"Kosinskaya basic comprehensive school"
Lesson using ICT
Solution of quadratic equations by 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 to generalize the facts being studied,
develop independence and creativity.
Equipment:
mathematical dictation (Presentation 1),
cards with multi-level tasks for independent work,
a table of formulas for solving quadratic equations (in the "To help with the lesson" corner),
printout of the "Old Problem" (number of students),
score-rating table on the board.
Overall plan:
Checking homework
Mathematical dictation.
oral exercises.
Solving strengthening exercises.
Independent work.
History reference.
During the classes.
Organizational moment.
Checking homework.
- Guys, what equations did we meet in the last lessons?
How can you solve quadratic equations?
- At home you had to solve 1 equation in two ways.
(The equation was given in 2 levels, designed for weak and strong students)
Let's check with me. How did you complete the task.
(on the blackboard, the teacher writes down the solution to the home assignment before the lesson)
Students check and conclude: incomplete quadratic equations are easier to solve by factoring or in the usual way, complete ones - by formula.
The teacher emphasizes: it is not in vain that the way to solve square. equations according to the formula is called universal.
Repetition.
Today in the lesson we will continue with you to solve quadratic equations. Our lesson will be unusual, because today not only I will evaluate you, but you yourself. In order to earn a good grade and be successful in self-study, you must earn as many points as possible. One point each, I think you have already earned by doing your homework.
- And now I want you to remember and repeat once again the definitions and formulas that we studied on this topic. (Students' answers are scored 1 point for a correct answer, and 0 points for an incorrect one)
- And now, guys, we will perform a mathematical dictation, carefully and quickly read the task on the computer monitor. (Presentation 1)
Students do the work, and use the key to evaluate their work.
Mathematical dictation.
A quadratic equation is an equation of the form ...
In the 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 the formula for calculating the root of a quadratic equation if there is only one root in the equation.
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 \u003d 0;
3. (2x - 1) (4 + x) \u003d 0;
4. (x – 0.1)x = 0;
5. x² + 5 = 0;
6. 9x² - 1 \u003d 0;
7. x² - 3x \u003d 0;
8. x + 2 = 0;
9. 16x² + 4 = 0;
10. 16x² - 4 \u003d 0;
11. 0.07x² \u003d 0.
Solution of exercises to consolidate the material.
Of the equations proposed on the PC monitor, they are performed independently (CD-7), when checking, students who have completed the calculations correctly raise their hands (1 point); at this time, weaker students solve one equation on the board and those who coped with the task on their own receive 1 point each.
Independent work in 2 versions.
Those who scored 5 or more points start independent work from No. 5.
Who scored 3 or less - from No. 1.
Option 1.
a) 3x² + 6x - 6 = 0, b) x² - 4x + 4 = 0, c) x² - x + 1 = 0.
No. 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 \u003d (-7²) - 4 5 2 \u003d 49 - 40 \u003d ...;
b) x² - x - 2 = 0,
D = b² - 4ac
D \u003d (-1) ² - 4 1 (-2) \u003d ...;
Number 3. Finish the equation
3x² - 5x - 2 = 0.
D = b² - 4ac
D \u003d (-5) ² - 4 3 (-2) \u003d 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.
No. 1. For each equation of the form ax² + bx + c = 0, write the values a, b, c.
a) 4x² - 8x + 6 = 0, b) x² + 2x - 4 = 0, c) x² - x + 2 = 0.
No. 2. Continue calculating the discriminant D of the quadratic equation ax² + bx + c = 0 using the formula D = b² - 4ac.
a) 5x² + 8x - 4 \u003d 0,
D = b² - 4ac
D \u003d 8² - 4 5 (- 4) \u003d 64 - 60 \u003d ...;
b) x² - 6x + 5 = 0,
D = b² - 4ac
D \u003d (-6) ² - 4 1 5 \u003d ...;
3 no. Finish the equation
x² - 6x + 5 = 0.
D = b² - 4ac
D \u003d (-6) ² - 4 1 5 \u003d 16.
x = ...
No. 4. Solve the equation.
a) (x + 4) (x - 6) = 0; b) 4x² - 5x + 1 = 0
No. 5. Convert the equation to a quadratic one 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.
Summary of the lesson.
Summing up the results of the score-rating table.
Historical reference and task.
