Defined by the formula $a((x)^(2))+bx+c$ $(a\ne 0).$ The numbers $a, b$ and $c$ are the coefficients of a quadratic trinomial, they are usually called: a - the leading one, b - second or average coefficient, c - free term. A function of the form y = ax 2 + bx + c is called a quadratic function.
All of these parabolas have their vertex at the origin; for a > 0 this is the lowest point of the graph ( smallest value functions), and for a< 0, наоборот, наивысшая точка (highest value functions). The Oy axis is the axis of symmetry of each of these parabolas.
As can be seen, for a > 0 the parabola is directed upward, for a< 0 - вниз.
There is a simple and convenient graphic method, which allows you to construct any number of points of the parabola y = ax 2 without calculations, if a point of the parabola other than the vertex is known. Let the point M(x 0 , y 0) lie on the parabola y = ax 2 (Fig. 2). If we want to construct an additional n points between points O and M, then we divide the segment ON of the abscissa axis into n + 1 equal parts and at the division points we draw perpendiculars to the Ox axis. We divide the segment NM into the same number of equal parts and connect the division points with rays to the origin of coordinates. The required points of the parabola lie at the intersection of perpendiculars and rays with the same numbers (in Fig. 2 the number of division points is 9).
The graph of the function y =ax 2 + bx + c differs from the graph y = ax 2 only in its position and can be obtained simply by moving the curve on the drawing. This follows from the representation of the quadratic trinomial in the form
from which it is easy to conclude that the graph of the function y = ax 2 + bx + c is a parabola y = ax 2, the vertex of which is moved to the point
and its axis of symmetry remained parallel to the Oy axis (Fig. 3). From the resulting expression for a quadratic trinomial, all its basic properties easily follow. The expression D = b 2 − 4ac is called the discriminant of the quadratic trinomial ax 2 + bx + c and the discriminant of the associated quadratic equation ax 2 + bx + c = 0. The sign of the discriminant determines whether the graph of a quadratic trinomial intersects the x-axis or lies on one side of it. Namely, if D< 0, то парабола не имеет общих точек с осью Ox, при этом: если a >0, then the parabola lies above the Ox axis, and if a< 0, то ниже этой оси (рис. 4). В случае D >0 the graph of a quadratic trinomial intersects the x-axis at two points x 1 and x 2, which are the roots of the quadratic equation ax 2 + bx + c = 0 and are equal, respectively
At D = 0 the parabola touches the Ox axis at the point
The properties of the quadratic trinomial form the basis for solving quadratic inequalities. Let's explain this with an example. Suppose we need to find all solutions to the inequality 3x 2 - 2x - 1< 0. Найдем дискриминант квадратного трехчлена, стоящего в левой части неравенства: D = 16. Так как D >0, then the corresponding quadratic equation 3x 2 − 2x − 1 = 0 has two different roots, they are determined by the formulas given earlier:
x 1 = −1/3 and x 2 = 1.
In the quadratic trinomial under consideration, a = 3 > 0, which means that the branches of its graph are directed upward and the values of the quadratic trinomial are negative only in the interval between the roots. So, all solutions to the inequality satisfy the condition
−1/3 < x < 1.
Various inequalities can be reduced to quadratic inequalities by the same substitutions by which various equations are reduced to quadratic ones.
Definition
Parabola called a graph quadratic function$y = ax^(2) + bx + c$, where $a \neq 0$.
Graph of the function $y = x^2$.
To schematically plot the graph of the function $y = x^2$, we will find several points that satisfy this equality. For convenience, we write down the coordinates of these points in the form of a table:
Graph of the function $y = ax^2$.
If the coefficient $a > 0$, then the graph $y = ax^2$ is obtained from the graph $y = x^2$ either by vertical stretching (for $a > 1$) or compression to the $x$ axis (for $0< a < 1$). Изобразим для примера графики $y = 2x^2$ и $y = \dfrac{x^2}{2}$:
| $y = 2x^2$ | $y = \dfrac(x^2)(2)$ |
|
|
If $a< 0$, то график функции $y = ax^2$ можно получить из графика $y = |a|x^2$, отразив его симметрично относительно оси $x$. Построим графики функций $y = - x^2$, $y = -2x^2$ и $y = - \dfrac{x^2}{2}$:
| $y = - x^2$ | $y = -2x^2$ | $y = - \dfrac(x^2)(2)$ |
|
|
|
Graph of a quadratic function.
