of the branches of the angle. This is possible in four cases (Fig. 16): the center of the circle can be at one of the points A A, B B, C C, D D. Since we need to find the smallest value of the parameter a a , we are interested in the abscissa of the point D D . Consider the right triangle D H M DHM. The distance from point D D to straight line H M HM is equal to the radius of the circle, therefore D H = 1 DH=1. So, D M = D H sin 60 ° = 2 3 DM=\dfrac(DH)(\textrm(sin)(60^(\circ))) = \dfrac(2)(\sqrt(3)) . The coordinates of the point M M are found as the coordinates of the intersection point of two lines y = 1 y=1 and y = - 3 x - 4 y=-\sqrt(3)x-4 (left side of the corner).
We get M (- 5 3) M(-\dfrac(5)(\sqrt(3))) . Then the abscissa of point D D is equal to - 5 3 - 2 3 = - 7 3 -\dfrac(5)(\sqrt(3))-\dfrac(2)(\sqrt(3))=-\dfrac(7)(\ sqrt(3)) .
Since the abscissa of the center of the circle is equal to a 3 a\sqrt(3) , it follows that a = - 7 3 a=-\dfrac(7)(3) .
ANSWER
A = - 7 3 a=-\dfrac(7)(3)
Find all values of the parameter a a , for each of which the system
$$\begin(cases) |4x+3y| \leq 12a,\\ x^2+y^2 \leq 14ax +6ay -57a^2+16a+64 \end(cases) $$
has exactly one solution.
Let us depict the sets of solutions to each of the inequalities on the plane x O y xOy .
In the second inequality, we select perfect squares:
x 2 - 14 a x + 49 + y 2 - 6 a y + 9 a 2 ≤ a 2 + 16 a + 64 ⇔ (x - 7 a) 2 + (y - 3 a) 2 ≤ (a + 8) 2 (19 ) x^2-14ax+49 + y^2-6ay + 9a^2 \leq a^2 + 16a + 64 \Leftrightarrow (x-7a)^2+(y-3a)^2 \leq (a+8 )^2 \:\:\:\: (19)
When a + 8 = 0 a+8=0 (a = - 8 a=-8), inequality (19) specifies a point with coordinates (7 a ; 3 a) (7a;3a), i.e. (- 56 ; - 24) (-56;-24) . For all other values of a a (19) defines a circle centered at the point (7 a ; 3 a) (7a;3a) of radius | a + 8 | |a+8| .
Let's consider the first inequality.
1) For negative a a it has no solutions. This means that the system has no solutions.
2) If a = 0 a=0, then we get the straight line 4 x + 3 y = 0 4x+3y=0. From the second inequality we get a circle with a center (0; 0) (0; 0) of radius 8. Obviously, there is more than one solution.
3) If $$a>0$$, then this inequality is equivalent to the double inequality - 12 a ≤ 4 x + 3 y ≤ 12 a -12a \leq 4x+3y \leq 12a . It defines a strip between two straight lines y = ± 4 a - 4 x 3 y=\pm 4a -\dfrac(4x)(3) , each of which is parallel to the straight line 4 x + 3 y = 0 4x+3y=0 (Fig. 17).
Since we are considering $$a>0$$, the center of the circle is located in the first quarter on the line y = 3 x 7 y = \dfrac(3x)(7) . Indeed, the coordinates of the center are x = 7 a x=7a , y = 3 a y=3a ; expressing a a and equating, we get x 7 = y 3 \dfrac(x)(7)=\dfrac(y)(3) , whence y = 3 x 7 y = \dfrac(3x)(7) . In order for the system to have exactly one solution, it is necessary and sufficient that the circle touches the straight line a 2 a_2 . This happens when the radius of the circle is equal to the distance from the center of the circle to the straight line a 2 a_2. According to the formula for the distance from a point to a line * (\^{*} получаем, что расстояние от точки (7 a ; 3 a) (7a;3a) до прямой 4 x + 3 y - 12 a = 0 4x+3y-12a=0 равно | 4 · 7 a + 3 · 3 a - 12 a | 4 2 + 3 2 = 5 a \dfrac{|4\cdot 7a + 3\cdot 3a -12a|}{\sqrt{4^2+3^2}} = 5\left|a\right| . Приравнивая к радиусу круга, получаем 5 a = | a + 8 | 5{a} = |a+8| . Так как $$a>0$$, опускаем модули и находим, что a = 2 a=2 .!}
ANSWER
A = 2 a=2
* {\^{*} Пусть даны точка M (x 0 ; y 0) M (x_0;y_0) и прямая l l , заданная уравнением a x + b y + c = 0 ax+by+c=0 . Тогда расстояние от точки M M до прямой l l определяется формулой ρ = | a x 0 + b x 0 + c | a 2 + b 2 \rho = \dfrac{|ax_0+bx_0+c|}{\sqrt{a^2+b^2}} . !}
At what values of the parameter a a does the system
$$\begin(cases) |x|+|y|=1,\\ |x+a|+|y+a|=1 \end(cases)$$ has no solutions?
