Consider now the quadratic equation
where is an unknown quantity, are the parameters (coefficients) of the equation.
The critical values of the parameter should include, first of all, the value At the specified value of the parameter, equation (1) takes the form
therefore, the order of the equation is reduced by one. Equation (2) is a linear equation and the method of its solution was considered earlier.
For other critical values of the parameters are determined by the discriminant of the equation. It is known that at , equation (1) has no roots; for it has a single root for equation (1) has two different roots and
one). Find all parameter values for which the quadratic equation
a) has two different roots;
b) has no roots;
c) has two equal roots.
Solution. This equation is quadratic by condition, and therefore Consider the discriminant of this equation
When the equation has two different roots, because
When the equation has no roots, because This quadratic equation cannot have two equal roots, because for and this contradicts the condition of the problem.
Answer: When the equation has two different roots.
When the equation has no roots.
2). Solve the equation. For each admissible value of the parameter, solve the equation
Solution. Consider first the case when
(in this case, the original equation becomes a linear equation). Thus, the value of the parameter and are its critical values. It is clear that for , the root of this equation is and for , its root is
If those. and then this equation is quadratic. Let's find its discriminant:
For all values, the discriminant takes non-negative values, and it vanishes at (these values of the parameter are also its critical values).
Therefore, if then this equation has a single root
In this case, the value of the parameter corresponds to the root
and the value corresponds to the root
If then the equation has two different roots. Let's find these roots.
Answer. If then if then if then
if then , .
3). Solve the equation. At what values of the parameter a does the equation have a unique solution?
Solution. This equation is equivalent to the system
The presence of a quadratic equation and the condition for the uniqueness of the solution will naturally lead to the search for the roots of the discriminant. However, the condition x ≠ -3 should attract attention. And the "subtle point" is that the quadratic equation of the system can have two roots! But only one of them must be equal to -3. We have
D= a 2 - 4 , hence D = 0 if a= ±2; x \u003d -3 - the root of the equation x 2 - a x +1 = 0 at
a= -10/3, and with this value a the second root of the quadratic equation is different
Answer. a= ±2 or a = -10/3.
4). Solve the equation. At what values of the parameter a the equation
(a- 2)x 2 + (4 - 2a) X+3 = 0 has a unique solution?
Solution. It is clear that it is necessary to start with the case a= 2. But at a = 2 The original equation has no solutions at all. If a a ≠ 2, then this equation is quadratic, and, it would seem, the desired values of the parameter are the roots of the discriminant. However, the discriminant vanishes when a = 2 or a = 5. Since we have established that a=2 does not fit, then
Answer, a = 5.
9). Solve the equation. At what values of the parameter a the equation Oh 2 - 4X + a+ 3 = 0 has more than one root?
Solution. At a= 0 the equation has a single root, which does not satisfy the condition. At a≠ 0 the original equation, being square, has two roots if its discriminant is 16 – 4 a 2 – 12a positive. From here we get -4<a<1.
However, the resulting interval (-4; 1) includes the number 0. Answer. -4<a<0 или 0<a<1.
ten). At what values of the parameter a the equation a(a+3)X 2 + (2a+6)X– 3a– 9 = 0 has more than one root?
Solution. Standard step - start with cases a= 0 and a= -3. At a= 0 the equation has a unique solution. It is curious that at a= -3 the solution of the equation is any real number. At a≠ -3 and a≠ 0, dividing both sides of this equation by a + 3, we get the quadratic equation Oh 2 + 2X- 3 = 0, whose discriminant is 4 (1 + Z a) is positive for a > ⅓. The experience of the previous examples suggests that from the interval
(-⅓ ;∞) you need to exclude the point a= 0, and don't forget to include a = -3.
Answer. a= -3, or - ⅓< а < 0, или а > 0.
11).Solve the equation :
Solution. First, note that for this equation is equivalent to an equation that has no solutions. If
Type equation f(x; a) = 0 is called variable equation X and parameter a.
Solve an equation with a parameter a This means that for every value a find values X satisfying this equation.
Example 1 Oh= 0
Example 2 Oh = a
Example 3
x + 2 = ax
x - ax \u003d -2
x (1 - a) \u003d -2
If 1 - a= 0, i.e. a= 1, then X 0 = -2 no roots
If 1 - a 0, i.e. a 1, then X =
Example 4
(a 2 – 1) X = 2a 2 + a – 3
(a – 1)(a + 1)X = 2(a – 1)(a – 1,5)
(a – 1)(a + 1)X = (1a – 3)(a – 1)
If a a= 1, then 0 X = 0
X- any real number
If a a= -1, then 0 X = -2
no roots
If a a 1, a-1 then X= (the only solution).
