Quadratic equations. Basic concepts

The lesson will introduce the concept of a quadratic equation, consider its two types: complete and incomplete. Special attention in the lesson will be paid to the varieties of incomplete quadratic equations; in the second half of the lesson, many examples will be considered.

Topic:Quadratic equations.

Lesson:Quadratic equations. Basic concepts

Definition.Quadratic equation is called an equation of the form

Fixed real numbers that define a quadratic equation. These numbers have specific names:

Senior coefficient (multiplier at);

Second coefficient (multiplier at);

Free term (number without variable multiplier).

Comment. It should be understood that the specified sequence of writing the terms in the quadratic equation is standard, but not mandatory, and in the case of their permutation, it is necessary to be able to determine the numerical coefficients not by their ordinal arrangement, but by belonging to variables.

Definition. The expression is called square trinomial.

Example 1. Quadratic equation given ... Its coefficients:

Senior coefficient;

Second coefficient (note that the coefficient is indicated with a front sign);

Free member.

Definition. If, then the quadratic equation is called unreduced, and if, then the quadratic equation is called given.

Example 2. Bring a quadratic equation ... Let's divide both parts of it into 2: .

Comment. As you can see from the previous example, by dividing by the leading coefficient, we did not change the equation, but changed its form (made it reduced), similarly it could be multiplied by some nonzero number. Thus, the quadratic equation is not given by a single triple of numbers, but they say that is specified up to a nonzero set of coefficients.

Definition.Reduced quadratic equation obtained from the unreduced by dividing by the leading coefficient, and it has the form:

The following designations are adopted:. Then reduced quadratic equation looks like:

Comment... In the reduced form of the quadratic equation, it can be seen that the quadratic equation can be set with just two numbers:.

Example 2 (continued). We indicate the coefficients that define the reduced quadratic equation ... ,. These coefficients are also indicated taking into account the sign. The same two numbers define the corresponding unreduced quadratic equation .

Comment... The corresponding unreduced and reduced quadratic equations are the same, i.e. have the same sets of roots.

Definition... Some of the coefficients in the unreduced form or in the reduced form of the quadratic equation may equal zero. In this case, the quadratic equation is called incomplete... If all the coefficients are nonzero, then the quadratic equation is called complete.

There are several types of incomplete quadratic equations.

If we have not yet considered the solution of the full quadratic equation, then we can easily solve the incomplete one with the methods we already know.

Definition.Solve Quadratic Equation- means to find all the values of the variable (the roots of the equation) at which the given equation turns into the correct numerical equality, or to establish that there are no such values.

Example 3. Consider an example of the specified type of incomplete quadratic equations. Solve the equation.

Solution. Let's take out the common factor. We are able to solve equations of this type according to the following principle: the product is equal to zero if and only if one of the factors is equal to zero, and the other exists for this value of the variable... In this way:

Answer.; .

Example 4. Solve the equation.

Solution. 1 way. Let us factorize by the difference of squares formula

, therefore, similar to the previous example, or.

Method 2. Move the free term to the right and extract the square root of both sides.

Answer. .

Example 5. Solve the equation.

Solution. Move the free term to the right, but , i.e. in the equation, a non-negative number is equated to a negative one, which does not make sense for any values of the variable, therefore, there are no roots.

Answer. There are no roots.

Example 6.Solve the equation.

Solution... Divide both sides of the equation by 7: .

Answer. 0.

Consider examples in which you first need to bring the quadratic equation to standard form, and then solve it.

Example 7... Solve the equation.

Solution... To reduce the quadratic equation to the standard form, it is necessary to transfer all terms in one direction, for example, to the left, and bring similar ones.

An incomplete quadratic equation is obtained, which we already know how to solve, we get that or .

Answer. .

Example 8 (text task)... The product of two consecutive natural numbers is twice the square of the smaller of them. Find these numbers.

Solution... As a rule, word problems are solved according to the following algorithm.

1) Compilation of a mathematical model... At this stage, it is necessary to translate the text of the problem into the language of mathematical symbols (make up an equation).

Let a certain first natural number be denoted by an unknown, then the next one after it (consecutive numbers) will be. The smaller of these numbers is a number, we write the equation according to the condition of the problem:

, where . The mathematical model is compiled.

