Square inequalities presentation. Solving quadratic inequalities, presentation
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/
0. We can solve the inequality graphically. To do this, p" title="(!LANG: Let's build a graph and determine for what values of x the function takes positive values. A square inequality is an inequality that can be reduced to the form ax 2 +bx+c > 0. We can solve the inequality using graphical method.To do this," class="link_thumb"> 3 !} Let's build a graph and determine at what values of x the function takes positive values. A quadratic inequality is an inequality that can be reduced to the form ax 2 +bx+c >0. We can solve the inequality graphically. To do this, consider the function 0. We can solve the inequality graphically. For this, p"> 0. We can solve the inequality graphically. To do this, consider the function "> 0. We can solve the inequality graphically. To do this, p" title="(!LANG: Let's build a graph and determine for what values of x the function takes positive values. A square inequality is an inequality that can be reduced to the form ax 2 +bx+c > 0. We can solve the inequality using graphical method.To do this,"> title="Let's build a graph and determine at what values of x the function takes positive values. A quadratic inequality is an inequality that can be reduced to the form ax 2 +bx+c >0. We can solve the inequality graphically. For this r"> !}
X Y 1 1 x 01 y a > 0 - the branches are directed upwards X x = 2 - the axis of symmetry Let's note the symmetrical points. Let's build a graph. 0 - the branches are directed upwards X x=2 - the axis of symmetry Let's note the symmetrical points. Let's build a graph. "> 0 - the branches are directed upwards Х x=2 - the axis of symmetry Let's note the symmetrical points. Let's build a graph."> 0 - the branches are directed upwards X x=2 - the axis of symmetry Let's note the symmetrical points. Let's build a graph." title="(!LANG:26.07.20154 X Y 1 1 x 01 y-5-8-2 a > 0 - branches are directed upwards X x=2 - axis of symmetry Let's mark the symmetrical points. Let's build a graph."> title="07/26/20154 X Y 1 1 x 01 y-5-8-2 а > 0 - the branches are directed upwards X x=2 is the axis of symmetry Note the symmetrical points. Let's build a graph."> !}
Let us determine at what values of x the function takes positive values X Y 1 1 X (the part of the graph lying above Ox). five
0 - the branches are directed upwards x=2 - the axis of symmetry Let's note the symmetrical points. What actions are needed? Intersection points with Ox." title="(!LANG: What actions were superfluous? Note the symmetrical points.What action is needed?Points of intersection with Ox." class="link_thumb"> 6 !} What actions are redundant? Y 1 1 X 5-1 x 01 y a > 0 - the branches are directed upward x=2 - the axis of symmetry Let's note the symmetrical points. What actions are needed? Intersection points with Oh. 0 - the branches are directed upwards x=2 - the axis of symmetry Let's note the symmetrical points. What actions are needed? Points of intersection with Ox."> 0 - branches are directed upwards x=2 - the axis of symmetry Let's note the symmetrical points. What actions are needed? Points of intersection with Ox."> 0 - the branches are directed upwards x=2 - the axis of symmetry Let's note the symmetrical points. What actions are needed? Intersection points with Ox." title="(!LANG: What actions were superfluous? Note the symmetrical points.What action is needed?Points of intersection with Ox."> title="What actions are redundant? 07/26/20156 Y 1 1 X 5-1 x 01 y-5-8-2 a > 0 - branches are directed upwards x=2 - axis of symmetry Note the symmetrical points. What actions are needed? Intersection points with Oh."> !}
0 - branches are directed upwards 1) Enter the function 3) Find the points of intersection with Ox: for this we solve the quadratic equation branches of a parabola. a > 0 - the branches are directed upwards 1) We introduce the function 3) Find the points of intersection with Ox: for this we solve the quadratic equation" class="link_thumb"> 7 !} 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 > 0 - the branches are directed upwards 1) We introduce the function 3) Find the points of intersection with Ox: for this we solve the quadratic equation 4) We schematically depict a parabola. 5) Let's look at the inequality sign, select the corresponding parts of the graph and the corresponding parts Ox. 6) 0 - the branches are directed upwards 1) We introduce the function 3) Find the points of intersection with Ox: for this we solve the quadratic equation "> 0 - the branches are directed upwards 1) We introduce the function 3) We find the points of intersection with Ox: for this we solve the quadratic equation 4) Schematically depict 5) Look at the inequality sign, select the corresponding parts of the graph and the corresponding parts Ox. 6) "> 0 - the branches are directed upwards 1) Introduce the function 3) Find the points of intersection with Ox: for this we solve the quadratic equation" title = "(! LANG:Algorithm for solving a quadratic inequality on the example of inequality 26.07.20157 X 5 26.07.2015 2) Determine the direction of the branches of the parabola. a > 0 - the branches are directed upwards 1) We introduce the function 3) Find the points of intersection with Ox: for this we solve the quadratic equation"> title="Algorithm for solving a quadratic inequality using the example of an inequality. 07/26/20157 X 5 07/26/2015 2) Determine the direction of the branches of the parabola. a > 0 - the branches are directed upwards 1) We introduce the function 3) Find the points of intersection with Ox: for this we solve the quadratic equation"> !}