This means that for every valid value a matches a single value X.
For example:
if a= 5, then X = = ;
if a= 0, then X= 3 etc.
Didactic material
1. Oh = X + 3
2. 4 + Oh = 3X – 1
3. a = +
at a= 1 there are no roots.
at a= 3 no roots.
at a = 1 X any real number except X = 1
at a = -1, a= 0 there are no solutions.
at a = 0, a= 2 no solutions.
at a = -3, a = 0, 5, a= -2 no solutions
at a = -With, With= 0 there are no solutions.
Quadratic equations with a parameter
Example 1 solve the equation
(a – 1)X 2 = 2(2a + 1)X + 4a + 3 = 0
At a = 1 6X + 7 = 0
When a 1 select those values of the parameter for which D goes to zero.
D = (2(2 a + 1)) 2 – 4(a – 1)(4a + 30 = 16a 2 + 16a + 4 – 4(4a 2 + 3a – 4a – 3) = 16a 2 + 16a + 4 – 16a 2 + 4a + 12 = 20a + 16
20a + 16 = 0
20a = -16
If a a < -4/5, то D < 0, уравнение имеет действительный корень.
If a a> -4/5 and a 1, then D > 0,
X = 
If a a= 4/5, then D = 0,
Example 2 At what values of the parameter a the equation
x 2 + 2( a + 1)X + 9a– 5 = 0 has 2 different negative roots?
D = 4( a + 1) 2 – 4(9a – 5) = 4a 2 – 28a + 24 = 4(a – 1)(a – 6)
4(a – 1)(a – 6) > 0
according to t. Vieta: X 1 + X 2 = -2(a + 1)
X 1 X 2 = 9a – 5
By condition X 1 < 0, X 2 < 0 то –2(a + 1) < 0 и 9a – 5 > 0
| Eventually | 4(a – 1)(a – 6) > 0 - 2(a + 1) < 0 9a – 5 > 0 |
a < 1: а > 6 a > - 1 a > 5/9 |
 (Rice. one) < a < 1, либо a > 6 |
Example 3 Find values a for which this equation has a solution.
x 2 - 2( a – 1)X + 2a + 1 = 0
D = 4( a – 1) 2 – 4(2a + 10 = 4a 2 – 8a + 4 – 8a – 4 = 4a 2 – 16a
4a 2 – 16 0
4a(a – 4) 0
a( a – 4)) 0
a( a – 4) = 0
a = 0 or a – 4 = 0
a = 4
(Rice. 2)
Answer: a 0 and a 4
Didactic material
1. At what value a the equation Oh 2 – (a + 1) X + 2a– 1 = 0 has one root?
2. At what value a the equation ( a + 2) X 2 + 2(a + 2)X+ 2 = 0 has one root?
3. For what values of a is the equation ( a 2 – 6a + 8) X 2 + (a 2 – 4) X + (10 – 3a – a 2) = 0 has more than two roots?
4. For what values of a equation 2 X 2 + X – a= 0 has at least one common root with equation 2 X 2 – 7X + 6 = 0?
5. For what values of a do the equations X 2 +Oh+ 1 = 0 and X 2 + X + a= 0 have at least one common root?
1. When a = - 1/7, a = 0, a = 1
2. When a = 0
3. When a = 2
4. When a = 10
5. When a = - 2
Exponential Equations with a Parameter
Example 1.Find all values a, for which the equation
9 x - ( a+ 2) * 3 x-1 / x +2 a*3 -2/x = 0 (1) has exactly two roots.
Solution. Multiplying both sides of equation (1) by 3 2/x, we obtain an equivalent equation
3 2(x+1/x) – ( a+ 2) * 3 x + 1 / x + 2 a = 0 (2)
Let 3 x+1/x = at, then equation (2) takes the form at 2 – (a + 2)at + 2a= 0, or
(at – 2)(at – a) = 0, whence at 1 =2, at 2 = a.
If a at= 2, i.e. 3 x + 1/x = 2 then X + 1/X= log 3 2 , or X 2 – X log 3 2 + 1 = 0.
This equation has no real roots because it D= log 2 3 2 – 4< 0.
If a at = a, i.e. 3 x+1/x = a then X + 1/X= log 3 a, or X 2 –X log 3 a + 1 = 0. (3)
Equation (3) has exactly two roots if and only if
D = log 2 3 2 – 4 > 0, or |log 3 a| > 2.
If log 3 a > 2, then a> 9, and if log 3 a< -2, то 0 < a < 1/9.