To graph the function $y = ax^2 + bx + c$, you need to isolate $ax^2 + bx + c$ from the square trinomial perfect square, that is, represent it in the form $a(x - x_0)^2 + y_0$. The graph of the function $y = a(x - x_0)^2 + y_0$ is obtained from the corresponding graph $y = ax^2$ by shifting by $x_0$ along the $x$ axis, and by $y_0$ along the $y$ axis. As a result, point $(0;0)$ will move to point $(x_0;y_0)$.
Definition
The top the parabola $y = a(x - x_0)^2 + y_0$ is the point with coordinates $(x_0;y_0)$.
Let's construct a parabola $y = 2x^2 - 4x - 6$. Selecting the complete square, we get $y = 2(x - 1)^2 - 8$.
| Let's plot $y = 2x^2$ | Let's move it to the right by 1 | And down by 8 |
|
|
|
The result is a parabola with its vertex at point $(1;-8)$.
The graph of the quadratic function $y = ax^2 + bx + c$ intersects the $y$ axis at the point $(0; c)$ and the $x$ axis at the points $(x_(1,2);0)$, where $ x_(1,2)$ are the roots of the quadratic equation $ax^2 + bx + c = 0$ (and if the equation has no roots, then the corresponding parabola does not intersect the $x$ axis).
For example, the parabola $y = 2x^2 - 4x - 6$ intersects the axes at the points $(0; -6)$, $(-1; 0)$ and $(3; 0)$.
Lesson: How to construct a parabola or quadratic function?
A parabola is a graph of a function described by the formula ax 2 +bx+c=0.
To build a parabola you need to follow a simple algorithm:
1) Parabola formula y=ax 2 +bx+c,
If a>0 then the branches of the parabola are directed up,
otherwise the branches of the parabola are directed down.
Free member c this point intersects the parabola with the OY axis;
2), it is found using the formula x=(-b)/2a, we substitute the found x into the parabola equation and find y;
3)Function zeros or, in other words, the points of intersection of the parabola with the OX axis, they are also called the roots of the equation. To find the roots we equate the equation to 0 ax 2 +bx+c=0;
Types of equations:
a) The complete quadratic equation has the form ax 2 +bx+c=0 and is solved by the discriminant;
b) Incomplete quadratic equation of the form ax 2 +bx=0. To solve it, you need to take x out of brackets, then equate each factor to 0:
ax 2 +bx=0,
x(ax+b)=0,
x=0 and ax+b=0;
c) Incomplete quadratic equation of the form ax 2 +c=0. To solve it, you need to move the unknowns to one side, and the knowns to the other. x =±√(c/a);
4) Find several additional points to construct the function.
And so now, using an example, we will analyze everything step by step:
Example #1:
y=x 2 +4x+3
c=3 means the parabola intersects OY at the point x=0 y=3. The branches of the parabola look upward since a=1 1>0.
a=1 b=4 c=3 x=(-b)/2a=(-4)/(2*1)=-2 y= (-2) 2 +4*(-2)+3=4- 8+3=-1 vertex is at point (-2;-1)
Let's find the roots of the equation x 2 +4x+3=0
Using the discriminant we find the roots
a=1 b=4 c=3
D=b 2 -4ac=16-12=4
x=(-b±√(D))/2a
x 1 =(-4+2)/2=-1
x 2 =(-4-2)/2=-3
Let's take several arbitrary points that are located near the vertex x = -2
x -4 -3 -1 0
y 3 0 0 3
Substitute instead of x into the equation y=x 2 +4x+3 values
y=(-4) 2 +4*(-4)+3=16-16+3=3
y=(-3) 2 +4*(-3)+3=9-12+3=0
y=(-1) 2 +4*(-1)+3=1-4+3=0
y=(0) 2 +4*(0)+3=0-0+3=3
It can be seen from the function values that the parabola is symmetrical with respect to the straight line x = -2
Example #2:
y=-x 2 +4x
c=0 means the parabola intersects OY at the point x=0 y=0. The branches of the parabola look down since a=-1 -1 Let's find the roots of the equation -x 2 +4x=0
Incomplete quadratic equation of the form ax 2 +bx=0. To solve it, you need to take x out of brackets, then equate each factor to 0.
x(-x+4)=0, x=0 and x=4.