The first equation of the system defines the square A B C D ABCD on the plane x O y xOy (to construct it, consider x ≥ 0 x\geq 0 and y ≥ 0 y\geq 0 . Then the equation takes the form x + y = 1 x+y=1 . We obtain a segment - part of the straight line x + y = 1 x+y=1, lying in the first quarter. Next, we reflect this segment relative to the O x Ox axis, and then reflect the resulting set relative to the O y Oy axis (see Fig. 18). The second equation defines the square P Q R S PQRS , equal to the square A B C D ABCD , but centered at the point (- a ; - a) (-a;-a) . In Fig. As an example, Fig. 18 shows this square for a = - 2 a=-2. The system has no solutions if these two squares do not intersect.
It is easy to see that if the segments P Q PQ and B C BC coincide, then the center of the second square is at the point (1; 1) (1;1). Those values of a a are suitable for us, at which the center is located “above” and “to the right”, i.e. $$a1$$.
ANSWER
A ∈ (- ∞ ; - 1) ∪ (1 ; + ∞) a\in (-\infty;-1)\bigcup(1;+\infty) .
Find all values of the parameter b b for which the system
$$\begin(cases) y=|b-x^2|,\\ y=a(x-b) \end(cases) $$
has at least one solution for any value of a a .
Let's consider several cases.
1) If $$b2) If b = 0 b=0 , then the system takes the form $$\begin(cases) y=x^2,\\ y=ax .\end(cases) $$
For any a a the pair of numbers (0 ; 0) (0;0) is a solution to this system, therefore b = 0 b=0 is suitable.
3) Let us fix some $$b>0$$. The first equation is satisfied by the set of points obtained from the parabola y = x 2 - b y=x^2-b by reflecting part of this parabola relative to the O x Ox axis (see Fig. 19a, b). The second equation specifies a family of straight lines (by substituting different values of a a , you can get all kinds of straight lines passing through the point (b ; 0) (b;0) , except for the vertical one), passing through the point (b ; 0) (b;0) . If the point (b ; 0) (b;0) lies on the segment [ - b ; b ] [-\sqrt(b);\sqrt(b)] . abscissa axis, then the straight line intersects the graph of the first function for any slope (Fig. 19a). Otherwise (Fig. 19b) in any case there will be a straight line that does not intersect this graph. Solving the inequality - b ≤ b ≤ b -\sqrt(b)\leq b \leq \sqrt(b) and taking into account that $$b>0$$, we obtain that b ∈ (0 ; 1 ] b \in ( 0;1] .
We combine the results: $$b \in $$.
ANSWER
$$b \in $$
Find all values of a a , for each of which the function f (x) = x 2 - | x - a 2 | - 3 x f(x) = x^2-|x-a^2|-3x has at least one maximum point.
Expanding the module, we get that
$$f(x) = \begin(cases) x^2-4x+a^2, \:\:\: x\geq a^2 ,\\ x^2-2x-a^2, \:\ :\: x\leq a^2 . \end(cases) $$
On each of the two intervals, the graph of the function y = f (x) y=f(x) is a parabola with branches upward.
Since parabolas with upward branches cannot have maximum points, the only possibility is that the maximum point is the boundary point of these intervals - the point x = a 2 x=a^2 . At this point there will be a maximum if the vertex of the parabola y = x 2 - 4 x + a 2 y=x^2-4x+a^2 falls on the interval $$x>a^2$$, and the vertex of the parabola y = x 2 - 2 x - a 2 y=x^2-2x-a^2 - for the interval $$x\lt a^2$$ (see Fig. 20). This condition is given by the inequalities and $$2 \gt a^2$$ and $$1 \lt a^2$$, solving which we find that a ∈ (- 2 ; 1) ∪ (1 ; 2) a\in (-\ sqrt(2);1)\bigcup(1;\sqrt(2)) .
