Linear equations with a parameter. Solving equations with a parameter in mathematics Solving equations with given parameters
1. Systems linear equations with parameter
Systems of linear equations with a parameter are solved by the same basic methods as ordinary systems of equations: the substitution method, the method of adding equations, and the graphical method. Knowing the graphical interpretation of linear systems makes it easy to answer the question about the number of roots and their existence.
Example 1
Find all values for the parameter a for which the system of equations has no solutions.
(x + (a 2 - 3) y \u003d a,
(x + y = 2.
Solution.
Let's look at several ways to solve this problem.
1 way. We use the property: the system has no solutions if the ratio of the coefficients in front of x is equal to the ratio of the coefficients in front of y, but not equal to the ratio of free terms (a/a 1 = b/b 1 ≠ c/c 1). Then we have:
1/1 \u003d (a 2 - 3) / 1 ≠ a / 2 or a system
(and 2 - 3 = 1,
(a ≠ 2.
From the first equation a 2 \u003d 4, therefore, taking into account the condition that a ≠ 2, we get the answer.
Answer: a = -2.
2 way. We solve by the substitution method.
(2 - y + (a 2 - 3) y \u003d a,
(x = 2 - y,
((a 2 - 3) y - y \u003d a - 2,
(x = 2 - y.
After taking the common factor y out of brackets in the first equation, we get:
((a 2 - 4) y \u003d a - 2,
(x = 2 - y.
The system has no solutions if the first equation has no solutions, that is
(and 2 - 4 = 0,
(a - 2 ≠ 0.
It is obvious that a = ±2, but taking into account the second condition, only the answer with a minus is given.
Answer: a = -2.
Example 2
Find all values for the parameter a for which the system of equations has an infinite number of solutions.
(8x + ay = 2,
(ax + 2y = 1.
Solution.
By property, if the ratio of the coefficients at x and y is the same, and is equal to the ratio of the free members of the system, then it has an infinite set of solutions (i.e., a / a 1 \u003d b / b 1 \u003d c / c 1). Hence 8/a = a/2 = 2/1. Solving each of the equations obtained, we find that a \u003d 4 is the answer in this example.
Answer: a = 4.
2. Systems of rational equations with a parameter
Example 3
(3|x| + y = 2,
(|x| + 2y = a.
Solution.
Multiply the first equation of the system by 2:
(6|x| + 2y = 4,
(|x| + 2y = a.
Subtract the second equation from the first, we get 5|х| = 4 – a. This equation will have a unique solution for a = 4. In other cases, this equation will have two solutions (for a< 4) или ни одного (при а > 4).
Answer: a = 4.
Example 4
Find all values of the parameter a for which the system of equations has a unique solution.
(x + y = a,
(y - x 2 \u003d 1.
Solution.
We will solve this system using the graphical method. So, the graph of the second equation of the system is a parabola, lifted up along the Oy axis by one unit segment. The first equation defines the set of lines parallel to the line y = -x (picture 1). The figure clearly shows that the system has a solution if the straight line y \u003d -x + a is tangent to the parabola at the point with coordinates (-0.5; 1.25). Substituting these coordinates into the equation of a straight line instead of x and y, we find the value of the parameter a:
1.25 = 0.5 + a;
Answer: a = 0.75.
Example 5
Using the substitution method, find out at what value of the parameter a, the system has a unique solution.
(ax - y \u003d a + 1,
(ax + (a + 2)y = 2.
Solution.
Express y from the first equation and substitute it into the second:
(y \u003d ah - a - 1,
(ax + (a + 2) (ax - a - 1) = 2.
We bring the second equation to the form kx = b, which will have a unique solution for k ≠ 0. We have:
ax + a 2 x - a 2 - a + 2ax - 2a - 2 \u003d 2;
a 2 x + 3ax \u003d 2 + a 2 + 3a + 2.
The square trinomial a 2 + 3a + 2 can be represented as a product of brackets
(a + 2)(a + 1), and on the left we take x out of brackets:
(a 2 + 3a) x \u003d 2 + (a + 2) (a + 1).
Obviously, a 2 + 3a must not be equal to zero, therefore,
a 2 + 3a ≠ 0, a(a + 3) ≠ 0, which means a ≠ 0 and ≠ -3.
Answer: a ≠ 0; ≠ -3.
Example 6
Using the graphical solution method, determine at what value of the parameter a, the system has a unique solution.
(x 2 + y 2 = 9,
(y - |x| = a.
Solution.
Based on the condition, we build a circle with a center at the origin of coordinates and a radius of 3 unit segments, it is this circle that sets the first equation of the system
x 2 + y 2 \u003d 9. The second equation of the system (y \u003d | x | + a) is a broken line. By using figure 2 we consider all possible cases of its location relative to the circle. It is easy to see that a = 3.
