In the term “quadratic equation,” the key word is “quadratic.” This means that the equation must necessarily contain a variable (that same x) squared, and there should not be xes to the third (or greater) power.
The solution of many equations comes down to solving exactly quadratic equations.
Let's learn to determine that this is a quadratic equation and not some other equation.
Example 1.
Let's get rid of the denominator and multiply each term of the equation by
Let's move everything to the left side and arrange the terms in descending order of powers of X
Now we can say with confidence that this equation is quadratic!
Example 2.
Multiply the left and right sides by:
This equation, although it was originally in it, is not quadratic!
Example 3.
Let's multiply everything by:
Scary? The fourth and second degrees... However, if we make a replacement, we will see that we have a simple quadratic equation:
Example 4.
It seems to be there, but let's take a closer look. Let's move everything to the left side:
See, it's reduced - and now it's a simple linear equation!
Now try to determine for yourself which of the following equations are quadratic and which are not:
Examples:
Answers:
- square;
- square;
- not square;
- not square;
- not square;
- square;
- not square;
- square.
Mathematicians conventionally divide all quadratic equations into the following types:
- Complete quadratic equations- equations in which the coefficients and, as well as the free term c, are not equal to zero (as in the example). In addition, among complete quadratic equations there are given- these are equations in which the coefficient (the equation from example one is not only complete, but also reduced!)
- Incomplete quadratic equations- equations in which the coefficient and or the free term c are equal to zero:
They are incomplete because they are missing some element. But the equation must always contain x squared!!! Otherwise, it will no longer be a quadratic equation, but some other equation.
Why did they come up with such a division? It would seem that there is an X squared, and okay. This division is determined by the solution methods. Let's look at each of them in more detail.
Solving incomplete quadratic equations
First, let's focus on solving incomplete quadratic equations - they are much simpler!
There are types of incomplete quadratic equations:
- , in this equation the coefficient is equal.
- , in this equation the free term is equal to.
- , in this equation the coefficient and the free term are equal.
1. i. Because we know how to extract Square root, then let's express from this equation
The expression can be either negative or positive. A squared number cannot be negative, because when multiplying two negative or two positive numbers, the result will always be a positive number, so: if, then the equation has no solutions.
And if, then we get two roots. There is no need to memorize these formulas. The main thing is that you must know and always remember that it cannot be less.
Let's try to solve some examples.
Example 5:
Solve the equation
Now all that remains is to extract the root from the left and right sides. After all, you remember how to extract roots?
Answer:
Never forget about roots with a negative sign!!!
Example 6:
Solve the equation
Answer:
Example 7:
Solve the equation
Oh! The square of a number cannot be negative, which means that the equation
no roots!
For such equations that have no roots, mathematicians came up with a special icon - (empty set). And the answer can be written like this:
Answer:
Thus, this quadratic equation has two roots. There are no restrictions here, since we did not extract the root.
Example 8:
Solve the equation
Let's take the common factor out of brackets:
Thus,
This equation has two roots.
Answer:
The simplest type of incomplete quadratic equations (although they are all simple, right?). Obviously, this equation always has only one root:
We will dispense with examples here.
Solving complete quadratic equations
We remind you that a complete quadratic equation is an equation of the form equation where
Solving complete quadratic equations is a little more difficult (just a little) than these.
Remember, Any quadratic equation can be solved using a discriminant! Even incomplete.
The other 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 a discriminant.
Solving quadratic equations using this method is very simple; the main thing is to remember the sequence of actions and a couple of formulas.
If, then the equation has a root. Special attention take a step. Discriminant () tells us the number of roots of the equation.
- If, then the formula in the step will be reduced to. Thus, the equation will only have 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's go back to our equations and look at some examples.
Example 9:
Solve the equation
Step 1 we skip.
Step 2.
We find the discriminant:
This means the equation has two roots.
Step 3.
Answer:
Example 10:
Solve the equation
The equation is presented in standard form, so Step 1 we skip.
Step 2.
We find the discriminant:
This means that the equation has one root.
Answer:
Example 11:
Solve the equation
The equation is presented in standard form, so Step 1 we skip.
Step 2.
We find the discriminant:
This means we will not be able to extract the root of the discriminant. There are no roots of the equation.
Now we know how to correctly write down such answers.