Problems on quadratic equations are found already in 499. In ancient India, public competitions in solving difficult problems were common. One of the ancient Indian books says: “As the sun outshines the stars with its brilliance, so a learned person will outshine the glory of another in public meetings, proposing and solving algebraic problems.” Often they were in poetic form. Here is one of the problems of the famous 12th century Indian mathematician Bhaskara:
Frisky flock of monkeys
After eating to your heart's content, I had fun
Them squared part eight
Having fun in the meadow.
And 12 by vines ...
They began to jump, hanging.
How many monkeys were
You tell me, in this flock?
VII. Homework.
It is proposed to solve this historical problem and arrange it on separate sheets, with a picture.
APPENDIX
No. F.I.
student Activities TOTAL
Homework Dictation Oral exercises Consolidation of the material
PC work Whiteboard work
1 Ivanov I.
2 Fedorov G.
3 Yakovleva Ya.
…
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"
We remind you that the complete quadratic equation is an equation of the form:
Solving full quadratic equations is a bit more complicated (just a little bit) than those given.
Remember, any quadratic equation can be solved using the discriminant!
Even incomplete.
The rest of the methods will help you do it faster, but if you have problems with quadratic equations, first master the solution using the discriminant.
1. Solving quadratic equations using the discriminant.
Solving quadratic equations in this way is very simple, the main thing is to remember the sequence of actions and a couple of formulas.
If, then the equation has 2 roots. Pay special attention to step 2.
The discriminant D tells us the number of roots of the equation.
- If, then the formula at the step will be reduced to. Thus, the equation will have only a root.
- If, then we will not be able to extract the root of the discriminant at the step. This indicates that the equation has no roots.
Let us turn to the geometric meaning of the quadratic equation.
The graph of the function is a parabola:
Let's go back to our equations and look at a few examples.
Example 9
Solve the Equation
Step 1 skip.
Step 2
Finding the discriminant:
So the equation has two roots.
Step 3
Answer:
Example 10
Solve the Equation
The equation is in standard form, so Step 1 skip.
Step 2
Finding the discriminant:
So the equation has one root.
Answer:
Example 11
Solve the Equation
The equation is in standard form, so Step 1 skip.
Step 2
Finding the discriminant:
This means that we will not be able to extract the root from the discriminant. There are no roots of the equation.
Now we know how to write down such answers correctly.
Answer: no roots
2. Solving quadratic equations using the Vieta theorem
If you remember, then there is such a type of equations that are called reduced (when the coefficient a is equal to):
Such equations are very easy to solve using Vieta's theorem:
The sum of the roots given quadratic equation is equal, and the product of the roots is equal.
You just need to choose a pair of numbers whose product is equal to the free term of the equation, and the sum is equal to the second coefficient, taken with the opposite sign.
Example 12
Solve the Equation
This equation is suitable for solution using Vieta's theorem, because .
The sum of the roots of the equation is, i.e. we get the first equation:
And the product is:
Let's create and solve the system:
- and. The sum is;
- and. The sum is;
- and. The amount is equal.
and are the solution of the system:
Answer: ; .
Example 13
Solve the Equation
Answer:
Example 14
Solve the Equation
The equation is reduced, which means:
Answer:
QUADRATIC EQUATIONS. AVERAGE LEVEL
What is a quadratic equation?
In other words, a quadratic equation is an equation of the form, where - unknown, - some numbers, moreover.
The number is called the highest or first coefficient quadratic equation, - second coefficient, a - free member.
Because if, the equation will immediately become linear, because will disappear.
In this case, and can be equal to zero. In this chair equation is called incomplete.
If all the terms are in place, that is, the equation - complete.
Methods for solving incomplete quadratic equations
To begin with, we will analyze the methods for solving incomplete quadratic equations - they are simpler.
The following types of equations can be distinguished:
I. , in this equation the coefficient and the free term are equal.
II. , in this equation the coefficient is equal.
III. , in this equation the free term is equal to.
Now consider the solution of each of these subtypes.
Obviously, this equation always has only one root:
A number squared cannot be negative, because when multiplying two negative or two positive numbers, the result will always be a positive number. That's why:
if, then the equation has no solutions;
if we have two roots
These formulas do not need to be memorized. The main thing to remember is that it cannot be less.
Examples of solving quadratic equations
Example 15
Answer:
Never forget about roots with a negative sign!