Let's take several arbitrary points that are located near the vertex x=2
x 0 1 3 4
y 0 3 3 0
Substitute instead of x into the equation y=-x 2 +4x values
y=0 2 +4*0=0
y=-(1) 2 +4*1=-1+4=3
y=-(3) 2 +4*3=-9+13=3
y=-(4) 2 +4*4=-16+16=0
It can be seen from the function values that the parabola is symmetrical about the straight line x = 2
Example No. 3
y=x 2 -4
c=4 means the parabola intersects OY at the point x=0 y=4. The branches of the parabola look upward since a=1 1>0.
a=1 b=0 c=-4 x=(-b)/2a=0/(2*(1))=0 y=(0) 2 -4=-4 the vertex is at point (0;-4 )
Let's find the roots of the equation x 2 -4=0
Incomplete quadratic equation of the form ax 2 +c=0. To solve it, you need to move the unknowns to one side, and the knowns to the other. x =±√(c/a)
x 2 =4
x 1 =2
x 2 =-2
Let's take several arbitrary points that are located near the vertex x=0
x -2 -1 1 2
y 0 -3 -3 0
Substitute instead of x into the equation y= x 2 -4 values
y=(-2) 2 -4=4-4=0
y=(-1) 2 -4=1-4=-3
y=1 2 -4=1-4=-3
y=2 2 -4=4-4=0
It can be seen from the function values that the parabola is symmetrical about the straight line x = 0
Subscribe to the channel on YOUTUBE to keep abreast of all the new products and prepare with us for exams.
Graph of a quadratic trinomial
2019-04-19
Square trinomial
We called a square trinomial an entire rational function of the second degree:
$y = ax^2 + bx + c$, (1)
where $a \neq 0$. Let us prove that the graph of a quadratic trinomial is a parabola obtained by parallel shifts (in the directions of the coordinate axes) from the parabola $y = ax^2$. To do this, we present expression (1) using simple identity transformations to mind
$y = a(x + \alpha)^2 + \beta$. (2)
The corresponding transformations, written below, are known as "exact square extraction":
$y = x^2 + bx + c = a \left (x^2 + \frac(b)(a) x \right) + c = a \left (x^2 + \frac(b)(a) x + \frac (b^2)(4a^2) \right) - \frac (b^2)(4a) + c = a \left (x + \frac(b)(2a) \right)^2 - \frac (b^2 - 4ac)(4a)$. (2")
We have reduced the quadratic trinomial to form (2); wherein
$\alpha = \frac(b)(2a), \beta = - \frac (b^2 - 4ac)(4a)$
(these expressions should not be memorized; it is more convenient to transform the trinomial (1) to the form (2) directly each time).
Now it is clear that the graph of the trinomial (1) is a parabola equal to the parabola $y = ax^2$ and obtained by shifting the parabola $y = ax^2$ in the directions of the coordinate axes by $\alpha$ and $\beta$ (taking into account the sign $\alpha$ and $\beta$) respectively. The vertex of this parabola is located at the point $(- \alpha, \beta)$, its axis is the straight line $x = - \alpha$. For $a > 0$, the vertex is the lowest point of the parabola, for $a
Let us now carry out a study of the quadratic trinomial, i.e., we will find out its properties depending on the numerical values of the coefficients $a, b, c$ in its expression (1).
In equality (2") we denote the value $b^2- 4ac$ by $d$:
$y = a \left (x + \frac(b)(2a) \right)^2 - \frac(d)(4a)$; (4)
$d = b^2 - 4ac$ is called the discriminant of a quadratic trinomial. The properties of the trinomial (1) (and the location of its graph) are determined by the signs of the discriminant $d$ and the leading coefficient $a$.