ANSWER
A ∈ (- 2 ; 1) ∪ (1 ; 2) a\in (-\sqrt(2);1)\bigcup(1;\sqrt(2))
Find all values of a a , for each of which the general solutions of the inequalities
y + 2 x ≥ a y+2x \geq a and y - x ≥ 2 a (20) y-x \geq 2a \:\:\:\:\:\:\:\: (20)
are solutions to the inequality
$$2y-x>a+3 \:\:\:\:\:\:\:\:\: (21)$$
To navigate the situation, it is sometimes useful to consider one parameter value. Let's make a drawing, for example, for a = 0 a=0 . Inequalities (20) (in fact, we are dealing with a system of inequalities (20)) are satisfied by the points of the angle B A C BAC (see Fig. 21) - points, each of which lies above both straight lines y = - 2 x y=-2x and y = x y =x (or on these lines). Inequality (21) is satisfied by points lying above the straight line y = 1 2 x + 3 2 y = \dfrac(1)(2)x + \dfrac(3)(2) . It can be seen that when a = 0 a=0 the condition of the problem is not satisfied.
What will change if we take a different value for the parameter a a ? Each of the lines will move and turn into a line parallel to itself, since the angular coefficients of the lines do not depend on a a. In order for the condition of the problem to be fulfilled, the entire angle B A C BAC must lie above the straight line l l . Since the angular coefficients of the straight lines A B AB and A C AC are greater in absolute value than the angular coefficient of the straight line l l , it is necessary and sufficient that the vertex of the angle lies above the straight line l l .
Solving a system of equations
$$\begin(cases) y+2x=a,\\ y-x=2a, \end(cases)$$
find the coordinates of point A (- a 3 ; 5 a 3) A(-\dfrac(a)(3);\dfrac(5a)(3)) . They must satisfy inequality (21), so $$\dfrac(10a)(3)+\dfrac(a)(3) > a+3$$, whence $$a>\dfrac(9)(8)$$ .
ANSWER
$$a>\dfrac(9)(8)$$
TO tasks with parameter This may include, for example, the search for solutions to linear and quadratic equations in general form, the study of the equation for the number of roots available depending on the value of the parameter.
Without giving detailed definitions, consider the following equations as examples:
y = kx, where x, y are variables, k is a parameter;
y = kx + b, where x, y are variables, k and b are parameters;
ax 2 + bx + c = 0, where x are variables, a, b and c are a parameter.
Solving an equation (inequality, system) with a parameter means, as a rule, solving an infinite set of equations (inequalities, systems).
Tasks with a parameter can be divided into two types:
A) the condition says: solve the equation (inequality, system) - this means, for all values of the parameter, find all solutions. If at least one case remains uninvestigated, such a solution cannot be considered satisfactory.
b) it is required to indicate the possible values of the parameter at which the equation (inequality, system) has certain properties. For example, it has one solution, has no solutions, has solutions belonging to the interval, etc. In such tasks, it is necessary to clearly indicate at what parameter value the required condition is satisfied.
The parameter, being an unknown fixed number, has a kind of special duality. First of all, it is necessary to take into account that the assumed popularity indicates that the parameter must be perceived as a number. Secondly, the freedom to manipulate the parameter is limited by its obscurity. For example, operations of dividing by an expression that contains a parameter or extracting the root of an even degree from such an expression require preliminary research. Therefore, care is required when handling the parameter.
For example, to compare two numbers -6a and 3a, you need to consider three cases:
1) -6a will be greater than 3a if a is a negative number;
2) -6a = 3a in the case when a = 0;
3) -6a will be less than 3a if a is a positive number 0.
The solution will be the answer.
Let the equation kx = b be given. This equation is a short form for an infinite number of equations with one variable.
When solving such equations there may be cases:
1. Let k be any real number not equal to zero and b be any number from R, then x = b/k.
2. Let k = 0 and b ≠ 0, the original equation will take the form 0 x = b. Obviously, this equation has no solutions.
3. Let k and b be numbers equal to zero, then we have the equality 0 x = 0. Its solution is any real number.
Algorithm for solving this type of equation:
1. Determine the “control” values of the parameter.
2. Solve the original equation for x for the parameter values that were determined in the first paragraph.
3. Solve the original equation for x for parameter values different from those chosen in the first paragraph.
4. You can write the answer in the following form:
1) for ... (parameter values), the equation has roots ...;
2) for ... (parameter values), there are no roots in the equation.
