Solve graphically equation x2. Graphical solution of quadratic equations

Hello. In this article, I will try to show you possible ways solutions quadratic equations using charts.

Suppose we need to solve the equation x 2 - 2x - 3 = 0. In this example, we will consider the options for solving a quadratic equation graphically.

1) We can represent our equation in the form x 2 \u003d 2x + 3. Next, we will build graphs of the functions y \u003d x 2 and y \u003d 2x + 3 in the same coordinate system. The graph y \u003d x 2 is shown in Figure 1, and both graphs in Figure 2.

Picture 1 Figure 2

The graphs intersect at two points, our equation has a solution x = -1 and x = 3.

2) But you can represent the equation in a different way, for example, x 2 - 2x \u003d 3 and plot the graphs of the functions y \u003d x 2 - 2x and y \u003d 3 in one coordinate system. You can see them in Figures 3 and 4. Figure 3 shows a graph y = x 2 - 2x, and Figure 4 shows both graphs y \u003d x 2 - 2x and y \u003d 3.

Figure 3 Figure 4

As we can see, these two graphs also intersect at two points, where x = -1 and x = 3. So answer: - 1; 3.

3) There is another version of the representation of this equation x 2 - 3 = 2x. And again we build graphs of the functions y \u003d x 2 - 3 and y \u003d 2x in the same coordinate system. The first y = x 2 - 3 in Figure 5 and both graphs in Figure 6.

Figure 5 Figure 6

Answer: - 1; 3.

4) You can build a parabola y \u003d x 2 - 2x - 3.

The top of the parabola x 0 \u003d - b / 2a \u003d 2/2 \u003d 1, y 0 \u003d 1 2 - 2 1 - 3 \u003d 1 - 2 - 3 \u003d - 4. This is the point (1; - 4). Then our parabola is symmetrical with respect to the line x = 1. If we take two points that are symmetrical with respect to the straight line x = 1, for example: x = - 2 and x = 4, then we will get two points through which the branches of the graph pass.

If x \u003d -2, then y \u003d (- 2) 2 - 2 (-2) - 3 \u003d 4 + 4 - 3 \u003d 5.

Similarly, x \u003d 4, y \u003d 4 2 - 2 4 - 3 \u003d 16 - 8 - 3 \u003d 5. The points obtained are (-2; 5); (1; 4) and (4; 5) we mark in on the plane and draw a parabola in Figure 7.

Figure 7

The parabola intersects the x-axis at points - 1 and 3. These are the roots of the equation x 2 - 2x - 3 = 0.

Answer: - 1 and 3.

5) And you can select the square of the binomial:

x 2 - 2x - 3= 0

(x 2 - 2x + 1) - 1 - 3= 0

(x -1) 2 - 4 = 0

Then plot the graphs of the functions y = (x - 1) 2 and y = 4 in the same coordinate system. The first graph y = (x - 1) 2 in Figure 8, and both graphs y = (x - 1) 2 and y = 4 in figure 9.

Figure 8 Figure 9

They also intersect at two points where x = -1, x = 3.

Answer: - 1; 3.

6) Since x \u003d 0 is not the root of the equation x 2 - 2x - 3 \u003d 0 (otherwise, the equality 0 2 - 2 0 -3 \u003d 0 would hold), then all terms of the equation can be divided by x. As a result, we get the equation x - 2 - 3 / x \u003d 0. Move 3 / x to the right and get the equation x - 2 \u003d 3 / x Then we can plot the graphs of the functions y \u003d 3 / x and y \u003d x - 2 in the same coordinate system .

Figure 10 shows the graph of the function y \u003d 3 / x, and in Figure 11 both graphs of the functions y \u003d 3 / x and y \u003d x - 2.

Figure 10 Figure 11

They also intersect at two points where x = -1, x = 3.

Answer: - 1; 3.

If you were careful, you noticed that no matter how you represent the equation in the form of two functions, you will always have the same answer (of course, you will not make mistakes when transferring expressions from one part of the equation to another and when plotting). Therefore, when solving an equation graphically, choose a way to represent the functions of a graph that is easier for you to construct. And one more note, if the roots of the equation are not integers, then the answer will not be accurate.