1) $a > 0, d 0$; since $a > 0$, then the graph is located above the vertex $O^( \prime)$; it lies in the upper half-plane ($y > 0$ - Fig. a.).
2) $a
3) $a > 0, d > 0$. The vertex $O^( \prime)$ lies below the $Ox$ axis, the parabola intersects the $Ox$ axis at two points $x_1, x_2$ (Fig. c.).
4) $a 0$. The vertex $O^( \prime)$ lies above the $Ox$ axis, the parabola again intersects the $Ox$ axis at two points $x_1, x_2$ (Fig. d).
5) $a > 0, d = 0$. The vertex lies on the $Ox$ axis itself, the parabola is located in the upper half-plane (Fig. e).
6) $a
Conclusions. If $d 0$), or lower (if $a
If $d > 0$, then the function is alternating (the graph partly lies below and partly above the $Ox$ axis). A square trinomial with $d > 0$ has two roots (zeros) $x_1, x_2$. For $a > 0$ it is negative in the interval between the roots (Fig. c) and positive outside this interval. At $a
Introductory remarks and simple examples
Example 1. For what values of a does the equation ax 2 + 2x + 1 = 0 have two different roots?
Solution.
This equation is quadratic with respect to the variable x for a№ 0 and has different roots when its discriminant
i.e. for a< 1.
In addition, when a = 0, the equation 2x + 1 = 0 is obtained, which has one root.
Thus, a O (– Ґ ; 0) AND (0; 1).
Rule 1. If the coefficient of x 2 of a polynomial of the second degree contains a parameter, it is necessary to analyze the case when it vanishes.
Example 2. The equation ax 2 + 8x + c = 0 has a single root equal to 1. What are a and c equal to?
Solution. Let's start solving the problem with special occasion a = 0, the equation is 8x + c = 0. This linear equation has a solution x 0 = 1 for c = – 8.
When a no. 0 quadratic equation has a single root if 
In addition, substituting the root x 0 = 1 into the equation, we get a + 8 + c = 0.
Solving the system of two linear equations, we find a = c = – 4.
Theorem 1.
For the reduced quadratic trinomial y = x 2 + px + q (assuming p 2і
4q)
sum of roots x 1 + x 2 = – p, product of roots x 1 x 2 = q, difference of roots is 
and the sum of the squares of the roots x 1 2 + x 2 2 = p 2 – 2q.
Theorem 2.
For a quadratic trinomial y = ax 2 + bx + c with two roots x 1 and x 2, we have
expansion ax 2 + bx + c = a(x – x 1)(x – x 2), for a trinomial with one root x 0 – expansion
ax 2 + bx + c = a(x – x 0) 2 .
Comment. Often, about quadratic equations with a discriminant equal to zero and having, accordingly, one root, they say that it has two coinciding roots (?). This is related to the factorization of the polynomial given in Theorem 2.(The correct way to say and understand in this case is “one root of multiple two.” – Ed.)
We will pay attention to this subtlety and highlight the case of a single root of multiplicity 2.
Example 3. In the equation x 2 + ax + 12 = 0, determine a in such a way that the difference between the roots of the equation is equal to one.
Solution. Root difference 
whence a = ± 7.
Example 4. For what a does the sum of the squares of the roots of the equation 2x 2 + 4x + a = 0 equal 6?
Solution. Let's write the equation in the form 
whence x 1 2 + x 2 2 = 4 – a = 6 and a = – 2.
Example 5. For all a, solve the equation ax 2 – 2x + 4 = 0.
Solution. If a = 0, then x = 2. If a№
0, then the equation becomes quadratic. Its discriminant
equal to D = 4 – 16a. If D< 0, т. е. a > ,
the equation has no solutions. If D = 0, i.e. a = ,
x = 4. If D > 0, i.e. a< ,
the equation has two roots
Location of the roots of the quadratic trinomial
The graph of a quadratic equation is a parabola, and the solutions to a quadratic equation are the abscissas of the points of intersection of this parabola with the Ox axis. The basis for solving all the problems in this section is the study of the features of the location of parabolas with given properties on the coordinate plane.