Example 1.
Solve the equation with the parameter |6 – x| = a.
Solution.
It is easy to see that a ≥ 0 here.
According to the rule of module 6 – x = ±a, we express x:
Answer: x = 6 ± a, where a ≥ 0.
Example 2.
Solve the equation a(x – 1) + 2(x – 1) = 0 with respect to the variable x.
Solution.
Let's open the brackets: aх – а + 2х – 2 = 0
Let's write the equation in standard form: x(a + 2) = a + 2.
If the expression a + 2 is not zero, i.e. if a ≠ -2, we have the solution x = (a + 2) / (a + 2), i.e. x = 1.
If a + 2 is equal to zero, i.e. a = -2, then we have the correct equality 0 x = 0, so x is any real number.
Answer: x = 1 for a ≠ -2 and x € R for a = -2.
Example 3.
Solve the equation x/a + 1 = a + x with respect to the variable x.
Solution.
If a = 0, then we transform the equation to the form a + x = a 2 + ax or (a – 1)x = -a(a – 1). The last equation for a = 1 has the form 0 x = 0, therefore x is any number.
If a ≠ 1, then the last equation will take the form x = -a.
This solution can be illustrated on the coordinate line (Fig. 1)
Answer: there are no solutions for a = 0; x – any number with a = 1; x = -a for a ≠ 0 and a ≠ 1.
Let's consider another way to solve equations with a parameter - graphically. This method is used quite often.
Example 4.
Depending on the parameter a, how many roots does the equation ||x| – 2| = a?
Solution.
To solve using the graphical method, we construct graphs of the functions y = ||x| – 2| and y = a (Fig. 2).
The drawing clearly shows possible cases of the location of the straight line y = a and the number of roots in each of them.
Answer: the equation will not have roots if a< 0; два корня будет в случае, если a >2 and a = 0; the equation will have three roots in the case of a = 2; four roots – at 0< a < 2.
Example 5.
At what a the equation 2|x| + |x – 1| = a has a single root?
Solution.
Let us depict the graphs of the functions y = 2|x| + |x – 1| and y = a. For y = 2|x| + |x – 1|, expanding the modules using the interval method, we obtain:
(-3x + 1, at x< 0,
y = (x + 1, for 0 ≤ x ≤ 1,
(3x – 1, for x > 1.
On Figure 3 It is clearly seen that the equation will have a single root only when a = 1.
Answer: a = 1.
Example 6.
Determine the number of solutions to the equation |x + 1| + |x + 2| = a depending on parameter a?
Solution.
Graph of the function y = |x + 1| + |x + 2| will be a broken line. Its vertices will be located at points (-2; 1) and (-1; 1) (Figure 4).
Answer: if the parameter a is less than one, then the equation will not have roots; if a = 1, then the solution to the equation is an infinite set of numbers from the interval [-2; -1]; if the values of parameter a are greater than one, then the equation will have two roots.
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Olga Otdelkina, 9th grade student
This topic is an integral part of the school algebra course. The purpose of this work is to study this topic in more depth, to identify the most rational solution that quickly leads to an answer. This essay will help other students understand the use of the graphical method for solving equations with parameters, learn about the origin and development of this method.
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Introduction2
Chapter 1. Equations with a parameter
History of the emergence of equations with parameter3
Vieta's theorem4
Basic concepts5
Chapter 2. Types of equations with parameters.
Linear equations6
Quadratic equations……………………………………………………………......7
Chapter 3. Methods for solving equations with a parameter
Analytical method….…………………………………………......8
Graphic method. History of origin….…………………………9
Algorithm for solving the graphical method..…………….....…………….10
Solution of the equation with modulus………………...…………………………….11
Practical part…………………...……………………………………12
Conclusion……………………………………………………………………………….19
References……………………………………………………………20
Introduction.
I chose this topic because it is an integral part of the school algebra course. In preparing this work, I set the goal of a deeper study of this topic, identifying the most rational solution that quickly leads to an answer. My essay will help other students understand the use of the graphical method for solving equations with parameters, learn about the origin and development of this method.
In modern life, the study of many physical processes and geometric patterns often leads to solving problems with parameters.
For solving such equations, the graphical method is very effective when you need to determine how many roots the equation has depending on the parameter α.
