Solution of quadratic inequalities presentation. The parabola does not intersect or touch the x-axis
Instructional card Solving quadratic inequalities
- 1. We introduce the corresponding function y = ax 2 + bx + c.
2. Determine the direction of the branches of the parabola y \u003d ax 2 + bx + c
(when a 0 branches are directed upwards; at a 0 branches are directed downwards).
3. Find the zeros of the function, i.e. solve the equation a 2 +bx+c=o.
4. If the equation has roots, then mark the roots on
coordinate line and schematically draw a parabola in accordance with the direction of the branches. If the equation is not
has roots, then we schematically draw a parabola in accordance with the direction of the branches.
5. We find a solution to the inequality, taking into account the meaning of the inequality sign.
Example 1 D 0
Solve the inequality -X 2 - 2x+3 0.
Example 1 D 0
Solve the inequality -X 2 - 2x+3 0.
- Let y = -x 2 - 2x + 3.
- a = -1 0, branches pointing down.
- Solve the equation -x 2 - 2x + 3 = 0
x=1 and x=-3.
4. Note the numbers 1 and -3 on the coordinate
straight line and build a sketch of the graph.
5. Because inequality sign ( ), then the solution
is a segment -3; 1 .
Answer: -3; 1 .
Example 2 D = 0
Solve the inequality 4x 2 +4x+1 0.
Example 2 D = 0
Solve the inequality 4x 2 +4x+1 0.
- Let f(x) = 4x 2 + 4x + 1 .
- a = 4 0 branches pointing upwards.
- Solve the equation 4x 2 + 4x + 1 = 0
X 1 = x 2 = -0,5.
4. The parabola touches the x-axis.
5. Because inequality sign ( ), then the solution
are all numbers except x = -0.5.
Answer: (- ; -0,5) (-0,5; + ).
- Inequality solution 4x 2 +4x+1 0 is span
(- ; + ).
- Inequality solution 4x 2 +4x+1 0 is just the number -0.5.
- Inequality 4x 2 +4x+1 0 has no solution.
Example 3 D 0
Solve the inequality -X 2 - 6x - 10 0.
Example 3 D 0
Solve the inequality -X 2 - 6x - 10 0.
- Let f(x) = -x 2 - 6x - 10.
- a = -1 0, branches pointing down.
- Equation -x 2 - 6x - 10 = 0 has no solution.
4. The parabola does not intersect the x-axis and does not touch it.
5. Because inequality sign ( ), then all numbers are its solution.
Answer: (- ; + ).
Example 3 D 0
Inequality -X 2 - 6x - 10 0 solutions not
Necessary skills and abilities for the successful solution of quadratic inequalities by a graphical method. 1) Be able to solve quadratic equations. 2) Be able to build a graph of a quadratic function and determine from the graph at what values of x the function takes positive, negative, non-positive, non-negative values. shah.ucoz.ru/load/8_klass/8_klass/postroenie_grafikov_vida_u_f_x_l_m_postroenie_grafika_kvadrati chnoj_funkcii/
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Algorithm for solving a quadratic inequality using the example of the inequality Х) Let us determine the direction of the branches of the parabola. a 
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0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => schedule no higher Oh. 6) In this case D=0. x= -2 – touch point. What changed? No higher Oh no there is one dot." title="(!LANG:26.07.201512 Х -2-2 26.07.2015 2) and >0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => schedule no higher Oh. 6) In this case D=0. x= -2 – touch point. What changed? No higher Oh no there is one point." class="link_thumb"> 12 !} X) a>0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => schedule no higher Oh. 6) In this case D=0. x= -2 – touch point. What changed? No higher Oh no there is one point. 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => schedule no higher Oh. 6) In this case D=0. x= -2 – touch point. What changed? Not higher than Ox there is one point. "> 0 - branches. 1) V. f. 3) Ox: 4) Let's schematically depict a parabola. 5) => graph not higher than Ox. 6) In this case D = 0. x = -2 - point of contact. What has changed? Not higher Oh no, there is one point. "> 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => schedule no higher Oh. 6) In this case D=0. x= -2 – touch point. What changed? No higher Oh no there is one dot." title="(!LANG:26.07.201512 Х -2-2 26.07.2015 2) and >0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => schedule no higher Oh. 6) In this case D=0. x= -2 – touch point. What changed? No higher Oh no there is one point."> title="07/26/201512 Х -2-2 07/26/2015 2) a>0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => schedule no higher Oh. 6) In this case D=0. x= -2 – touch point. What changed? No higher Oh no there is one point."> !}
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This presentation can be used to explain the topic "Square inequalities". Textbook Algebra Grade 9. Authors: G.B. Dorofeev, S.B. Suvorova, E.A. Bunimovich, L.V. Kuznetsova, S. S. Minaeva. With the help of animation effects, the concept of quadratic inequality is introduced in an accessible form. The presentation provides an algorithm for solving a quadratic inequality, an example of solving an algorithm, a slide for oral work on a finished drawing of a function graph.