Example 6. For what a do the roots of the equation x 2 – 2ax + a 2 – a – 6 = 0 have different signs?
Solution (Fig. 1).
A quadratic equation either has no solutions (the graph is a parabola of type D), or has one or two positive roots (parabola C), or has one or two negative roots (parabola A), or has roots of different signs (parabola B).
It is easy to understand that the last type of parabolas, unlike others, is characterized by the fact that f(0)< 0. Таким образом, f(0) = a 2 – a – 6 < 0, откуда 0 < a < .
This solution allows for a generalization, which we will formulate as the following rule.
Rule 2. In order for the equation ax 2 + bx + c = 0
had two different roots x 1 and x 2 such that x 1< M < x 2 , необходимо и достаточно, чтобы a f(M) < 0.
Example 7. For what a does the equation x 2 – 2ax + a 2 – a – 6 = 0 have two different roots of the same sign?
Solution. We are interested in parabolas of type A and C (see Fig. 1). They are characterized by the fact that
whence a O (– 6; – 2) AND (3; + Ґ ).
Example 8. For what a does the equation x 2 – 2ax + a 2 – a – 6 = 0 have two different positive roots?
Solution. We are interested in the type C parabolas in Fig. 1.
For the equation to have roots, we require
Since both roots of the equation must be positive by condition, the abscissa of the vertex of the parabola lying between the roots is positive: x 0 = a > 0.
Vertex ordinate f(x 0)< 0 в силу того, что мы потребовали существование корней, поэтому если, кроме того, потребовать выполнение условия f(x 0) >0, then, due to the continuity of the function under study, there is a point x 1 ABOUT (0; x 0) such that f(x 1) = 0. Obviously, this is a smaller root of the equation.
So, f(0) = a 2 – a – 6 > 0, and, putting all the conditions together, we get the system
with the solution a O (3; + Ґ ).
Example 9. For what a does the equation x 2 – 2ax + a 2 – a – 6 have two different negative roots?
Solution. Having studied the type A parabolas in Fig. 1, we get the system
whence a O (– 6; – 2).
Let us generalize the solution to the previous problems in the form of the following rule.
Rule 3. In order for the equation ax 2 + bx + c = 0 to have two different roots x 1 and x 2, each of which is greater (less than) M, it is necessary and sufficient that
Example 10. The function f(x) is given by the formula
Find all values of the parameter a for which the equation f(x) = 0 has at least one solution.
Solution. All possible solutions to a given equation are obtained as solutions to a quadratic equation
x 2 – (4a + 14)x + 4a 2 + 33a + 59 = 0
with the additional condition that at least one (obviously larger) root x 2 i a.
Naturally, for the equation to have roots, it must be = – 5(a + 2) і
0,
whence a Ј – 2.
The graph of the left side of the selected equation is a parabola, the abscissa of the vertex of which is x 0 = 2a + 7. The solution to the problem is given by two types of parabolas (Fig. 2).
A: x 0 i a, from where a i – 7. In this case, the larger root of the polynomial is x 2 i x 0 i a.
B: x 0< a, f(a)
Ј
0, from where 
.
In this case also the larger root of the polynomial is x 2 i a.
Finally 
.
Three solutions to one inequality
Example 11. Find all values of the parameter a for which the inequality x 2 – 2ax + a 2 + 2a – 3 > 0
performed:
1) for all values of x;
2) for all positive values of x;
3) for all values of x O [– 1; 1].
Solution.
First way.
1) Obviously, this inequality holds for all x when the discriminant is negative, i.e.
= a 2 – (a 2 + 2a – 3) = – 2a + 3< 0,
whence a >.
2) To better understand what is required in the problem statement, let’s use a simple technique: draw some parabolas on the coordinate plane, and then take and close the left half-plane relative to the Oy axis. The part of the parabola that remains visible must be above the Ox axis.
The condition of the problem is satisfied in two cases (see Fig. 3):
< 0, откуда a > ;
B: both roots (maybe one, but double) of the equation x 2 – 2ax + a 2 + 2a – 3 = 0 are to the left of the origin. According to rule 3, this condition is equivalent to the system of inequalities Dі 0, x 0 Ј 0 and f(0) і 0.