Algorithm for solving a quadratic inequality using the example of the inequality Х) Let us determine the direction of the branches of the parabola. but 
Branches, parabola not Oh. How can the parabola y \u003d ax 2 + bx + c be located depending on the behavior of the coefficient a and the discriminant? 1)a>0 D>0 Branches, two points with Ox. X 2) a 0 X 3) a>0 D=0 X 4) a 0 X 5) a>0 D 0 D>0 Branches, two points with Ox. X 2) a 0 X 3) a>0 D=0 X 4) a 0 X 5) a>0 D 0 D>0 Branches, two points with Ox. X 2) a 0 X 3) a>0 D=0 X 4) a 0 X 5) a>0 D 0 D>0 Branches, two points with Ox. X 2) a 0 X 3) a>0 D=0 X 4) a 0 X 5) a>0 D 0 D>0 Branches, two points with Ox. X 2) a 0 X 3) a>0 D=0 X 4) a 0 X 5) a>0 D 
0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => graph not below Oh. 6) In this case D=0. x= -2 – touch point." title="(!LANG:26.07.201510 X -2-2 26.07.2015 2) and >0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => graph not below Oh. 6) In this case D=0. x= -2 – touch point." class="link_thumb">
10
!} X) a>0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => graph not below Oh. 6) In this case D=0. x= -2 – touch point. 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => graph not below Oh. 6) In this case D=0. x \u003d -2 - the point of contact. "\u003e 0 - branches. 1) V. f. 3) Ox: 4) Let's schematically depict a parabola. 5) \u003d\u003e the graph is not lower than Ox. 6) In this case, D \u003d 0. x \u003d -2 – touch point."> 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => graph not below Oh. 6) In this case D=0. x= -2 – touch point." title="(!LANG:26.07.201510 X -2-2 26.07.2015 2) and >0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => graph not below Oh. 6) In this case D=0. x= -2 – touch point.">
title="07/26/201510 X -2-2 07/26/2015 2) a >0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => graph not below Oh. 6) In this case D=0. x= -2 – touch point.">
!}
0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => chart above Oh. 6) In this case D=0. x= -2 – touch point. What has changed?" title="(!LANG:26.07.201511 Х -2-2 26.07.2015 2) and >0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => chart above Oh. 6) In this case D=0. x= -2 – touch point. What changed?" class="link_thumb"> 11 !} X) a>0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => chart above Oh. 6) In this case D=0. x= -2 – touch point. What changed? 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => chart above Oh. 6) In this case D=0. x= -2 – touch point. What has changed?"> 0 - branches. 1) V. f. 3) Ox: 4) Let's schematically depict a parabola. 5) => the graph above Ox. 6) In this case, D \u003d 0. x \u003d -2 - the point of contact. What has changed?" > 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => chart above Oh. 6) In this case D=0. x= -2 – touch point. What has changed?" title="(!LANG:26.07.201511 Х -2-2 26.07.2015 2) and >0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => chart above Oh. 6) In this case D=0. x= -2 – touch point. What changed?"> title="07/26/201511 Х -2-2 07/26/2015 2) a>0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => chart above Oh. 6) In this case D=0. x= -2 – touch point. What changed?"> !}
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."> !}
0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => chart below Oh. 6) In this case D=0. x= -2 – touch point. What changed? Ø There are no dots below Ox." title="(!LANG:26.07.201513 Х -2-2 26.07.2015 2) and >0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => chart below Oh. 6) In this case D=0. x= -2 – touch point. What changed? Ø Below Oh, there is not a single dot." class="link_thumb"> 13 !} X) a>0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => chart below Oh. 6) In this case D=0. x= -2 – touch point. What changed? Ø Below Oh, there is not a single dot. 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => chart below Oh. 6) In this case D=0. x= -2 – touch point. What changed? Ø Below Ox there is not a single point. "> 0 - branches. 1) V. f. 3) Ox: 4) Let's schematically depict a parabola. 5) => graph below Ox. 6) In this case D = 0. x = - 2 - point of contact. What has changed? Ø There is not a single point below Ohh. "> 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => chart below Oh. 6) In this case D=0. x= -2 – touch point. What changed? Ø There are no dots below Ox." title="(!LANG:26.07.201513 Х -2-2 26.07.2015 2) and >0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => chart below Oh. 6) In this case D=0. x= -2 – touch point. What changed? Ø Below Oh, there is not a single dot."> title="07/26/201513 Х -2-2 07/26/2015 2) a>0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => chart below Oh. 6) In this case D=0. x= -2 – touch point. What changed? Ø Below Oh, there is not a single dot."> !}
X) a>0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => graph not below Oh. 6) Do not intersect with Oh. 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => graph not below Oh. 6) Not points of intersection with Ox."> 0 - branches. 1) V. f. 3) Ox: 4) Let's schematically depict a parabola. 5) \u003d\u003e graph not lower than Ox. 6) No points of intersection with Ox."> 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => graph not below Oh. 6) Not intersection points with Ox." title = "(!LANG: 07/26/201514 X 07/26/2015 2) and > 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => graph not below Oh. 6) Do not intersect with Oh.">
title="07/26/201514 X 07/26/2015 2) a >0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => graph not below Oh. 6) Do not intersect with Oh.">
!}
Inequalities in the course of algebra occupy an important place. They are assigned not a 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.