Answer: no roots
2. Solving quadratic equations using Vieta's theorem.
If you remember, there is a type of equation that is called reduced (when the coefficient a is equal to):
Such equations are very easy to solve using Vieta’s theorem:
Sum of roots given quadratic equation is equal, and the product of the roots is equal.
Example 12:
Solve the equation
This equation can be solved using Vieta's theorem because .
The sum of the roots of the equation is equal, i.e. we get the first equation:
And the product is equal to:
Let's compose and solve the system:
- And. The amount is equal to;
- And. The amount is equal to;
- And. The amount is equal.
and are the solution to the system:
Answer: ; .
Example 13:
Solve the equation
Answer:
Example 14:
Solve the equation
The equation is given, 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 - the unknown, - some numbers, and.
The number is called the highest or first coefficient quadratic equation, - second coefficient, A - free member.
Why? Because if the equation immediately becomes 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 is complete.
Solutions to various types of quadratic equations
Methods for solving incomplete quadratic equations:
First, let's look at methods for solving incomplete quadratic equations - they are simpler.
We can distinguish the following types of equations:
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 let's look at the solution to each of these subtypes.
Obviously, this equation always has only one root:
A squared number cannot be negative, because when you multiply 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
There is no need to memorize these formulas. The main thing to remember is that it cannot be less.
Examples:
Solutions:
Answer:
Never forget about roots with a negative sign!
The square of a number cannot be negative, which means that the equation
no roots.
To briefly write down that a problem has no solutions, we use the empty set icon.
Answer:
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:
Let's factor the left side of the equation and find the roots:
Answer:
Methods for solving complete quadratic equations:
1. Discriminant
Solving quadratic equations 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 a discriminant! Even incomplete.
Did you notice the root from the discriminant in the formula for roots? 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 roots:
- If, then the equation has the same roots, and 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 different numbers of roots possible? Let's turn to geometric sense quadratic equation. The graph of the function is a parabola:
In a special case, which is a quadratic equation, . This means that the roots of a quadratic equation are the points of intersection with the abscissa axis (axis). A parabola may not intersect the axis at all, or may intersect it at one (when the vertex 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 upward, and if, then downward.
Examples:
Solutions:
Answer:
Answer: .
Answer:
This means there are no solutions.
Answer: .
2. Vieta's theorem
It is very easy to use Vieta's theorem: 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.
It is important to remember that Vieta's theorem can only be applied in reduced quadratic equations ().
Let's look at a few examples:
Example #1:
Solve the equation.
Solution:
This equation can be solved using Vieta's theorem because . Other coefficients: ; .
The sum of the roots of the equation is:
And the product is equal to:
Let's select pairs of numbers whose product is equal and check whether their sum is equal:
- And. The amount is equal to;
- And. The amount is equal to;
- And. The amount is equal.
and are the solution to the system:
Thus, and are the roots of our equation.
Answer: ; .
Example #2:
Solution:
Let's select pairs of numbers that give in the product, and then check whether their sum is equal:
and: they give in total.
and: they give in total. To obtain, it is enough to simply change the signs of the supposed roots: and, after all, the product.
Answer:
Example #3:
Solution:
The free term of the equation is negative, and therefore the product of the roots is a negative number. This is only possible if one of the roots is negative and the other is positive. Therefore the sum of the roots is equal to differences of their modules.
Let us select pairs of numbers that give in the product, and whose difference is equal to:
and: their difference is equal - does not fit;
and: - not suitable;
and: - not suitable;
and: - suitable. All that remains is to remember that one of the roots is negative. Since their sum must be equal, the root with the smaller modulus must be negative: . We check:
Answer:
Example #4:
Solve the equation.
Solution:
The equation is given, which means:
The free term is negative, and therefore the product of the roots is negative. And this is only possible when one root of the equation is negative and the other is positive.
Let's select pairs of numbers whose product is equal, and then determine which roots should have a negative sign:
Obviously, only the roots and are suitable for the first condition:
Answer:
Example #5:
Solve the equation.
Solution:
The equation is given, 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 have a minus sign.
Let us select pairs of numbers whose product is equal to:
Obviously, the roots are the numbers and.
Answer:
Agree, it’s very convenient to come up with roots orally, instead of counting this nasty discriminant. Try to use Vieta's theorem as often as possible.