Problems with parameters are of purely mathematical interest, contribute to the intellectual development of students, and serve as good material for practicing skills. They have diagnostic value, since they can be used to test knowledge of the main branches of mathematics, the level of mathematical and logical thinking, initial research skills and promising opportunities for successfully mastering a mathematics course in higher educational institutions.
My essay discusses frequently encountered types of equations, and I hope that the knowledge I gained in the process of work will help me when passing school exams, becauseequations with parametersare rightfully considered one of the most difficult problems in school mathematics. It is precisely these tasks that are included in the list of tasks in the Unified State Examination.
History of the emergence of equations with a parameter
Problems on equations with a parameter were already encountered in the astronomical treatise “Aryabhattiam”, compiled in 499 by the Indian mathematician and astronomer Aryabhatta. Another Indian scientist, Brahmagupta (7th century), outlined a general rule for solving quadratic equations reduced to a single canonical form:
αx 2 + bx = c, α>0
The coefficients in the equation, except for the parameter, can also be negative.
Quadratic equations by al-Khwarizmi.
In algebraic treatise al-Khwarizmi gives a classification of linear and quadratic equations with parameter a. The author counts 6 types of equations, expressing them as follows:
1) “Squares are equal to roots,” i.e. αx 2 = bx.
2) “Squares are equal to numbers”, i.e. αx 2 = c.
3) “The roots are equal to the number,” i.e. αx = c.
4) “Squares and numbers are equal to roots,” i.e. αx 2 + c = bx.
5) “Squares and roots are equal to the number”, i.e. αx 2 + bx = c.
6) “Roots and numbers are equal to squares,” i.e. bx + c = αx 2 .
Formulas for solving quadratic equations according to al-Khwarizmi in Europe were first set forth in the “Book of Abacus,” written in 1202 by the Italian mathematician Leonardo Fibonacci.
The derivation of the formula for solving a quadratic equation with a parameter in general form is available from Vieta, but Vieta recognized only positive roots. Italian mathematicians Tartaglia, Cardano, Bombelli were among the first in the 12th century. In addition to positive ones, negative roots are also taken into account. Only in the 17th century. Thanks to the works of Girard, Descartes, Newton and other scientists, the method of solving quadratic equations took on its modern form.
Vieta's theorem
A theorem expressing the relationship between the parameters, coefficients of a quadratic equation and its roots, named after Vieta, was first formulated by him in 1591 as follows: “If b + d multiplied by α minus α 2 , is equal to bc, then α is equal to b and equal to d.”
To understand Vieta, we should remember that α, like any vowel letter, meant the unknown (our x), while the vowels b, d are coefficients for the unknown. In the language of modern algebra, the above Vieta formulation means:
If there is
(α + b)x - x 2 = αb,
That is, x 2 - (α -b)x + αb =0,
then x 1 = α, x 2 = b.
By expressing the relationship between the roots and coefficients of equations with general formulas written using symbols, Vieta established uniformity in methods for solving equations. However, the symbolism of Viet is still far from its modern form. He did not recognize negative numbers and therefore, when solving equations, he considered only cases where all the roots were positive.
Basic Concepts
Parameter - an independent variable, the value of which is considered a fixed or arbitrary number, or a number belonging to the interval specified by the condition of the problem.
Equation with parameter— mathematicalequation, the appearance and solution of which depends on the values of one or more parameters.
Decide equation with parameter means for each valuefind the values of x that satisfy this equation, and also:
- 1. Investigate at what parameter values the equation has roots and how many there are for different parameter values.
- 2. Find all expressions for the roots and indicate for each of them those parameter values at which this expression actually determines the root of the equation.
Consider the equation α(x+k)= α +c, where α, c, k, x are variable quantities.
System of permissible values of variables α, c, k, xis any system of variable values in which both the left and right sides of this equation take real values.
Let A be the set of all admissible values of α, K the set of all admissible values of k, X the set of all admissible values of x, C the set of all admissible values of c. If for each of the sets A, K, C, X we select and fix, respectively, one value α, k, c, and substitute them into the equation, then we obtain an equation for x, i.e. equation with one unknown.
The variables α, k, c, which are considered constant when solving an equation, are called parameters, and the equation itself is called an equation containing parameters.
The parameters are denoted by the first letters of the Latin alphabet: α, b, c, d, ..., k, l, m, n, and the unknowns are denoted by the letters x, y, z.
Two equations containing the same parameters are called equivalent if:
a) they make sense for the same parameter values;
b) every solution to the first equation is a solution to the second and vice versa.