Example 16
The square of a number cannot be negative, which means that the equation
no roots.
To briefly write that the problem has no solutions, we use the empty set icon.
Answer:
Example 17
So, this equation has two roots: and.
Answer:
Let's take the common factor out of brackets:
The product is equal to zero if at least one of the factors is equal to zero. This means that the equation has a solution when:
So, this quadratic equation has two roots: and.
Example:
Solve the equation.
Solution:
We factorize the left side of the equation and find the roots:
Answer:
Methods for solving complete quadratic equations
1. Discriminant
Solving quadratic equations in this way is easy, the main thing is to remember the sequence of actions and a couple of formulas. Remember, any quadratic equation can be solved using the discriminant! Even incomplete.
Did you notice the root of the discriminant in the root formula?
But the discriminant can be negative.
What to do?
We need to pay special attention to step 2. The discriminant tells us the number of roots of the equation.
- If, then the equation has a root:
- If, then the equation has the same root, but in fact, one root:
Such roots are called double roots.
- If, then the root of the discriminant is not extracted. This indicates that the equation has no roots.
Why are there different numbers of roots?
Let us turn to the geometric meaning of the quadratic equation. The graph of the function is a parabola:
In a particular case, which is a quadratic equation, .
And this means that the roots of the quadratic equation are the points of intersection with the x-axis (axis).
The parabola may not cross the axis at all, or it may intersect it at one (when the top of the parabola lies on the axis) or two points.
In addition, the coefficient is responsible for the direction of the branches of the parabola. If, then the branches of the parabola are directed upwards, and if - then downwards.
4 examples of solving quadratic equations
Example 18
Answer:
Example 19
Answer: .
Example 20
Answer:
Example 21
This means there are no solutions.
Answer: .
2. Vieta's theorem
Using Vieta's theorem is very easy.
All you need is pick up such a pair of numbers, the product of which is equal to the free term of the equation, and the sum is equal to the second coefficient, taken with the opposite sign.
It is important to remember that Vieta's theorem can only be applied to given quadratic equations ().
Let's look at a few examples:
Example 22
Solve the equation.
Solution:
This equation is suitable for solution using Vieta's theorem, because . Other coefficients: ; .
The sum of the roots of the equation is:
And the product is:
Let's select such pairs of numbers, the product of which is equal, and check if their sum is equal:
- and. The sum is;
- and. The sum is;
- and. The amount is equal.
and are the solution of the system:
Thus, and are the roots of our equation.
Answer: ; .
Example 23
Solution:
We select such pairs of numbers that give in the product, and then check whether their sum is equal:
and: give in total.
and: give in total. To get it, you just need to change the signs of the alleged roots: and, after all, the work.
Answer:
Example 24
Solution:
The free term of the equation is negative, and hence the product of the roots is a negative number. This is possible only if one of the roots is negative and the other is positive. So the sum of the roots is differences of their modules.
We select such pairs of numbers that give in the product, and the difference of which is equal to:
and: their difference is - not suitable;
and: - not suitable;
and: - not suitable;
and: - suitable. It remains only to remember that one of the roots is negative. Since their sum must be equal, then the root, which is smaller in absolute value, must be negative: . We check:
Answer:
Example 25
Solve the equation.
Solution:
The equation is reduced, which means:
The free term is negative, and hence the product of the roots is negative. And this is possible only when one root of the equation is negative and the other is positive.
We select such pairs of numbers whose product is equal, and then determine which roots should have a negative sign:
Obviously, only roots and are suitable for the first condition:
Answer:
Example 26
Solve the equation.
Solution:
The equation is reduced, which means:
The sum of the roots is negative, which means that at least one of the roots is negative. But since their product is positive, it means both roots are minus.
We select such pairs of numbers, the product of which is equal to:
Obviously, the roots are the numbers and.
Answer:
Agree, it is very convenient - to invent roots orally, instead of counting this nasty discriminant.
Try to use Vieta's theorem as often as possible!
But the Vieta theorem is needed in order to facilitate and speed up finding the roots.
To make it profitable for you to use it, you must bring the actions to automatism. And for this, solve five more examples.
But don't cheat: you can't use the discriminant! Only Vieta's theorem!
5 examples of Vieta's theorem for self-study
Example 27
Task 1. ((x)^(2))-8x+12=0
According to Vieta's theorem:
As usual, we start the selection with the product:
Not suitable because the amount;
: the amount is what you need.