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You have already met with quadratic equations in the 7th grade algebra course. Recall that a quadratic equation is an equation of the form ax 2 + bx + c \u003d 0, where a, b, c are any numbers (coefficients), and a. Using our knowledge of some functions and their graphs, we are now able, without waiting for the systematic study of the topic "Quadricular Equations", to solve some quadratic equations, and different ways; we will consider these methods using the example of one quadratic equation.

Example. Solve the equation x 2 - 2x - 3 = 0.
Solution.
I way . Let's build a graph of the function y \u003d x 2 - 2x - 3, using the algorithm from § 13:

1) We have: a \u003d 1, b \u003d -2, x 0 \u003d \u003d 1, y 0 \u003d f (1) = 1 2 - 2 - 3 \u003d -4. This means that the point (1; -4) is the vertex of the parabola, and the straight line x = 1 is the axis of the parabola.

2) Take two points on the x-axis that are symmetrical about the axis of the parabola, for example, the points x \u003d -1 and x \u003d 3.

We have f(-1) = f(3) = 0. Let's construct points (-1; 0) and (3; 0) on the coordinate plane.

3) Through the points (-1; 0), (1; -4), (3; 0) we draw a parabola (Fig. 68).

The roots of the equation x 2 - 2x - 3 \u003d 0 are the abscissas of the points of intersection of the parabola with the x-axis; so the roots of the equation are: x 1 \u003d - 1, x 2 - 3.

II way. We transform the equation to the form x 2 \u003d 2x + 3. Let us construct graphs of the functions y - x 2 and y \u003d 2x + 3 in one coordinate system (Fig. 69). They intersect at two points A(-1; 1) and B(3; 9). The roots of the equation are the abscissas of points A and B, which means that x 1 \u003d - 1, x 2 - 3.


III way . Let's transform the equation to the form x 2 - 3 = 2x. Let us construct graphs of the functions y \u003d x 2 - 3 and y \u003d 2x in one coordinate system (Fig. 70). They intersect at two points A (-1; - 2) and B (3; 6). The roots of the equation are the abscissas of points A and B, therefore x 1 \u003d - 1, x 2 \u003d 3.

IV way. Let's transform the equation to the form x 2 -2x 4-1-4 \u003d 0
and beyond
x 2 - 2x + 1 = 4, i.e. (x - IJ = 4.
Let's build a parabola y \u003d (x - 1) 2 and a straight line y \u003d 4 in one coordinate system (Fig. 71). They intersect at two points A(-1; 4) and B(3; 4). The roots of the equation are the abscissas of points A and B, therefore x 1 \u003d -1, x 2 \u003d 3.

V way. Dividing term by term both sides of the equation by x, we get


We construct a hyperbola and a straight line y \u003d x - 2 in the same coordinate system (Fig. 72).

They intersect at two points A (-1; -3) and B(3; 1). The roots of the equation are the abscissas of points A and B, therefore, x 1 \u003d - 1, x 2 \u003d 3.

So, we solved the quadratic equation x 2 - 2x - 3 \u003d 0 graphically in five ways. Let's analyze the essence of these methods.

I way. Build a graph of the function at the point of its intersection with the x-axis.

II way. They transform the equation to the form ax 2 \u003d -bx - c, build a parabola y \u003d ax 2 and a straight line y \u003d -bx - c, find their intersection points (the roots of the equation are the abscissas of the intersection points, if, of course, there are any).

III way. Transform the equation to the form ax 2 + c \u003d - bx, build a parabola y - ax 2 + c and a straight line y \u003d -bx (it passes through the origin); find their points of intersection.

IV way. Applying the full square selection method, the equation is converted to the form

Build a parabola y \u003d a (x + I) 2 and a straight line y \u003d - m, parallel to the x axis; find the points of intersection of the parabola and the line.

V way. Transform the equation to the form


They build a hyperbola (this is a hyperbola provided that ) and a straight line y \u003d - ax - b; find their points of intersection.

Note that the first four methods are applicable to any equations of the form ax 2 + bx + c \u003d 0, and the fifth - only to those with c. In practice, you can choose whichever way you feel best fits the given equation, or which you like best (or more understandable).

