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 "Solution square 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.

Graphical method for solving quadratic inequalities Algebra Grade 8

Definition Square inequalities are inequalities of the form ax 2 + b x + c> 0, ax 2 + b x + c

According to the graph of the function y \u003d x 2 - 6 x + 8, determine at what values ​​x a) y \u003d 0, b) y > 0, c) y 0 at x 4 y

Algorithm for solving a quadratic inequality Find the roots of a square trinomial ax 2 + b x + c Mark the found roots on the x-axis and determine where the branches of the parabola are directed (up or down), serving as a graph of the function y \u003d ax 2 + b x + c; sketch a graph. Using the obtained geometric model, determine at what intervals of the x-axis the ordinates of the graph are positive (negative); include these gaps in the answer.

Example 1 Solve the inequality: x 2 - 9  0 x 2 - 9 \u003d 0, x 2 \u003d 9, x 1.2 \u003d  3, mark the roots on the Ox axis The branches of the parabola are directed upwards (a \u003d 1, 1> 0) We draw graph sketch We are looking for x values ​​at which the points of the parabola lie above or on the Ox axis (the y sign of the inequality is not strict “ ≥ ”) Answer: x  - 3, x  3 - 3 3 x  x  - 3 x  3

Example 2 Solve the inequality: - x 2 - x +12 > 0 - x 2 - x +12 \u003d 0, x 1 \u003d - 4, x 2 \u003d 3 - 4 - 4

Example 3 Solve the inequality: x 2 + 9 > 0 x 2 + 9 \u003d 0, x 2 \u003d - 9, - 9 0) Draw a sketch of the graph We are looking for x values ​​for which the graph of the function is located above the Ox axis. Answer: x is any number (or (- ∞; + ∞)) . x All points of the parabola lie above the x-axis. The inequality holds for any value of x

Example 4 Solve the inequality: x 2 + 9 0) Draw a sketch of the graph We are looking for x values ​​at which the graph of the function is located below the Ox axis. Answer: there are no solutions x There are no points on the parabola that lie below the x-axis. The inequality has no solutions.

Example 5 Solve the inequality: - 4x 2 + 12x-9  0 - 4x 2 + 12x-9 \u003d 0, D \u003d 0, x \u003d 1.5

Example 6 Solve the inequality: - 4x 2 +12x-9\u003e 0 - 4x 2 +12x-9 \u003d 0, D \u003d 0, x \u003d 1.5

Example 7 Solve the inequality: - 4x 2 + 12x-9  0 - 4x 2 + 12x-9 \u003d 0, D \u003d 0, x \u003d 1.5

Example 8 Solve the inequality: - 4x 2 + 12x-9

On the topic: methodological developments, presentations and notes

1. Demonstration material for systematization and generalization of knowledge on the above topic, made in the form multimedia presentation with video and sound, which will allow you to use it both in the lesson and for ...

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 solutions.

Answer: (-  ; +  ).

Example 3 D  0

Inequality -X 2 - 6x - 10  0 solutions not

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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Definition Quadratic inequalities are inequalities of the form ax 2 + bx + c> 0, ax 2 + bx + c 0, ax 2 +bx+c"> 0, ax 2 +bx+c"> 0, ax 2 +bx+c" title="(!LANG:Definition Quadratic inequalities are inequalities of the form ax 2 +bx+c>0 , ax 2 + bx + c"> 0, ах 2 +bх+c" title="Definition Quadratic inequalities are inequalities of the form ax 2 + bx + c> 0, ax 2 + bx + c"> !}

According to the graph of the function y \u003d x 2 - 6x +8, determine at what values ​​x a) y \u003d 0, b) y\u003e 0, c) y0 at x 4 y 0, c) y0 at x 4 y"> 0, c) y0 at x 4 y"> 0, c) y0 at x 4 y" title="(!LANG: According to the graph of the function y= x 2 - 6x +8 determine for what values ​​x a) y=0, b) y>0, c) y0 at x 4 y"> title="According to the graph of the function y \u003d x 2 - 6x +8, determine at what values ​​x a) y \u003d 0, b) y\u003e 0, c) y0 at x 4 y"> !}

Algorithm for solving a quadratic inequality 1. Find the roots of the square trinomial ax 2 + bx + c 2. Mark the found roots on the x-axis and determine where the branches of the parabola are directed (up or down), serving as a graph of the function y \u003d ax 2 + bx + c; sketch a graph. 3. Using the obtained geometric model, determine at what intervals of the x-axis the ordinates of the graph are positive (negative); include these gaps in the answer.