Answer: ; .
Example 28
Task 2.
And again, our favorite Vieta theorem: the sum should work out, but the product is equal.
But since it should be not, but, we change the signs of the roots: and (in total).
Answer: ; .
Example 29
Task 3.
Hmm... Where is it?
It is necessary to transfer all the terms into one part:
The sum of the roots is equal to the product.
Yes, stop! The equation is not given.
But Vieta's theorem is applicable only in the given equations.
So first you need to bring the equation.
If you can’t bring it up, drop this idea and solve it in another way (for example, through the discriminant).
Let me remind you that to bring a quadratic equation means to make the leading coefficient equal to:
Then the sum of the roots is equal, and the product.
It's easier to pick up here: after all - a prime number (sorry for the tautology).
Answer: ; .
Example 30
Task 4.
The free term is negative.
What's so special about it?
And the fact that the roots will be of different signs.
And now, during the selection, we check not the sum of the roots, but the difference between their modules: this difference is equal, but the product.
So, the roots are equal and, but one of them is with a minus.
Vieta's theorem tells us that the sum of the roots is equal to the second coefficient with the opposite sign, that is.
This means that the smaller root will have a minus: and, since.
Answer: ; .
Example 31
Task 5.
What needs to be done first?
That's right, give the equation:
Again: we select the factors of the number, and their difference should be equal to:
The roots are equal and, but one of them is minus. Which? Their sum must be equal, which means that with a minus there will be a larger root.
Answer: ; .
Summarize
- Vieta's theorem is used only in the given quadratic equations.
- Using the Vieta theorem, you can find the roots by selection, orally.
- If the equation is not given or no suitable pair of factors of the free term was found, then there are no integer roots, and you need to solve it in another way (for example, through the discriminant).
3. Full square selection method
If all the terms containing the unknown are represented as terms from the formulas of abbreviated multiplication - the square of the sum or difference - then after the change of variables, the equation can be represented as an incomplete quadratic equation of the type.
For example:
Example 32
Solve the equation: .
Solution:
Answer:
Example 33
Solve the equation: .
Solution:
Answer:
In general, the transformation will look like this:
This implies: .
Doesn't it remind you of anything?
It's the discriminant! That's exactly how the discriminant formula was obtained.
QUADRATIC EQUATIONS. BRIEFLY ABOUT THE MAIN
Quadratic equation is an equation of the form, where is the unknown, are the coefficients of the quadratic equation, is the free term.
Complete quadratic equation- an equation in which the coefficients are not equal to zero.
Reduced quadratic equation- an equation in which the coefficient, that is: .
Incomplete quadratic equation- an equation in which the coefficient and or free term c are equal to zero:
- if the coefficient, the equation has the form: ,
- if a free term, the equation has the form: ,
- if and, the equation has the form: .
1. Algorithm for solving incomplete quadratic equations
1.1. An incomplete quadratic equation of the form, where, :
1) Express the unknown: ,
2) Check the sign of the expression:
- if, then the equation has no solutions,
- if, then the equation has two roots.
1.2. An incomplete quadratic equation of the form, where, :
1) Let's take the common factor out of brackets: ,
2) The product is equal to zero if at least one of the factors is equal to zero. Therefore, the equation has two roots:
1.3. An incomplete quadratic equation of the form, where:
This equation always has only one root: .
2. Algorithm for solving complete quadratic equations of the form where
2.1. Solution using the discriminant
1) Let's bring the equation to the standard form: ,
2) Calculate the discriminant using the formula: , which indicates the number of roots of the equation:
3) Find the roots of the equation:
- if, then the equation has a root, which are found by the formula:
- if, then the equation has a root, which is found by the formula:
- if, then the equation has no roots.
2.2. Solution using Vieta's theorem
The sum of the roots of the reduced quadratic equation (an equation of the form, where) is equal, and the product of the roots is equal, i.e. , a.