However, when solving this system, the first inequality can be omitted, since even if some value a does not satisfy the condition Dі 0, then it automatically falls into the solution of point A. Thus, we solve the system
whence a Ј – 3.
Combining the solutions of points A and B, we get
answer: 
3) The condition of the problem is satisfied in three cases (see Fig. 4):
A: the graph of the function y = x 2 – 2ax + a 2 + 2a – 3 lies above the Ox axis, i.e. D< 0, откуда a > ;
B: both roots (maybe one of multiple 2) of the equation x 2 – 2ax + a 2 + 2a – 3 = 0 are to the left of – 1. This condition is equivalent, as we know from rule 3, to the system of inequalities Dі 0, x 0< – 1, f(– 1) > 0;
C: both roots of the equation x 2 – 2ax + a 2 + 2a – 3 = 0 are to the right of 1.
This condition is equivalent to D i 0, x 0 > 1, f(1) > 0.
However, in points B and C, as well as in solving the previous problem, the inequality associated with the discriminant can be omitted.
Accordingly, we obtain two systems of inequalities
Having considered all cases, we get the result: a >
in point
in C.
The answer to the problem is the union of these three sets.
Second way. In order for the condition of each of the three points of the problem to be fulfilled, the smallest value of the function
y = x 2 – 2ax + a 2 + 2a – 3 on each of the corresponding intervals must be positive.
1) The vertex of the parabola y = x 2 – 2ax + a 2 + 2a – 3 is at the point (a; 2a – 3), therefore the smallest value of the function on the entire number line is 2a – 3, and a > .
2) on the semi-axis x i 0 the smallest value of the function is f(0) = a 2 + 2a – 3, if a< 0, и f(a) = 2a – 3, если a
і
0. Analyzing both cases, we get 
3) The smallest on the segment [– 1; 1] function value is
Since the smallest value must be positive, we obtain systems of inequalities
The solution to these three systems is a set
Third way. 1) Vertex of the parabola y = x 2 – 2ax + a 2 + 2a – 3
is located at point (a; 2a – 3). Let us draw a set on the coordinate plane that is formed by the vertices of all parabolas for different a (Fig. 5).
This is the line y = 2x – 3. Let us recall that each point on this line has its own parameter value, and from each point on this line a parabola “comes out”, corresponding given value parameter. Parabolas that are entirely above the Ox axis are characterized by the condition 2a – 3 > 0.
2) The solutions to this point are all the solutions to the first point, and, in addition, parabolas for which a are negative, and f(0) = a 2 + 2a – 3і 0.
3) From Fig. 5 it is clear that we are interested in parabolas for which either a is negative and f(– 1) = a 2 + 4a – 2 > 0,
or a is positive and f(1) = a 2 – 2 > 0.
Equations and inequalities that reduce to quadratic ones
Example 12. For what values of a does the equation 2x 4 – 2ax 2 + a 2 – 2 = 0 have no solutions?
Solution. Making the substitution y = x 2, we obtain the quadratic equation f(y) = 2y 2 – 2ay + a 2 – 2 = 0.
The resulting equation has no solution when D< 0. Кроме того, первоначальное уравнение не имеет решений, когда корни уравнения f(y) = 0 отрицательны.
These conditions can be written as a set
where 
Example 13. For each value of the parameter a, solve the equation cos x sin 2x = asin 3x.
Solution. Since 2cos x sin 2x = sin x + sin 3x and sin 3x = 3sin x – 4sin 3 x,
then the equation will be written as sin x (sin 2 x (4a – 2) – (3a – 2)) = 0.
From here we obtain solutions x = p n, n O Z for any a. The equation 
has solutions 
not coinciding with the solutions of the first equation, only under the condition 
The latter restrictions are equivalent
Answer: x = p n, n O Z for any a; Besides,
Example 14. Find all values of the parameter a, for each of which the inequality
a 2 + 2a – sin 2 x – 2acos x > 2 holds for any number x.