Types of equations with parameters
Equations with parameters are: linear and square.
1) Linear equation. General view:
α x = b, where x is unknown;α, b - parameters.
For this equation, the special or control value of the parameter is the one at which the coefficient of the unknown becomes zero.
When solving a linear equation with a parameter, cases are considered when the parameter is equal to its special value and different from it.
A special value of the parameter α is the valueα = 0.
1.If, and ≠0, then for any pair of parametersα and b it has a unique solution x = .
2.If, and =0, then the equation takes the form:0 x = b . In this case the value b = 0 is a special parameter value b.
2.1. At b ≠ 0 the equation has no solutions.
2.2. At b =0 the equation will take the form:0 x =0.
The solution to this equation is any real number.
Quadratic equation with parameter.
General view:
α x 2 + bx + c = 0
where parameter α ≠0, b and c - arbitrary numbers
If α =1, then the equation is called a reduced quadratic equation.
The roots of a quadratic equation are found using the formulas
Expression D = b 2 - 4 α c is called a discriminant.
1. If D> 0, the equation has two different roots.
2. If D< 0 — уравнение не имеет корней.
3. If D = 0, the equation has two equal roots.
Methods for solving equations with a parameter:
- Analytical - a method of direct solution, repeating standard procedures for finding the answer in an equation without parameters.
- Graphic - depending on the conditions of the problem, the position of the graph of the corresponding quadratic function in the coordinate system is considered.
Analytical method
Solution algorithm:
- Before you begin solving a problem with parameters using the analytical method, you need to understand the situation for a specific numerical value of the parameter. For example, take the value of the parameter α =1 and answer the question: is the value of the parameter α =1 required for this task.
Example 1. Solve relatively X linear equation with parameter m:
According to the meaning of the problem (m-1)(x+3) = 0, that is, m= 1, x = -3.
Multiplying both sides of the equation by (m-1)(x+3), we get the equation
We get
Hence, at m= 2.25.
Now we need to check whether there are any values of m for which
the value of x found is -3.
solving this equation, we find that x is equal to -3 with m = -0.4.
Answer: with m=1, m =2.25.
Graphic method. History of origin
The study of common dependencies began in the 14th century. Medieval science was scholastic. With this nature, there was no room left for the study of quantitative dependencies; it was only about the qualities of objects and their connections with each other. But among the scholastics a school arose that argued that qualities can be more or less intense (the dress of a person who has fallen into a river is wetter than that of someone who has just been caught in the rain)
The French scientist Nikolai Oresme began to depict intensity with the lengths of segments. When he placed these segments perpendicular to a certain straight line, their ends formed a line, which he called the “line of intensity” or the “line of the upper edge” (graph of the corresponding functional dependence). Oresme even studied “planar” and “physical” qualities, i.e. functions , depending on two or three variables.
Oresme's important achievement was his attempt to classify the resulting graphs. He identified three types of qualities: Uniform (with constant intensity), uniform-uneven (with a constant rate of change in intensity) and uneven-uneven (all others), as well as the characteristic properties of the graphs of such qualities.
To create a mathematical apparatus for studying the graphs of functions, the concept of a variable was needed. This concept was introduced into science by the French philosopher and mathematician Rene Descartes (1596-1650). It was Descartes who came up with the ideas about the unity of algebra and geometry and the role of variables; Descartes introduced a fixed unit segment and began to consider the relationships of other segments to it.
Thus, graphs of functions over the entire period of their existence have gone through a number of fundamental transformations, which led them to the form to which we are accustomed. Each stage or stage in the development of graphs of functions is an integral part of the history of modern algebra and geometry.
The graphical method of determining the number of roots of an equation depending on the parameter included in it is more convenient than the analytical one.
Solving algorithm by graphical method
Graph of a function - a set of points at whichabscissaare valid argument values, A ordinates- corresponding valuesfunctions.
Algorithm for graphically solving equations with a parameter:
- Find the domain of definition of the equation.
- We express α as a function of x.
- In the coordinate system we build a graph of the functionα (x) for those values of x that are included in the domain of definition of this equation.
- Finding the intersection points of a lineα =с, with the graph of the function
α(x). If the line α =с crosses the graphα (x), then we determine the abscissas of the intersection points. To do this, it is enough to solve the equation c = α (x) relative to x.
- Write down the answer
Solving equations with modulus
When solving equations with modulus containing a parameter, graphically, it is necessary to construct graphs of functions and consider all possible cases for different values of the parameter.