2.3. Full square solution
This video tutorial shows you how to solve a quadratic equation. The solution of quadratic equations usually begins to be studied in a comprehensive school, grade 8. The roots of a 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, which are real numbers. First, you need to determine the discriminant using the formula D=b2-4ac. After that, it remains to calculate the roots of the quadratic equation using a well-known formula. Now let's try to solve a specific example. Let's take x2+x-12=0 as the initial equation, i.e. coefficient a=1, b=1, c=-12. According to the well-known formula, you can determine the discriminant. Then, using the formula for finding the roots of the equation, we calculate them. In our case, the discriminant will be equal to 49. The fact that the value of the discriminant is a positive number tells us that this quadratic equation will have two roots. After simple calculations, we get that x1=-4, x2=3. Thus, we solved the quadratic equation by calculating its roots Video lesson “Solving quadratic equations (grade 8). We find the 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 complicated here. The ability to solve them is 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 solution methods, we note that all quadratic equations can be divided into three classes:
- Have no roots;
- They have exactly one root;
- They have two different roots.
This is an important difference between quadratic and linear equations, where the root always exists and is unique. How to determine how many roots an equation has? There is a wonderful thing for this - discriminant.
Discriminant
Let the quadratic equation ax 2 + bx + c = 0 be given. Then the discriminant is simply the number D = b 2 − 4ac .
This formula must be known by heart. Where it comes from is not important 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 people think. Take a look at the examples and you will understand everything yourself:
A task. How many roots do quadratic equations have:
- x 2 - 8x + 12 = 0;
- 5x2 + 3x + 7 = 0;
- x 2 − 6x + 9 = 0.
We write 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 the same way:
a = 5; b = 3; c = 7;
D \u003d 3 2 - 4 5 7 \u003d 9 - 140 \u003d -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 equal to zero - the root will be one.
Note that coefficients have been written out for each equation. Yes, it's long, yes, it's tedious - but you won't mix up the odds 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 solved equations - in general, not so many.
The roots of a quadratic equation
Now let's move on to the solution. If the discriminant D > 0, the roots can be found using the formulas:
The 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;
- x2 + 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 again has two roots. Let's 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 negative coefficients are substituted into the formula. Here, again, the technique described above will help: look at the formula literally, paint each step - and get rid of mistakes very soon.
Incomplete quadratic equations
It happens that the quadratic equation is somewhat different from what is given in the definition. For example:
- x2 + 9x = 0;
- x2 − 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. the coefficient of the variable x or the free element is equal to zero.
Of course, a very difficult case is possible when both of these coefficients are equal to zero: b \u003d c \u003d 0. In this case, the equation takes the form ax 2 \u003d 0. Obviously, such an equation has a single root: x \u003d 0.
Let's consider other cases. Let b \u003d 0, then we get an incomplete quadratic equation of the form ax 2 + c \u003d 0. Let's slightly transform it:
Since the arithmetic square root exists only from a non-negative number, the last equality only makes sense when (−c / a ) ≥ 0. Conclusion:
- If an incomplete quadratic equation of the form ax 2 + c = 0 satisfies the inequality (−c / a ) ≥ 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 - there are no complex calculations at all in incomplete quadratic equations. In fact, it is not even necessary to remember the inequality (−c / a ) ≥ 0. It is enough to express the value of x 2 and see what is 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 factorize the polynomial:
Taking the common factor out of the bracketThe product is equal to zero when at least one of the factors is equal to zero. This is where the roots come from. In conclusion, we will analyze several of these equations:
A task. Solve quadratic equations:
- x2 − 7x = 0;
- 5x2 + 30 = 0;
- 4x2 − 9 = 0.
x 2 − 7x = 0 ⇒ x (x − 7) = 0 ⇒ x 1 = 0; x2 = −(−7)/1 = 7.
5x2 + 30 = 0 ⇒ 5x2 = -30 ⇒ x2 = -6. There are no roots, because the 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 \u003d -1.5.
Class: 8
Consider the standard (studied in the school mathematics course) and non-standard methods for solving quadratic equations.
1. Decomposition of the left side of the quadratic equation into linear factors.
Consider examples:
3) x 2 + 10x - 24 = 0.
6(x 2 + x - x) = 0 | : 6
x 2 + x - x - \u003d 0;
x(x - ) + (x - ) = 0;
x(x - ) (x + ) = 0;
= ; – .Answer: ; – .
For independent work:
Solve quadratic equations using the method of factoring the left side of a quadratic equation into linear factors.
| a) x 2 - x \u003d 0; d) x 2 - 81 = 0; g) x 2 + 6x + 9 = 0; |
b) x 2 + 2x \u003d 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 selection of a full square.