Answer: a = 3.
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To tasks with parameter include, for example, the search for solutions to linear and quadratic equations in general view, 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 parameters.
To solve an equation (inequality, system) with a parameter means, as a rule, to solve an infinite set of equations (inequalities, systems).
Tasks with a parameter can be conditionally 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 unexplored, such a solution cannot be considered satisfactory.
b) it is required to indicate the possible values of the parameter for which the equation (inequality, system) has certain properties. For example, it has one solution, has no solutions, has solutions that belong to the interval, etc. In such tasks, it is necessary to clearly indicate at what value of the parameter the required condition is satisfied.
The parameter, being an unknown fixed number, has, as it were, a special duality. First of all, it must be taken into account that the alleged fame suggests that the parameter must be perceived as a number. Secondly, the freedom to handle a parameter is limited by its unknown. So, for example, the operations of dividing by an expression in which there is a parameter or extracting a root of an even degree from a similar expression require preliminary research. Therefore, care must be taken in handling the parameter.
For example, to compare two numbers -6a and 3a, three cases need to be considered:
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 decision will be the answer.
Let the equation kx = b be given. This equation is shorthand for an infinite set of equations in one variable.
When solving such equations, there may be cases:
1. Let k be any real number non-zero and b is 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.
The algorithm for solving this type of equations:
1. Determine the "control" values of the parameter.
2. Solve the original equation for x with the values of the parameter that were determined in the first paragraph.
3. Solve the original equation for x with parameter values that differ from those selected in the first paragraph.
4. You can write down the answer in the following form:
1) when ... (parameter value), the equation has roots ...;
2) when ... (parameter value), 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 here a ≥ 0.
By the rule of modulo 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: ax - a + 2x - 2 \u003d 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 \u003d -2, then we have the correct equality 0 x \u003d 0, therefore x is any real number.
Answer: x \u003d 1 for a ≠ -2 and x € R for a \u003d -2.
Example 3
Solve the equation x/a + 1 = a + x with respect to the variable x.
Solution.
If a \u003d 0, then we transform the equation to the form a + x \u003d a 2 + ax or (a - 1) x \u003d -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 at a = 1; x \u003d -a with a ≠ 0 and a ≠ 1.
Graphic method
Consider another way to solve equations with a parameter - graphical. This method is used quite often.
Example 4
How many roots, depending on the parameter a, does the equation ||x| – 2| = a?
Solution.
To solve by a graphical method, we construct graphs of functions y = ||x| – 2| and y = a (Fig. 2).
The drawing clearly shows the possible cases of the location of the line y = a and the number of roots in each of them.
Answer: the equation will have no roots if a< 0; два корня будет в случае, если a >2 and a = 0; the equation will have three roots in the case a = 2; four roots - at 0< a < 2.
Example 5
For which a the equation 2|x| + |x – 1| = a has a single root?
Solution.
Let's draw graphs of functions y = 2|x| + |x – 1| and y = a. For y = 2|x| + |x - 1|, expanding the modules by the gap method, we get:
(-3x + 1, at x< 0,
y = (x + 1, for 0 ≤ x ≤ 1,
(3x - 1, for x > 1.
On the Figure 3 it is clearly seen that the equation will have a unique root only when a = 1.
Answer: a = 1.
Example 6
Determine the number of solutions of the equation |x + 1| + |x + 2| = a depending on the parameter a?
Solution.
Graph of the function y = |x + 1| + |x + 2| will be a broken line. Its vertices will be located at the points (-2; 1) and (-1; 1) (picture 4).
Answer: if the parameter a is less than one, then the equation will have no roots; if a = 1, then the solution of the equation is an infinite set of numbers from the interval [-2; -one]; if the values of the parameter a are greater than one, then the equation will have two roots.
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For what values of the parameter $a$ does the inequality $()-x^2 + (a + 2)x - 8a - 1 > 0$ have at least one solution?
Solution
We reduce this inequality to a positive coefficient for $x^2$:
$()-x^2 + (a + 2)x - 8a - 1 > 0 \quad \Leftrightarrow \quad x^2 - (a + 2)x + 8a + 1< 0 .$
Calculate the discriminant: $D = (a + 2)^2 - 4(8a + 1) = a^2 + 4a + 4 - 32a - 4 = a^2 - 28a$. For this inequality to have a solution, it is necessary that at least one point of the parabola lies below the $x$ axis. Since the branches of the parabola are directed upwards, this requires that the square trinomial on the left side of the inequality has two roots, that is, its discriminant is positive. We come to the need to solve the quadratic inequality $a^2 - 28a > 0$. The square trinomial $a^2 - 28a$ has two roots: $a_1 = 0$, $a_2 = 28$. Therefore, the inequality $a^2 - 28a > 0$ is satisfied by the intervals $a \in (-\infty; 0) \cup (28; + \infty)$.