This video tutorial explains how to solve a quadratic equation. The solution of quadratic equations is usually started in a comprehensive school, grade 8. The roots of the quadratic equation are found using a special formula. Let a quadratic equation of the form ax2 + bx + c = 0 be given, where x is an unknown, a, b and c are coefficients that are real numbers. First, you need to determine the discriminant by the formula D = b2-4ac. After that, it remains to calculate the roots of the quadratic equation using the well-known formula. Now let's try to solve a specific example. We take x2 + x-12 = 0 as the initial equation, i.e. coefficient a = 1, b = 1, c = -12. A well-known formula can be used to determine the discriminant. Then, using the formula for finding the roots of the equation, we calculate them. In our case, the discriminant will be 49. The fact that the value of the discriminant is a positive number tells us that this quadratic equation will have two roots. After some simple calculations, we get that x1 = -4, x2 = 3. Thus, we have solved the quadratic equation by calculating its roots Video lesson “Solving quadratic equations (grade 8). We find roots by the formula "you can watch online at any time for free. Good luck to you!

Quadratic equations are studied in grade 8, so there is nothing difficult here. The ability to solve them is absolutely essential.

A quadratic equation is an equation of the form ax 2 + bx + c = 0, where the coefficients a, b and c are arbitrary numbers, and a ≠ 0.

Before studying specific methods for solving, we note that all quadratic equations can be conditionally divided into three classes:

Have no roots;
Have exactly one root;
They have two distinct roots.

This is an important difference between quadratic and linear equations, where the root always exists and is unique. How do you determine how many roots an equation has? There is a wonderful thing for this - discriminant.

Discriminant

Let a quadratic equation ax 2 + bx + c = 0 be given. Then the discriminant is just the number D = b 2 - 4ac.

You need to know this formula by heart. Where it comes from - it doesn't matter now. Another thing is important: by the sign of the discriminant, you can determine how many roots a quadratic equation has. Namely:

If D< 0, корней нет;
If D = 0, there is exactly one root;
If D> 0, there will be two roots.

Please note: the discriminant indicates the number of roots, and not at all their signs, as for some reason many believe. Take a look at the examples - and you yourself will understand everything:

Task. How many roots do quadratic equations have:

x 2 - 8x + 12 = 0;

5x 2 + 3x + 7 = 0;

x 2 - 6x + 9 = 0.

Let us write down the coefficients for the first equation and find the discriminant:
a = 1, b = −8, c = 12;
D = (−8) 2 - 4 1 12 = 64 - 48 = 16

So the discriminant is positive, so the equation has two different roots. We analyze the second equation in a similar way:
a = 5; b = 3; c = 7;
D = 3 2 - 4 5 7 = 9 - 140 = −131.

The discriminant is negative, there are no roots. The last equation remains:
a = 1; b = −6; c = 9;
D = (−6) 2 - 4 1 9 = 36 - 36 = 0.

The discriminant is zero - there will be one root.

Note that coefficients have been written for each equation. Yes, it’s long, yes, it’s boring - but you won’t mix up the coefficients and don’t make stupid mistakes. Choose for yourself: speed or quality.

By the way, if you “fill your hand”, after a while you will no longer need to write out all the coefficients. You will perform such operations in your head. Most people start doing this somewhere after 50-70 equations are solved - in general, not that much.

Quadratic Roots

Now let's move on to the solution. If the discriminant D> 0, the roots can be found by the formulas:

Basic formula for the roots of a quadratic equation

When D = 0, you can use any of these formulas - you get the same number, which will be the answer. Finally, if D< 0, корней нет — ничего считать не надо.

x 2 - 2x - 3 = 0;

15 - 2x - x 2 = 0;

x 2 + 12x + 36 = 0.

First equation:
x 2 - 2x - 3 = 0 ⇒ a = 1; b = −2; c = −3;
D = (−2) 2 - 4 1 (−3) = 16.

D> 0 ⇒ the equation has two roots. Let's find them:

Second equation:
15 - 2x - x 2 = 0 ⇒ a = −1; b = −2; c = 15;
D = (−2) 2 - 4 (−1) 15 = 64.