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0 1. x 2 - x +12 = 0, x 1 = - 4, x 2 = 3 2 - x +12\u003e 0 1. x 2 - x +12 \u003d 0, x 1 \u003d - 4, x 2 \u003d 3" class="link_thumb"> 6 !} Example 2 Solve the inequality: x 2 - x +12 > 0 1. x 2 - x +12 \u003d 0, x 1 \u003d - 4, x 2 \u003d 3 2. The branches of the parabola are directed downward (a \u003d - 1, -1) 5 .Answer: - 4 - 4 0 1. x 2 - x +12 = 0, x 1 = - 4, x 2 = 3 2. The branches of the parabola are directed downwards (a = - 1, -1 "> 0 1. x 2 - x +12 = 0, x 1 = - 4, x 2 = 3 2. The branches of the parabola are directed downwards (a = - 1, -1) 5. Answer: - 4 - 4 0 1. x 2 - x +12 = 0, x 1 = - 4, x 2 = 3 2 - x +12 = 0, x 1 = - 4, x 2 = 3 2. The branches of the parabola are directed downwards (a = - 1, -1"> title="Example 2 Solve the inequality: x 2 - x +12 > 0 1. x 2 - x +12 \u003d 0, x 1 \u003d - 4, x 2 \u003d 3 2. The branches of the parabola are directed downward (a \u003d - 1, -1">!}

0 1.x 2 + 9 = 0, x 2 = 9, 9 0) 3. Draw a sketch of the graph 4. We are looking for x values ​​for which the function graph is located above the axis "title =" (!LANG: Example 3 Solve the inequality: x 2 + 9 > 0 1.x 2 + 9 = 0, x 2 = 9, 9 0)" class="link_thumb"> 7 !} Example 3 Solve the inequality: x\u003e 0 1.x \u003d 0, x 2 \u003d 9, 9 0) 3. Draw a sketch of the graph 4. We are looking for x values ​​for which the graph of the function is located above the Ox axis. 5. Answer: x - any number (or (-; +)). x All points of the parabola lie above the x-axis. The inequality holds for any value of x 0 1.x 2 + 9 \u003d 0, x 2 \u003d 9, 9 0) 3. Draw a sketch of the graph 4. We are looking for x values ​​for which the graph of the function is located above the axis "\u003e 0 1.x 2 + 9 \u003d 0, x 2 \u003d 9, 9 0) 3. Draw a sketch of the graph 4. Find the x values ​​for which the graph of the function is located above the Ox axis 5. Answer: x is any number (or (-; +)) x All points of the parabola lie above the Ox axis. The inequality is satisfied for any value x "> 0 1.x 2 + 9 = 0, x 2 = 9, 9 0) 3. Draw a sketch of the graph 4. We are looking for x values ​​for which the function graph is located above the axis" title = "(!LANG : Example 3 Solve the inequality: x 2 + 9 > 0 1. x 2 + 9 \u003d 0, x 2 \u003d 9, 9 0)"> title="Example 3 Solve the inequality: x 2 + 9 > 0 1. x 2 + 9 = 0, x 2 = 9, 9 0)"> !}

0) 3. Draw a sketch of the graph 4. We are looking for x values ​​for which the function graph is located below oc" title="(!LANG: Example 4 Solve the inequality: x 2 + 9 0) 3. We draw a sketch of the graph 4. We are looking for x values ​​for which the graph of the function is located below" class="link_thumb"> 8 !} Example 4 Solve the inequality: x 0) 3. Draw a sketch of the graph 4. We are looking for x values ​​for which the graph of the function is located below the Ox axis. 5. Answer: there are no solutions x There are no points on the parabola that lie below the x-axis. The inequality has no solutions. 0) 3. Draw a sketch of the graph 4. We are looking for x values ​​at which the function graph is located below os "\u003e 0) 3. We draw a sketch of the graph 4. We are looking for x values ​​at which the function graph is below the Ox axis. 5. Answer: there are no solutions x On There are no points in the parabola lying below the Ox axis. The inequality has no solutions."> 0) 3. Draw a sketch of the graph 4. We are looking for x values ​​for which the graph of the function is located below the axis" title = "(!LANG: Example 4 Solve the inequality: x 2 + 9 0) 3. Draw a sketch of the graph 4. We are looking for x values ​​for which the graph of the function is located below the axis"> title="Example 4 Solve the inequality: x 2 + 9 0) 3. Draw a sketch of the graph 4. We are looking for x values ​​for which the graph of the function is located below the axis"> !}

Example 5 Solve the inequality: - 4x 2 + 12x x 2 + 12x-9 = 0, D = 0, x = 1.5 2. The branches of the parabola are directed downwards (a = 4, 4

Example 6 Solve the inequality: - 4x 2 + 12x-9> x 2 + 12x-9 = 0, D = 0, x = 1.5 2. The branches of the parabola are directed downwards (a = 4, 4 0 1.- 4x 2 + 12x-9 \u003d 0, D \u003d 0, x \u003d 1.5 \u003d 0, x \u003d 1.5 2. The branches of the parabola are directed downwards (a \u003d 4, 4 "\u003e 0 1.- 4x 2 + 12x-9 \u003d 0, D \u003d 0, x \u003d 1.5 2. The branches of the parabola are directed downwards (a = 4, 4" title="(!LANG: Example 6 Solve the inequality: - 4x 2 +12x-9>0 1.- 4x 2 +12x-9=0, D = 0, x=1.5 2 .The branches of the parabola are directed downwards (a \u003d 4, 4"> title="Example 6 Solve the inequality: - 4x 2 +12x-9>0 1.- 4x 2 +12x-9=0, D = 0, x=1.5 2. The branches of the parabola are directed downwards (a = 4, 4"> !}

Example 7 Solve the inequality: - 4x 2 + 12x x 2 + 12x-9 = 0, D = 0, x = 1.5 2. The branches of the parabola are directed downwards (a = 4, 4