Solution. Let us transform the inequality to the form cos 2 x – 2acos x + a 2 + 2a – 3 > 0
and make the replacement t = cos x. It is important to note that the parameter t ranges from – 1 to 1, so the problem can be reformulated as follows: find all a such that
t 2 – 2at + a 2 + 2a – 3 > 0
holds for all t ABOUT [- 1; 1]. We have already solved this problem earlier.
Example 15. Determine for what values of a the equation log 3 (9 x + 9a 3) = x has solutions and find them.
Solution. Let's transform the equation to the form 9 x – 3 x + 9a 3 = 0
and, making the replacement y = 3 x, we get y 2 – y + 9a 3 = 0.
If the discriminant is negative, the equation has no solutions. When the discriminant
D = 1 – 36a 3 = 0, the equation has a single root,
and x = – log 3 2. Finally, when the discriminant is positive, i.e.,
the original equation has one root 
,
and if, in addition, expression 1 is positive,
then the equation also has a second root 
.
So, we finally get 
,
there are no solutions for the remaining a.
Example 16. For each value of the parameter a, solve the equation sin 4 x + cos 4 x + sin 2 x + a = 0.
Solution. Because
Let's rewrite the equation in the form sin 2 x – 2sin x – 2a – 2 = 0.
Let y = sin 2x, then y 2 – 2y – 2a – 2 = 0 (| y | J 1).
The graph of the function on the left side of the equation is a parabola with a vertex whose abscissa is y 0 = 1; the value of the function at the point y = – 1 is 1 – 2a; the discriminant of the equation is 8a + 12. This means that the larger root y 2 of the equation y 2 – 2y – 2a – 2 = 0, even if it exists, is greater than 1, and the corresponding equation sin 2x = y 2 has no solutions. 3. For what values of a does the equation 2x 2 + (3a + 1)x + a 2 + a + 2 = 0 have at least one root?
4. The equation ax 2 + bx + 5 = 0 has a single root equal to 1. What are a and b equal to?
5. For what values of the parameter a are the roots of the quadratic equation 5x 2 – 7x + a = 0 related as 2 to 5?
6. In the equation ax 2 + 8x + 3 = 0, determine a so that the difference between the roots of the equation is equal to one.
7. For what a does the sum of the squares of the roots of the equation x 2 – 2ax + 2(a + 1) = 0 equal 20?
8. For what b and c does the equation c + bx – 2x 2 = 0 have one positive and one negative root?
9. Find all values of the parameter a for which one root of the equation x 2 – (a + 1)x + 2 = 0 is greater than a, and the other is less than a.
10. Find all values of the parameter a for which the equation x 2 + (a + 1)x + 2 = 0 has two different roots of the same sign.
11. For what values of a are all the resulting roots of the equation (a – 3)x 2 – 2ax + 6a = 0 positive?
12. For what a are all the resulting roots of the equation (1 + a)x 2 – 3ax + 4a = 0 greater than 1?
13. Find all values of the parameter a for which both different roots of the equation x 2 + x + a = 0 are greater than a.
14. For what values of a are both roots of the equation 4x 2 – 2x + a = 0 contained between – 1 and 1?
15. For what values of a does the equation x 2 + 2(a – 1)x + a + 5 = 0 have at least one positive root?
16. The function f(x) is given by the formula
Find all values of the parameter a for which the equation f(x) = 0 has at least one solution.
17. For what a is the inequality (a 2 – 1)x 2 + 2(a – 1)x + 2 > 0 true for all x?
18. For what values of the parameter a does the inequality ax 2 + 2x > 1 – 3a hold for all positive x?
19. For what values of a does the equation x 4 + (1 – 2a)x 2 + a 2 – 1 = 0 have no solutions?
20. For what values of the parameter a does the equation 2x 4 – 2ax 2 + a2 – 2 = 0 have one or two solutions?
21. For each value of a, solve the equation acos x cos 2x = cos 3x.
22. Find all values of the parameter a, for each of which the inequality cos 2 x + 2asin x – 2a< a 2
– 4 выполняется для любого числа x.
23. For all a, solve the equation log 2 (4 x + a) = x.
24. For each value of the parameter a, solve the equation sin 2 x + asin 2 2x = sin.