D> 0 ⇒ the equation has two roots again. Find them

\ [\ begin (align) & ((x) _ (1)) = \ frac (2+ \ sqrt (64)) (2 \ cdot \ left (-1 \ right)) = - 5; \\ & ((x) _ (2)) = \ frac (2- \ sqrt (64)) (2 \ cdot \ left (-1 \ right)) = 3. \\ \ end (align) \]

Finally, the third equation:
x 2 + 12x + 36 = 0 ⇒ a = 1; b = 12; c = 36;
D = 12 2 - 4 · 1 · 36 = 0.

D = 0 ⇒ the equation has one root. Any formula can be used. For example, the first one:

As you can see from the examples, everything is very simple. If you know the formulas and be able to count, there will be no problems. Most often, errors occur when substituting negative coefficients in the formula. Here, again, the technique described above will help: look at the formula literally, describe each step - and very soon you will get rid of mistakes.

Incomplete quadratic equations

It happens that the quadratic equation is somewhat different from what is given in the definition. For example:

x 2 + 9x = 0;
x 2 - 16 = 0.

It is easy to see that one of the terms is missing in these equations. Such quadratic equations are even easier to solve than standard ones: they do not even need to calculate the discriminant. So, let's introduce a new concept:

The equation ax 2 + bx + c = 0 is called an incomplete quadratic equation if b = 0 or c = 0, i.e. coefficient at variable x or free element is equal to zero.

Of course, a very difficult case is possible when both of these coefficients are equal to zero: b = c = 0. In this case, the equation takes the form ax 2 = 0. Obviously, such an equation has a single root: x = 0.

Let's consider the rest of the cases. Let b = 0, then we get an incomplete quadratic equation of the form ax 2 + c = 0. Let's transform it a little:

Since the arithmetic square root exists only from a non-negative number, the last equality makes sense only for (−c / a) ≥ 0. Conclusion:

If the inequality (−c / a) ≥ 0 holds in an incomplete quadratic equation of the form ax 2 + c = 0, there will be two roots. The formula is given above;
If (−c / a)< 0, корней нет.

As you can see, the discriminant was not required - in incomplete quadratic equations there are no complicated calculations at all. In fact, it is not even necessary to remember the inequality (−c / a) ≥ 0. It is enough to express the value x 2 and see what stands on the other side of the equal sign. If there is a positive number, there will be two roots. If negative, there will be no roots at all.

Now let's deal with equations of the form ax 2 + bx = 0, in which the free element is equal to zero. Everything is simple here: there will always be two roots. It is enough to factor out the polynomial:

Bracketing a common factor

The product is equal to zero when at least one of the factors is equal to zero. From here are the roots. In conclusion, we will analyze several such equations:

Task. Solve quadratic equations:

x 2 - 7x = 0;

5x 2 + 30 = 0;

4x 2 - 9 = 0.

x 2 - 7x = 0 ⇒ x (x - 7) = 0 ⇒ x 1 = 0; x 2 = - (- 7) / 1 = 7.

5x 2 + 30 = 0 ⇒ 5x 2 = −30 ⇒ x 2 = −6. There are no roots, tk. a square cannot be equal to a negative number.

4x 2 - 9 = 0 ⇒ 4x 2 = 9 ⇒ x 2 = 9/4 ⇒ x 1 = 3/2 = 1.5; x 2 = −1.5.

Class: 8

Consider the standard (studied in the school course of mathematics) and non-standard techniques for solving quadratic equations.

1. Decomposition of the left side of the quadratic equation into linear factors.

Let's consider some examples:

3) x 2 + 10x - 24 = 0.

6 (x 2 + x - x) = 0 | : 6

x 2 + x - x - = 0;

x (x -) + (x -) = 0;

x (x -) (x +) = 0;

= ; – .

Answer: ; -.

For independent work:

Solve quadratic equations by linear factoring the left side of a quadratic equation.

a) x 2 - x = 0;

d) x 2 - 81 = 0;

g) x 2 + 6x + 9 = 0;

b) x 2 + 2x = 0;

e) 4x 2 - = 0;

h) x 2 + 4x + 3 = 0;

c) 3x 2 - 3x = 0;

f) x 2 - 4x + 4 = 0;

i) x 2 + 2x - 3 = 0.

a) 0; one

b) -2; 0

c) 0; one

2. The method of selecting a complete square.

Let's consider some examples:

For independent work.