For example, │х│= a,
Answer: if a < 0, то нет корней, a > 0, then x = a, x = - a, if a = 0, then x = 0.
Problem solving.
Problem 1. How many roots does the equation have?| | x | - 2 | =a depending on parameter a?
Solution. In the coordinate system (x; y) we will construct graphs of the functions y = | | x | - 2 | and y = a . Graph of the function y = | | x | - 2 | shown in the figure.
Graph of the function y =α a = 0).
From the graph it can be seen that:
If a = 0, then straight line y = a coincides with the Ox axis and has the graph of the function y = | | x | - 2 | two common points; this means that the original equation has two roots (in this case, the roots can be found: x 1,2
= + 2).
If 0<
a
< 2, то прямая y =
α
has with the graph of the function y = | | x | - 2 | four common points and, therefore, the original equation has four roots.
If a = 2, then the line y = 2 has three common points with the graph of the function. Then the original equation has three roots.
If a > 2, then straight line y = a will have two points with the graph of the original function, that is, this equation will have two roots.
Answer: if a < 0, то корней нет;
if a = 0, a > 2, then there are two roots;
if a = 2, then there are three roots;
if 0<
a
< 2, то четыре корня.
Problem 2. How many roots does the equation have?| x 2 - 2| x | - 3 | =a depending on parameter a?
Solution. In the coordinate system (x; y) we will construct graphs of the functions y = | x 2 - 2| x | - 3 | and y = a.
Graph of the function y = | x 2 - 2| x | - 3 | shown in the figure. Graph of the function y =α is a straight line parallel to Ox or coinciding with it (when a = 0).
From the graph you can see:
If a = 0, then straight line y = a coincides with the Ox axis and has the graph of the function y = | x2 - 2| x | - 3 | two common points, as well as the straight line y = a will have with the graph of the function y = | x 2
- 2| x | - 3 | two common points at a > 4. So, for a = 0 and a > 4 the original equation has two roots.
If 0<
a< 3, то прямая y =
a
has with the graph of the function y = | x 2
- 2| x | - 3 | four common points, as well as the straight line y= a will have four common points with the graph of the constructed function at a = 4. So, at 0<
a
< 3,
a
= 4 the original equation has four roots.
If a = 3, then straight line y = a intersects the graph of a function at five points; therefore, the equation has five roots.
If 3<
a< 4, прямая y =
α
intersects the graph of the constructed function at six points; This means that for these parameter values the original equation has six roots.
If a < 0, уравнение корней не имеет, так как прямая y =
α
does not intersect the graph of the function y = | x 2 - 2| x | - 3 |.
Answer: if a < 0, то корней нет;
if a = 0, a > 4, then there are two roots;
if 0<
a
< 3,
a
= 4, then four roots;
if a = 3, then five roots;
if 3<
a
< 4, то шесть корней.
Problem 3. How many roots does the equation have?
depending on parameter a?
Solution. Let us construct a graph of the function in the coordinate system (x; y)
but first let's present it in the form:
The lines x = 1, y = 1 are asymptotes of the graph of the function. Graph of the function y = | x | + a obtained from the graph of the function y = | x | displacement by a units along the Oy axis.
Function graphs intersect at one point at a > - 1; This means that equation (1) for these parameter values has one solution.
When a = - 1, a = - 2 graphs intersect at two points; This means that for these parameter values, equation (1) has two roots.
At - 2<
a< - 1,
a
< - 2 графики пересекаются в трех точках; значит, уравнение (1) при этих значениях параметра имеет три решения.
Answer: if a > - 1, then one solution;
if a = - 1, a = - 2, then there are two solutions;
if - 2<
a
< - 1,
a
< - 1, то три решения.
Comment. When solving the problem equation, special attention should be paid to the case when a = - 2, since the point (- 1; - 1) does not belong to the graph of the functionbut belongs to the graph of the function y = | x | + a.
Problem 4. How many roots does the equation have?
x + 2 = a | x - 1 |
depending on parameter a?
Solution. Note that x = 1 is not a root of this equation, since the equality 3 = a 0 cannot be true for any parameter value a . Let's divide both sides of the equation by | x - 1 |(| x - 1 |0), then the equation takes the formIn the coordinate system xOy we will plot the function
The graph of this function is shown in the figure. Graph of the function y = a is a straight line parallel to the Ox axis or coinciding with it (if a = 0).