Comment . Despite the abundance of ways to graphically solve quadratic equations, the confidence that any quadratic equation we
we can solve graphically, no. Let, for example, you need to solve the equation x 2 - x - 3 \u003d 0 (we will specially take an equation similar to what was in
considered example). Let's try to solve it, for example, in the second way: we transform the equation to the form x 2 \u003d x + 3, we construct a parabola y \u003d x 2 and
straight line y \u003d x + 3, they intersect at points A and B (Fig. 73), which means that the equation has two roots. But what are these roots, we use the drawing
we can’t say - points A and B do not have such “good” coordinates as in the above example. Now consider the equation
x 2 - 16x - 95 = 0. Let's try to solve it, say, in the third way. Let's transform the equation to the form x 2 - 95 = 16x. Here you need to build a parabola
y \u003d x 2 - 95 and a straight line y \u003d 16x. But the limited size of the notebook sheet does not allow this, because the parabola y \u003d x 2 must be lowered 95 cells down.

So, graphical methods for solving a quadratic equation are beautiful and pleasant, but they do not give a 100% guarantee of solving any quadratic equation. We will take this into account later.

Sometimes equations are solved graphically. To do this, you need to transform the equation so (if it is not already presented in a transformed form) that there are expressions to the left and right of the equal sign, for which you can easily draw graphs of functions. For example, given the following equation:
x² - 2x - 1 = 0

If we have not yet studied the solution of quadratic equations in an algebraic way, then we can try to do this either by factoring or graphically. To solve such an equation graphically, we represent it in this form:
x² = 2x + 1

From such a representation of the equation, it follows that it is required to find such values ​​x for which the left side will be equal to the right.

As you know, the graph of the function y \u003d x² is a parabola, and y \u003d 2x + 1 is a straight line. The x coordinate of the points of the coordinate plane lying both on the first graph and on the second one (that is, the intersection points of the graphs) are precisely those x values ​​for which the left side of the equation will be equal to the right. In other words, the x coordinates of the intersection points of the graphs are the roots of the equation.

Graphs can intersect at several points, at one point, not intersect at all. It follows that an equation may have several roots, or one root, or none at all.

Let's look at a simpler example:
x² - 2x = 0 or x² = 2x

Let's draw the graphs of the functions y = x² and y = 2x:

As can be seen from the drawing, the parabola and the straight line intersect at points (0; 0) and (2; 4). The x coordinates of these points are respectively equal to 0 and 2. Hence, the equation x² - 2x \u003d 0 has two roots - x 1 \u003d 0, x 2 \u003d 2.

Let's check this by solving the equation by taking the common factor out of brackets:
x² - 2x = 0
x(x - 2) = 0

Zero on the right side can be obtained either with x equal to 0 or 2.

The reason why we did not graphically solve the equation x² - 2x - 1 = 0 is that in most equations the roots are real (fractional) numbers, and it is difficult to accurately determine the value of x on the graph. Therefore, for most equations, the graphical method of solving is not the best. However, knowing this way gives a deeper understanding of the relationship between equations and functions.

:
- x^2 = 2x

Solution.
The graphical solution of equations comes down to the fact that you need to build functions that are on both sides of the equal sign in the equation, and find their intersection points. The abscissas of these points will be the roots of the given equation.
So we have an equation:

This equation consists of two functions that are equal to each other:

Let's build first function. To do this, we will conduct a small analysis of it.
The function is quadratic, therefore, its graph will be . The x square is preceded by a minus sign, which means that the function is directed downwards with branches. The function is even, since it is quadratic. The function does not have any coefficients and free members, which means that its vertex will be at the origin.
Find several points through which the function passes. To do this, instead of the variable x, we substitute the values ​​1, -1, 2 and -2.
, - point (-1; -1)
, - point (1; -1)
, - dot (-2; -4)
, - dot (2; -4)
Let's put all the points on the plane and draw a smooth curve through them.
Let's build second function. The function is , therefore, two points are enough to construct it. Let's find these points as the intersection points of the function with the coordinate axes.
With the x-axis: y = 0. Substitute the value at into the equation:

With the y-axis: x = 0.

We got only one point (0; 0). To find the second one, substitute an arbitrary value for the variable x, for example, 1.

Second point - (1; 2)
Let's plot these two points on the same coordinate plane and draw a straight line through them.
Now you need to lower the perpendiculars to the Ox axis from the intersection points of the function graphs and get the points 0 and -2.
These values ​​are the result of the graphical solution of the original equation.

 

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