Solve quadratic equations using the full square selection method.

3. Solution of quadratic equations by the formula.

ax 2 + in + c = 0, (a | 4a

4a 2 x 2 + 4av + 4ac = 0;

2ax + 2ax 2b + 2 - 2 + 4ac = 0;

2 = 2 - 4ac; = ±;

Let's look at some examples.

For independent work.

Solve quadratic equations using the formula x 1.2 =.

4. Solving quadratic equations using Vieta's theorem (forward and backward)

x 2 + px + q = 0 - reduced quadratic equation

by Vieta's theorem.

If that equation has two identical roots in sign and it depends on the coefficient.

If p then .

For instance:

If then the equation has two roots of different sign, and the root with the largest absolute value will be if p and will be if p.

For instance:

For independent work.

Without solving the quadratic equation, use the inverse Vieta's theorem to determine the signs of its roots:

a, b, k, l - different roots;

c, d, h - negative;

d, f, g, u, m - positive;

5. Solution of quadratic equations by the “transfer” method.

For independent work.

Solve quadratic equations using the flip method.

6. Solution of quadratic equations using the properties of its coefficients.

I. ax 2 + bx + c = 0, where a 0

1) If a + b + c = 0, then x 1 = 1; x 2 =

Proof:

ax 2 + bx + c = 0 |: a

x 2 + x + = 0.

By Vieta's theorem

By condition a + b + c = 0, then b = -a - c. Then we get

From this it follows that x 1 = 1; x 2 =. Q.E.D.

2) If a - b + c = 0 (or b = a + c), then x 1 = - 1; x 2 = -

Proof:

By Vieta's theorem

By condition a - b + c = 0, i.e. b = a + c. Then we get:

Therefore, x 1 = - 1; x 2 = -.

Let's look at some examples.

1) 345 x 2 - 137 x - 208 = 0.

a + b + c = 345 - 137 - 208 = 0

x 1 = 1; x 2 = =

2) 132 x 2 - 247 x + 115 = 0.

a + b + c = 132 -247 -115 = 0.

x 1 = 1; x 2 = =

Answer: 1;

For independent work.

Applying the properties of the coefficients of the quadratic equation, solve the equations

II. ax 2 + bx + c = 0, where a 0

x 1.2 =. Let b = 2k, i.e. even. Then we get

x 1,2 = = = =

Let's consider an example:

3x 2 - 14x + 16 = 0.

D 1 = (-7) 2 - 3 16 = 49 - 48 = 1

x 1 = = 2; x 2 =

Answer: 2;

For independent work.

a) 4x 2 - 36x + 77 = 0

b) 15x 2 - 22x - 37 = 0

c) 4x 2 + 20x + 25 = 0

d) 9x 2 - 12x + 4 = 0

Answers:

III. x 2 + px + q = 0

x 1,2 = - ± 2 - q

Let's consider an example:

x 2 - 14x - 15 = 0

x 1.2 = 7 = 7

x 1 = -1; x 2 = 15.

Answer: -1; 15.

For independent work.

a) x 2 - 8x - 9 = 0

b) x 2 + 6x - 40 = 0

c) x 2 + 18x + 81 = 0

d) x 2 - 56x + 64 = 0

7. Solving a quadratic equation using graphs.

a) x 2 - 3x - 4 = 0

Answer: -1; 4

b) x 2 - 2x + 1 = 0

c) x 2 - 2x + 5 = 0

Answer: no solutions

For independent work.

Solve quadratic equations graphically:

8. Solving quadratic equations using a compass and a ruler.

ax 2 + bx + c = 0,

x 2 + x + = 0.

x 1 and x 2 are roots.

Let A (0; 1), C (0;

By the secant theorem:

ОВ · ОД = ОА · ОS.

Therefore, we have:

x 1 x 2 = 1 OS;

OS = x 1 x 2

К (; 0), where = -

F (0;) = (0;) =)

1) Construct point S (-;) - the center of the circle and point A (0; 1).

2) Draw a circle with radius R = SA /

3) The abscissas of the points of intersection of this circle with the x-axis are the roots of the original quadratic equation.

There are 3 possible cases:

1) R> SK (or R>).