Consider examples:
For independent work.
Solve quadratic equations using the full square method.
3. Solution of quadratic equations by formula.
ax 2 + in + c \u003d 0, (a | 4a
4a 2 x 2 + 4ab + 4ac = 0;
2ax + 2ax 2v + in 2 - in 2 + 4ac \u003d 0;
2 \u003d in 2 - 4ac; =±;Consider examples.
For independent work.
Solve quadratic equations using the formula x 1,2 =.
4. Solving quadratic equations using the Vieta theorem (direct and inverse)
x 2 + px + q = 0 - reduced quadratic equation
If then the equation has two identical roots in sign and it depends on the coefficient.
If p, then 
.
If p, then 
.
For example:
If then the equation has two roots of different sign, and the larger root will be if p and will be if p.
For example:
For independent work.
Without solving the quadratic equation, use the inverse Vieta theorem to determine the signs of its roots:
a, b, j, l - various roots;
c, e, h – negative;
d, f, g, i, m – positive;
5. Solution of quadratic equations by the “transfer” method.
For independent work.
Solve quadratic equations using the "flip" method.
6. Solving quadratic equations using the properties of its coefficients.
I. ax 2 + bx + c = 0, where a 0
1) If a + b + c \u003d 0, then x 1 \u003d 1; x 2 =
Proof:
ax 2 + bx + c = 0 |: a
x 2 + x + = 0.
According to Vieta's theorem 
By condition a + b + c = 0, then b = -a - c. Next, we get
It follows from this that x 1 =1; x 2 = . Q.E.D.
2) If a - b + c \u003d 0 (or b \u003d a + c), then x 1 \u003d - 1; x 2 \u003d -
Proof:
According to Vieta's theorem 
By condition a - b + c \u003d 0, i.e. b = a + c. Next we get:
Therefore, x 1 \u003d - 1; x 2 \u003d -.
Consider examples.
1) 345 x 2 - 137 x - 208 = 0.
a + b + c \u003d 345 - 137 - 208 \u003d 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.
Using the properties of the coefficients of a 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 = = = =
Consider an example:
3x 2 - 14x + 16 = 0.
D 1 \u003d (-7) 2 - 3 16 \u003d 49 - 48 \u003d 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
Consider an example:
x 2 - 14x - 15 = 0
x 1.2 = 7 = 7
x 1 \u003d -1; x 2 = 15.
Answer: -1; 15.
For independent work.
a) x 2 - 8x - 9 \u003d 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 \u003d 0
Answer: -1; four
b) x 2 - 2x + 1 = 0
c) x 2 - 2x + 5 = 0
Answer: no solution
For independent work.
Solve quadratic equations graphically:
8. Solving quadratic equations with a compass and straightedge.
ax2 + bx + c = 0,
x 2 + x + = 0.
x 1 and x 2 are roots.
Let A(0; 1), C(0;
According to the secant theorem:
OV · OD = OA · OS.
Therefore we have:
x 1 x 2 = 1 OS;
OS = x 1 x 2
K(; 0), where = -
F(0; ) = (0; ) = )
1) Construct the point S(-; ) - the center of the circle and the 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.
3 cases are possible:
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 x-axis in anguish B 1 (x 1; 0), where x 1 is the root of the quadratic equation
ax2 + bx + c = 0.
3) R< SK (или R < ).
The circle has no common points with the x-axis, i.e. there are 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.
Let's draw a circle (S; AS), where A(0; 1).
9. Solving quadratic equations using a nomogram
For the solution, four-digit mathematical tables of V.M. Bradys (Plate XXII, p. 83).
The nomogram allows, without solving the quadratic equation x 2 + px + q = 0, to determine the roots of the equation by its coefficients. For example:
5) z2 + 4z + 3 = 0.
Both roots are negative. Therefore, we will make a replacement: z 1 = - t. We get a new equation:
t 2 - 4t + 3 = 0.
t 1 \u003d 1; t2 = 3
z 1 \u003d - 1; z 2 \u003d - 3.
Answer: - 3; - one
6) If the coefficients p and q are out of scale, then perform the substitution z \u003d k t and solve the equation using the nomogram: z 2 + pz + q \u003d 0.
k 2 t 2 + p kt + q = 0. |: k 2
k is taken with the expectation that inequalities take place:
For independent work.
y 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.