Answer.$a \in (-\infty; 0) \cup (28; + \infty)$.
For what values of the parameter $a$ does the equation $(a-2)x^2-2ax+a+3=0$ have at least one root, and all the roots are positive?
Solution
Let $a=2$. Then the equation takes the form $() - 4x +5 = 0$ , whence we get that $x=\dfrac(5)(4)$ is a positive root.
Now let $a\ne 2$. It turns out a quadratic equation. Let us first determine for what values of the parameter $a$ the given equation has roots. It is necessary that its discriminant be non-negative. That is:
$ D = 4a^2 - 4(a-2)(a+3) =() -4a+24\geqslant 0\Leftrightarrow a\leqslant 6.$
The roots must be positive by condition, therefore, from the Vieta theorem we get the system:
$ \begin(cases)x_1 + x_2 = \dfrac(2a)(a - 2)>0,\\ x_1x_2 = \dfrac(a + 3)(a - 2)> 0,\\a\leqslant 6\end (cases) \quad \Leftrightarrow \quad \begin(cases)a\in(- \infty;0)\cup(2; +\infty), \\ a\in(- \infty;-3)\cup( 2; +\infty), \\ a\in(-\infty;6] \end(cases)\quad\Leftrightarrow \quad a\in(-\infty;-3)\cup(2;6]. $
We combine the answers, we get the desired set: $a\in(-\infty;-3)\cup$.
Answer.$a\in(-\infty;-3)\cup$.
For what values of the parameter $a$ does the inequality $ax^2 + 4ax + 5 \leqslant 0$ have no solutions?
Solution
- If $a = 0$, then this inequality degenerates into the inequality $5 \leqslant 0$ , which has no solutions. Therefore, the value $a = 0$ satisfies the condition of the problem.
- If $a > 0$, then the graph of the square trinomial on the left side of the inequality is a parabola with upward branches. We calculate $\dfrac(D)(4) = 4a^2 - 5a$. The inequality has no solutions if the parabola is located above the x-axis, that is, when the square trinomial has no roots ($D< 0$). Решим неравенство $4a^2 - 5a < 0$. Корнями квадратного трёхчлена $4a^2 - 5a$ являются числа $a_1 = 0$ и $a_2 = \dfrac{5}{4}$, поэтому $D < 0$ при $0 < a < \dfrac{5}{4}$. Значит, из положительных значений $a$ подходят числа $a \in \left(0; \dfrac{5}{4}\right)$.
- If $a< 0$, то график квадратного трехчлена в левой части неравенства - парабола с ветвями, направленными вниз. Значит, обязательно найдутся значения $х$, для которых трёхчлен отрицателен. Следовательно, все значения $a < 0$ не подходят.
Answer.$a \in \left$ lies between the roots, so there must be two roots (hence $a\ne 0$). If the branches of the parabola $y = ax^2 + (a + 3)x - 3a$ point upwards, then $y(-1)< 0$ и $y(1) < 0$; если же они направлены вниз, то $y(-1) >0$ and $y(1) > 0$.
Case I. Let $a > 0$. Then
$\left\( \begin(array)(l) y(-1)=a-(a+3)-3a=-3a-3<0 \\ y(1)=a+(a+3)-3a=-a+3<0 \\ a>0 \end(array) \right. \quad \Leftrightarrow \quad \left\( \begin(array)(l) a>-1 \\ a>3 \\ a>0 \end(array) \right.\quad \Leftrightarrow \quad a>3. $
That is, in this case it turns out that all $a > 3$ fit.
Case II. Let $a< 0$. Тогда
$\left\( \begin(array)(l) y(-1)=a-(a+3)-3a=-3a-3>0 \\ y(1)=a+(a+3)-3a =-a+3>0 \\ a<0 \end{array} \right.\quad \Leftrightarrow \quad \left\{ \begin{array}{l} a<-1 \\ a<3 \\ a<0 \end{array} \right.\quad \Leftrightarrow \quad a<-1.$
That is, in this case, it turns out that all $a< -1$.
Answer.$a\in (-\infty ;-1)\cup (3;+\infty)$
Find all values of the parameter $a$, for each of which the system of equations
$ \begin(cases) x^2+y^2 = 2a, \\ 2xy=2a-1 \end(cases) $
has exactly two solutions.
Solution
Subtract the second from the first: $(x-y)^2 = 1$. Then
$ \left[\begin(array)(l) x-y = 1, \\ x-y = -1 \end(array)\right. \quad \Leftrightarrow \quad \left[\begin(array)(l) x = y+1, \\ x = y-1. \end(array)\right. $
Substituting the obtained expressions into the second equation of the system, we obtain two quadratic equations: $2y^2 + 2y - 2a + 1 = 0$ and $2y^2 - 2y - 2a + 1 =0$. The discriminant of each of them is equal to $D = 16a-4$.