The circle intersects the x axis at the point B (x 1; 0) and D (x 2; 0), where x 1 and x 2 are the roots of the quadratic equation ax 2 + bx + c = 0.

2) R = SK (or R =).

The circle touches the ox axis in anguish B 1 (x 1; 0), where x 1 is the root of the quadratic equation

ax 2 + bx + c = 0.

3) R< SK (или R < ).

The circle has no points in common with the ox axis, i.e. no solutions.

1) x 2 - 2x - 3 = 0.

Center S (-;), i.e.

x 0 = = - = 1,

y 0 = = = - 1.

(1; - 1) is the center of the circle.

Draw a circle (S; AS), where A (0; 1).

9. Solving quadratic equations using a nomogram

For the solution, they use the four-digit mathematical tables of V.M. Bradis (table XXII, p. 83).

The nomogram allows, without solving the quadratic equation x 2 + px + q = 0, by its coefficients to determine the roots of the equation. For instance:

5) z 2 + 4z + 3 = 0.

Both roots are negative. Therefore, we make the change: z 1 = - t. We get a new equation:

t 2 - 4t + 3 = 0.

t 1 = 1; t 2 = 3

z 1 = - 1; z 2 = - 3.

Answer: - 3; - one

6) If the coefficients p and q are outside the scale, then the substitution z = k · t is performed and the equation is solved using the nomogram: z 2 + pz + q = 0.

k 2 t 2 + p kt + q = 0. |: k 2

k is taken with the expectation that inequalities take place:

For independent work.

2 + 6y - 16 = 0.

y 2 + 6y = 16, | + 9

y 2 + 6y + 9 = 16 + 9

y 1 = 2, y 2 = -8.

Answer: -8; 2

For independent work.

Solve geometrically the equation y 2 - 6y - 16 = 0.

Municipal educational institution
"Kosinskaya basic secondary school"

Lesson using ICT

Solving quadratic equations using the formula.

Developer:
Cherevina Oksana Nikolaevna
mathematic teacher

Target:
fix the solution of quadratic equations by the formula,
contribute to the development of students' desire and need for generalization of the studied facts,
develop independence and creativity.

Equipment:
mathematical dictation (Presentation 1),
cards with multilevel assignments for independent work,
a table of formulas for solving quadratic equations (in the corner "To help with the lesson"),
a printout of the "Old-Time Problem" (number of students),
point-rating table on the board.

Overall plan:
Homework check
Mathematical dictation.
Oral exercises.
Solution of strengthening exercises.
Independent work.
History reference.

During the classes.
Organizational moment.

Homework check.
- Guys, what equations did we meet in the past lessons?
- What methods can be used to solve quadratic equations?
- At home, you had to solve 1 equation in two ways.
(The equation was given in 2 levels, calculated for weak and strong students)
- Let's check with me. how did you handle the assignment.
(on the blackboard, the teacher makes a note of the solution to the home assignment before the lesson)
Pupils check and conclude: incomplete quadratic equations are easier to solve by factorization or in the usual way, complete ones by a formula.
The teacher emphasizes: it is not in vain that the way to solve apt. equations by the formula are called universal.

Repetition.

Today in the lesson we will continue to work with you on solving quadratic equations. Our lesson will be unusual, because today not only I will evaluate you, but you yourself. You must earn as many points as possible to earn a good grade and do well in independent work. One point at a time, I think you've already earned by completing your homework.
- And now I want you to remember and once again repeat the definitions and formulas that we studied on this topic. (Students' answers are evaluated by 1 point for a correct answer, and 0 points for a wrong one)
- And now, guys, we will complete a mathematical dictation, carefully and quickly read the task on the computer monitor. (Presentation 1)
Students get the job done and use the key to evaluate their performance.

Mathematical dictation.

A quadratic equation is an equation of the form ...
In a quadratic equation, the 1st coefficient is ..., the 2nd coefficient is ..., the free term is ...
A quadratic equation is called reduced if ...
Write a formula for calculating the discriminant of a quadratic equation
Write a formula for calculating the root of a quadratic equation if the root in the equation is one.
Under what condition does a quadratic equation have no roots?

(self-test using a PC, for each correct answer - 1 point).