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Quadratic inequalities Mathematics teacher MOU secondary school №57 Astrakhan Bunina N.V.
y 0 y >0 Y=0 x y 2 - 3 1 y=x+x-6 2 With x= -3 and x= 2 With -3 2 With x= -3 and x= 2 x+x-6= 0 At -3 0 y=0 y 0 2 2 2 Inequalities of the form ax+ bx+c ≥ 0, ax+ bx+c > 0 or ax + bx+c ≤0, ax+ bx+c
Algorithm for solving a quadratic inequality Consider the function y \u003d ax 2 + bx + c Find the zeros of the function (solve the equation Determine the direction of the parabola branches Schematically plot the function. Given the inequality sign, write out the answer. ax 2 + bx + c \u003d 0
D >0 D =0 D 0 a
x 2.5 1 Solve the inequality 2x -7x + 5 0 the branches of the parabola are directed upwards Answer: (1; 2.5) 1 . 2x -7x+5 = 0 D=b-4ac=(-7)-4*2*5=9 x =1, x = 2.5 1 2 2 2 2 Example
1 3 y x y= x - 2x - 3 2 Solve the inequality a) x - 2x - 3 >0 2 b) x - 2x - 3≥ 0 2 c) x - 2x - 3
Solve the inequality - 4x + 2x≥0 2 1. - 4x + 2x \u003d 0 2 4x -2x \u003d 0 2 2x (2x -1) \u003d 0 X \u003d 0 x \u003d 0.5 1 2 2. a
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Inequalities in the course of algebra occupy an important place. They were not assigned small part in the content of the entire course of algebra. Thanks to the ability to solve various kinds of inequalities, one can have success in many other sciences. In order for the material taught in the lesson to be better absorbed, it is recommended to use different visualizations, including presentations.
slides 1-2 (Presentation topic "Solving quadratic inequalities. Part 1", example)
This presentation is intended for a lesson explaining new material, which is included in the system of lessons in the section "Inequalities". Before proceeding to the study of this topic "Solving square inequalities", students should receive the necessary amount of knowledge about what inequality is, the properties of numerical inequalities, how linear inequalities are solved. Presentations on these topics are available on this resource.
At the very beginning of the presentation, the author invites students to get acquainted with the concept of square inequalities. He defines them as an inequality of the form ax2+bx+c>0, where a>0. In order to learn how to solve such inequalities, it is enough to know how they look. Therefore, the author immediately suggests studying the ways of solving to consider immediately with examples. And the first such example demonstrates that you need to consider the function that is on the left side of the inequality. You need to build a schedule for it. Since the task is divided into four subparagraphs, and all these inequalities differ only in sign, it is enough for all these cases to have one graph. It is now up to decisions to be made.
For the first case, you need to find all values of the function that take only positive values. On the graph, this will correspond to all points of the graph that lie strictly above the x-axis. In order to determine the solutions of the second case, it is necessary to consider all points of the graph of this function that lie strictly below the x-axis. Since the inequality sign is strictly less than zero. The third case differs from the first only in that the function can also take on the value zero, so zero is also added to the solution of the first case.
slides 3-4 (examples)
Similarly, the fourth case, which is related to the second. It has the same solutions including zero. In this example, just, the author shows how the solutions of the inequality are written correctly in different cases. that is, in which case the bracket is round, in which it is square.
Here is a second example that shows a slightly different way of solving a quadrant inequality. Here it is already necessary to plot the graph of the function not in the coordinate system, but on a straight line, where the points of intersection of the graph with the abscissa axis should be marked. And then, looking at the inequality sign, you should determine what part of the graph is required as solutions, which lies below or above this line. In this case, sections of the graph that lie below the straight line are taken.
Therefore, the solution interval will be double. On the same slide there is another example, which shows the case when the graph does not intersect the straight line, but only touches it at one point. But since, according to the condition, the sign is less than or equal to zero, the section that is located below the straight line should be selected. But there are no such sections, the entire graph lies above. But since zero is allowed in the condition, the only solution is the value of the variable equal to 0.5.
slides 5-6 (solution algorithm, theorem)
Then the author comes to an algorithm for solving quadratic inequalities. It consists of three items. According to the first point, a quadratic equation should be solved by equating the quadratic trinomial to zero. Then mark the obtained roots on a straight line, which is the x axis, and draw a parabola through these points by hand, taking into account the direction of the branches. And then, using this model, find all solutions to the inequality.
And at the end of the presentation, the author proposes to consider a theorem that relates the number of solutions to an inequality from the sign of the discriminant of a trinomial. This means that with a negative discriminant and a positive first coefficient, the inequality ax2 + bx + c, which is greater than or equal to zero, has no solutions, and if it is greater than zero, then the solutions are all real values of the variable x.
This presentation can become an indispensable part of the lesson on the topic "Solving square inequalities." But this presentation is only the first part. Therefore, this topic should be continued. And you can also find a presentation that will be a continuation of this one here. At the request of the teacher, you can add your own examples to the presentation.