Note that it cannot happen that the pair of roots of the first of the quadratic equations coincides with the pair of roots of the second quadratic equation, since the sum of the roots of the first is equal to $-1$, and the second is 1.
This means that each of these equations must have one root, then the original system will have two solutions. That is $D = 16a - 4 = 0$.
Answer.$a=\dfrac(1)(4)$
Find all values of the parameter $a$ for each of which the equation $4x-|3x-|x+a||=9|x-3|$ has two roots.
Solution
Let's rewrite the equation in the form:
$ 9|x-3|-4x+|3x-|x+a|| = 0.$
Consider the function $f(x) = 9|x-3|-4x+|3x-|x+a||$.
For $x\geqslant 3$ the first modulus is expanded with a plus sign, and the function becomes: $f(x) = 5x-27+|3x-|x+a||$. It is obvious that with any expansion of the modules, as a result, a linear function with the coefficient $k\geqslant 5-3-1=1>0$ will be obtained, that is, this function grows without bound on this interval.
Consider now the interval $x<3$. В этом случае первый модуль раскрывается с минусом, и функция принимает следующий вид: $f(x) = - 13x+27+|3x-|x+a||$. При любом раскрытии модулей в итоге будет получаться линейная функция с коэффициентом $k\leqslant - 13+3+1 = - 9<0$, то есть на этом промежутке функция убывает.
So, we got that $x=3$ is the minimum point of this function. And this means that in order for the original equation to have two solutions, the value of the function at the minimum point must be less than zero. That is, the inequality takes place: $f(3)<0$.
$ 12-|9-|3+a||>0 \quad \Leftrightarrow \quad |9-|3+a||< 12 \quad \Leftrightarrow \quad -12 < 9-|3+a| < 12 \quad \Leftrightarrow \quad$
$\Leftrightarrow\quad |3+a|< 21 \quad \Leftrightarrow \quad - 21 < 3+a < 21 \quad \Leftrightarrow \quad -24
Answer.$a \in (-24; 18)$ For what values of the parameter $a$ does the equation $5^(2x)-3\cdot 5^x+a-1=0$ have a single root? Let's make a change: $t = 5^x > 0$. Then the original equation takes the form of a quadratic equation: $t^2-3t+a-1 =0$. The original equation will have a single root if this equation has one positive root or two roots, one of which is positive, the other is negative. The discriminant of the equation is: $D = 13-4a$. This equation will have one root if the resulting discriminant is equal to zero, that is, for $a = \dfrac(13)(4)$. In this case, the root $t=\dfrac(3)(2) > 0$, so the given value of $a$ is suitable. If there are two roots, one positive and one non-positive, then $D = 13-4a > 0$, $x_1+x_2 = 3 > 0$ and $x_1x_2 = a - 1 \leqslant 0$. That is, $a\in(-\infty;1]$ Answer.$a\in(-\infty;1]\cup\left\(\dfrac(13)(4)\right\)$ Find all values of parameter $a$ for which the system $ \begin(cases)\log_a y = (x^2-2x)^2, \\ x^2+y=2x\end(cases) $ has exactly two solutions. Let's transform the system to the following form: $ \begin(cases) \log_a y = (2x-x^2)^2, \\ y = 2x-x^2. \end(cases) $ Since the parameter $a$ is at the base of the logarithm, the following restrictions are imposed on it: $a>0$, $a \ne 1$. Since the variable $y$ is the argument of the logarithm, then $y > 0$. Combining both equations of the system, we pass to the equation: $\log_a y = y^2$. Depending on what values the $a$ parameter takes, two cases are possible: Answer.$a\in(0;1)$ Consider the case when $a > 1$. Since for large values of $t$ the graph of the function $f(t) = a^t$ lies above the straight line $g(t) = t$, the only common point can only be a point of contact. Let $t_0$ be the touch point. At this point, the derivative to $f(t) = a^t$ is equal to one (the tangent of the slope of the tangent), in addition, the values of both functions are the same, that is, the system takes place: $ \begin(cases) a^(t_0)\ln a = 1, \\ a^(t_0) = t_0 \end(cases) \quad \Leftrightarrow \quad \begin(cases) a^(t_0) = \dfrac (1)(\ln a), \\ a^(\tau) = \tau \end(cases) $ Whence $t_0 = \dfrac(1)(\ln a)$. $ a^(\frac(1)(\ln a))\ln a = 1 \quad \Leftrightarrow \quad a^(\log_a e) =\frac(1)(\ln a) \quad \Leftrightarrow \quad a = e^(\frac(1)(e)). $ At the same time, other common points of the line and exponential function obviously not. Answer.$a \in (0;1] \cup \left\(e^(e^(-1))\right\)$Solution
Solution