Oral exercises. (on the back of the board)
- How many roots does each equation have? (the task is also estimated at 1 point)
1. (x - 1) (x +11) = 0;
2. (x - 2) ² + 4 = 0;
3. (2x - 1) (4 + x) = 0;
4. (x - 0.1) x = 0;
5.x² + 5 = 0;
6. 9x² - 1 = 0;
7.x² - 3x = 0;
8.x + 2 = 0;
9.16x² + 4 = 0;
10.16x² - 4 = 0;
11.0.07x² = 0.

The solution of exercises to consolidate the material.

Of the equations proposed on the PC monitor, they are performed independently (CD-7), when checking, the students who completed the calculations raise their hands correctly (1 point); at this time, weaker students solve one equation on the blackboard and those who coped with the task on their own receive 1 point.

Independent work in 2 versions.
Those who scored 5 or more points start independent work from No. 5.
Who scored 3 or less - from number 1.

Option 1.

a) 3x² + 6x - 6 = 0, b) x² - 4x + 4 = 0, c) x² - x + 1 = 0.

# 2. Continue calculating the discriminant D of the quadratic equation ax² + bx + c = 0 using the formula D = b² - 4ac.

a) 5x² - 7x + 2 = 0,
D = b² - 4ac
D = (-7²) - 4 5 2 = 49 - 40 =…;
b) x² - x - 2 = 0,
D = b² - 4ac
D = (-1) ² - 4 1 (-2) = ...;

No. 3. Finish solving the equation
3x² - 5x - 2 = 0.
D = b² - 4ac
D = (-5) ² - 4 3 (-2) = 49.
x = ...

No. 4. Solve the equation.

a) (x - 5) (x + 3) = 0; b) x² + 5x + 6 = 0

a) (x-3) ^ 2 = 3x-5; b) (x + 4) (2x-1) = x (3x + 11)

No. 6. Solve the equation x2 + 2√2 x + 1 = 0
No. 7. At what value of a does the equation x² - 2ax + 3 = 0 have one root?

Option 2.

# 1. For each equation of the form ax² + bx + c = 0, enter the values a, b, c.

a) 4x² - 8x + 6 = 0, b) x² + 2x - 4 = 0, c) x² - x + 2 = 0.

# 2. Continue calculating the discriminant D of the quadratic equation ax² + bx + c = 0 using the formula D = b² - 4ac.

a) 5x² + 8x - 4 = 0,
D = b² - 4ac
D = 8² - 4 5 (- 4) = 64 - 60 =…;

b) x² - 6x + 5 = 0,
D = b² - 4ac
D = (-6) ² - 4 1 5 =…;

3 #. Finish solving the equation
x² - 6x + 5 = 0.
D = b² - 4ac
D = (-6) ² - 4 1 5 = 16.
x = ...

No. 4. Solve the equation.

a) (x + 4) (x - 6) = 0; b) 4x² - 5x + 1 = 0

No. 5. Square the equation and solve it:

a) (x-2) ^ 2 = 3x-8; b) (3x-1) (x + 3) + 1 = x (1 + 6x)

No. 6. Solve the equation x2 + 4√3 x + 12 = 0

No. 7. At what value of a does the equation x² + 3ax + a = 0 have one root.

Lesson summary.
Summing up the results of the score-rating table.

Historical background and task.
Problems for quadratic equations have been encountered as early as 499. In ancient India, public competition for solving difficult problems was common. One of the ancient Indian books says: "As the sun eclipses the stars with its brilliance, so a learned man will eclipse the glory of another in popular assemblies, proposing and solving algebraic problems." They were often in poetic form. Here is one of the tasks of the famous 12th century Indian mathematician Bhaskara:
Frisky flock of monkeys
I ate my fill of fun,
Part eighth squared
I was amusing myself in the clearing.
And 12 vines ...
They began to jump while hanging.
How many monkeys were there
You tell me, in this pack?

Vii. Homework.
It is proposed to solve this historical problem and arrange it on separate sheets with a drawing.

APPENDIX

No. Full name
student Activities TOTAL
Homework Dictation Oral exercises Strengthening the material
PC work Whiteboard work
1 Ivanov I.
2 Fedorov G.
3 Yakovleva J.
…

The maximum number is 22-23 points.
Minimum - 3-5 points

3-10 points - score "3",
11-20 points - score "4",
21-23 points - score "5"