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.
0). 9 \u003d 0, x 2 \u003d 9, x 1.2 \u003d 3, mark the roots on the Ox axis 2. The branches of the parabola are directed upwards (a \u003d 1, 1> 0) parabolas lie above or on the x-axis" class="link_thumb"> 5 !} Example 1 Solve the inequality: x 2 - x 2 - 9 \u003d 0, x 2 \u003d 9, x 1.2 \u003d 3, mark the roots on the Ox axis 2. The branches of the parabola are directed upwards (a = 1, 1> 0) sketch of the graph 4. 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) 5. Answer: x - 3, x x x - 3 x 3 0) 3. Draw a sketch of the graph 4. We are looking for x values at which the points of the parabola lie above or on the Ox axis (the y sign is "\u003e 0) 3. We draw a sketch of the graph 4. We are looking for x values at which the points of the parabola lie above or on axis Ox (the y sign of the inequality is not strict) 5. Answer: x - 3, x 3 - 3 3 x x - 3 x 3 "> 0) 3. Draw a sketch of the graph 4. We are looking for x values at which the parabola points lie above or on x-axis (sign y ne" title="(!LANG: Example 1 Solve the inequality: x 2 - 9 0 1.x 2 - 9 = 0, x 2 = 9, x 1,2 = 3, mark the roots on the x-axis 2. The branches of the parabola are directed upwards (a \u003d 1, 1> 0) 3. We draw a sketch of the graph 4. We are looking for x values at which the points of the parabola lie above or on the Ox axis (the y sign is not"> title="Example 1 Solve the inequality: x 2 - 9 0 1.x 2 - 9 = 0, x 2 = 9, x 1.2 = 3, mark the roots on the Ox axis ) 3. Draw a sketch of the graph 4. We are looking for x values at which the points of the parabola lie above or on the Ox axis (the y sign is not"> !}
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\u003e 0 1. x 2 - x +12 \u003d 0, x 1 \u003d - 4, x 2 \u003d 3">!}
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 at 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
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 the 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
On the topic: methodological developments, presentations and notes
Methodical manual: "The system of exercises. Inequalities and systems of inequalities".
This manual proposes a system of exercises with solutions on the topic: "Inequalities and systems of inequalities" for students in grades 10-11 ....
Reduction of a logarithmic inequality to a system of rational inequalities
In this development, we consider a standard method for solving a logarithmic inequality based on a variable. The standard solution method involves parsing d...
Control and generalizing lesson "Solving inequalities and systems of inequalities with one variable"
Control and generalizing lesson "Solving inequalities and systems of inequalities with one variable". The purpose of the lesson: generalization, systematization and testing of knowledge, skills and abilities in ...
This lesson is a reinforcing lesson on the topic "Solving inequalities and systems of inequalities" in grade 8. A presentation was created to help the teacher ....
Topic 6. ALGEBRAIC INEQUALITIES. SQUARE INEQUALITIES. RATIONAL INEQUALITIES OF HIGHER DEGREES. FRACTIONAL-RATIONAL INEQUALITIES. Theory. Key methods for solving problems. Exercises.
Final control on topics No. 6,7: “Algebraic inequalities. Square inequalities. Rational inequalities of higher degrees. Fractional-rational inequalities. Inequalities with the modulus. Irrational inequalities»
Dear Colleagues, An urgent task today is the qualitative preparation of students for the state final certification (GIA) and the unified state exam (USE) in mathematics, ...
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 = - 4, x 2 \u003d 3 The branches of the parabola are directed downwards (a \u003d - 1, -1 ”) Answer: - 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 of a multimedia presentation with video and sound, which will allow it to be used both in the lesson and for ...