Answer: 0< a < 1/9, a > 9.
Example 2. At what values of a equation 2 2x - ( a - 3) 2 x - 3 a= 0 has solutions?
To given equation has solutions, it is necessary and sufficient that the equation t 2 – (a - 3) t – 3a= 0 has at least one positive root. Let's find the roots using Vieta's theorem: X 1 = -3, X 2 = a = >
a is a positive number.
Answer: when a > 0
Didactic material
1. Find all values of a for which the equation
25 x - (2 a+ 5) * 5 x-1 / x + 10 a* 5 -2/x = 0 has exactly 2 solutions.
2. For what values of a does the equation
2 (a-1) x? + 2 (a + 3) x + a \u003d 1/4 has a single root?
3. For what values of the parameter a the equation
4 x - (5 a-3) 2 x +4 a 2 – 3a= 0 has a unique solution?
Logarithmic Equations with a Parameter
Example 1 Find all values a, for which the equation
log 4x (1 + Oh) = 1/2 (1)
has a unique solution.
Solution. Equation (1) is equivalent to the equation
1 + Oh = 2X at X > 0, X 1/4 (3)
X = at
au 2 - at + 1 = 0 (4)
The (2) condition from (3) is not satisfied.
Let a 0, then au 2 – 2at+ 1 = 0 has real roots if and only if D = 4 – 4a 0, i.e. at a 1. To solve inequality (3), we construct graphs of functions Galitsky M.L., Moshkovich M.M., Shvartsburd S.I. In-depth study of the course of algebra and mathematical analysis. - M.: Enlightenment, 1990
1. Task.
At what values of the parameter a the equation ( a - 1)x 2 + 2x + a- 1 = 0 has exactly one root?
1. Decision.
At a= 1 equation has the form 2 x= 0 and obviously has a single root x= 0. If a No. 1, then this equation is quadratic and has a single root for those values of the parameter for which the discriminant square trinomial zero. Equating the discriminant to zero, we obtain an equation for the parameter a
4a 2 - 8a= 0, whence a= 0 or a = 2.
1. Answer: the equation has a single root at a O(0; 1; 2).
2. Task.
Find all parameter values a, for which the equation has two different roots x 2 +4ax+8a+3 = 0.
2. Decision.
The equation x 2 +4ax+8a+3 = 0 has two distinct roots if and only if D =
16a 2 -4(8a+3) > 0. We get (after reduction by a common factor of 4) 4 a 2 -8a-3 > 0, whence
2. Answer:
| a O (-Ґ ; 1 - | C 7 2 |
) AND (1 + | C 7 2 |
; Ґ ). |
3. Task.
It is known that
f 2 (x) = 6x-x 2 -6.
a) Graph the function f 1 (x) at a = 1.
b) At what value a function graphs f 1 (x) and f 2 (x) have a single common point?
3. Solution.
3.a. Let's transform f 1 (x) in the following way
The graph of this function a= 1 is shown in the figure on the right.
3.b. We immediately note that the function graphs y =
kx+b and y = ax 2 +bx+c
(a No. 0) intersect at a single point if and only if the quadratic equation kx+b =
ax 2 +bx+c has a single root. Using View f 1 of 3.a, we equate the discriminant of the equation a = 6x-x 2 -6 to zero. From Equation 36-24-4 a= 0 we get a= 3. Doing the same with equation 2 x-a = 6x-x 2 -6 find a= 2. It is easy to verify that these parameter values satisfy the conditions of the problem. Answer: a= 2 or a = 3.
4. Task.
Find all values a, under which the set of solutions of the inequality x 2 -2ax-3a i 0 contains the segment .
4. Solution.
The first coordinate of the vertex of the parabola f(x) =
x 2 -2ax-3a is equal to x 0 =
a. From the properties of a quadratic function, the condition f(x) i 0 on the interval is equivalent to the totality of three systems
has exactly two solutions?
5. Decision.
Let's rewrite this equation in the form x 2 + (2a-2)x - 3a+7 = 0. This is a quadratic equation, it has exactly two solutions if its discriminant is strictly greater than zero. Calculating the discriminant, we get that the condition for having exactly two roots is the fulfillment of the inequality a 2 +a-6 > 0. Solving the inequality, we find a < -3 или a> 2. The first of the inequalities is obviously solutions in natural numbers does not have, and the smallest natural solution of the second is the number 3.
5. Answer: 3.
6. Task (10 cells)
Find all values a, for which the graph of the function or, after obvious transformations, a-2 = |
2-a| . The last equation is equivalent to the inequality a i 2.
6. Answer: a O )