But Vieta’s theorem is needed in order to facilitate and speed up finding the roots. In order for you to benefit from using it, you must bring the actions to automaticity. And for this, solve five more examples. But don't cheat: you can't use a discriminant! Only Vieta's theorem:
Solutions to tasks for independent work:
Task 1. ((x)^(2))-8x+12=0
According to Vieta's theorem:
As usual, we start the selection with the piece:
Not suitable because the amount;
: the amount is just what you need.
Answer: ; .
Task 2.
And again our favorite Vieta theorem: the sum must be equal, and the product must be equal.
But since it must be not, but, we change the signs of the roots: and (in total).
Answer: ; .
Task 3.
Hmm... Where is that?
You need to move all the terms into one part:
The sum of the roots is equal to the product.
Okay, stop! The equation is not given. But Vieta's theorem is applicable only in the given equations. So first you need to give an equation. If you can’t lead, give up this idea and solve it in another way (for example, through a discriminant). Let me remind you that to give a quadratic equation means to make the leading coefficient equal:
Great. Then the sum of the roots is equal to and the product.
Here it’s as easy as shelling pears to choose: after all, it’s a prime number (sorry for the tautology).
Answer: ; .
Task 4.
The free member is negative. What's special about this? And the fact is that the roots will have different signs. And now, during the selection, we check not the sum of the roots, but the difference in their modules: this difference is equal, but a product.
So, the roots are equal to and, but one of them is 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: ; .
Task 5.
What should you do 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 to and, but one of them is minus. Which? Their sum should be equal, which means that the minus will have a larger root.
Answer: ; .
Let me summarize:
- Vieta's theorem is used only in the quadratic equations given.
- Using Vieta's theorem, you can find the roots by selection, orally.
- If the equation is not given or no equation is found suitable pair multipliers of the free term, which means there are no whole roots, and you need to solve it in another way (for example, through a discriminant).
3. Method for selecting a complete square
If all terms containing the unknown are represented in the form of terms from abbreviated multiplication formulas - the square of the sum or difference - then after replacing variables, the equation can be presented in the form of an incomplete quadratic equation of the type.
For example:
Example 1:
Solve the equation: .
Solution:
Answer:
Example 2:
Solve the equation: .
Solution:
Answer:
IN general view the transformation will look like this:
This implies: .
Doesn't remind you of anything? This is a discriminatory thing! That's exactly how we got the discriminant formula.
QUADRATIC EQUATIONS. BRIEFLY ABOUT THE MAIN THINGS
Quadratic equation- this is an equation of the form, where - the unknown, - the coefficients of the quadratic equation, - 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 the free term c are equal to zero:
- if the coefficient, the equation looks like: ,
- if there is a free term, the equation has the form: ,
- if and, the equation looks like: .
1. Algorithm for solving incomplete quadratic equations
1.1. An incomplete quadratic equation of the form, where, :
1) Let's 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 discriminant
1) Let's bring the equation to standard form: ,
2) Let's 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 roots, 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 (equation of the form where) is equal, and the product of the roots is equal, i.e. , A.
2.3. Solution by the method of selecting a complete square
If a quadratic equation of the form has roots, then it can be written in the form: .
Well, the topic is over. If you are reading these lines, it means you are very cool.
Because only 5% of people are able to master something on their own. And if you read to the end, then you are in this 5%!
Now the most important thing.
You have understood the theory on this topic. And, I repeat, this... this is just super! You are already better than the vast majority of your peers.
The problem is that this may not be enough...
For what?
For successfully passing the Unified State Exam, for entering college on a budget and, MOST IMPORTANTLY, for life.
I won’t convince you of anything, I’ll just say one thing...
People who received a good education, earn much more than those who did not receive it. This is statistics.
But this is not the main thing.
The main thing is that they are MORE HAPPY (there are such studies). Perhaps because many more opportunities open up before them and life becomes brighter? Don't know...
But think for yourself...
What does it take to be sure to be better than others on the Unified State Exam and ultimately be... happier?
GAIN YOUR HAND BY SOLVING PROBLEMS ON THIS TOPIC.
You won't be asked for theory during the exam.
You will need solve problems against time.
And, if you haven’t solved them (A LOT!), you’ll definitely make a stupid mistake somewhere or simply won’t have time.
It's like in sports - you need to repeat it many times to win for sure.
Find the collection wherever you want, necessarily with solutions, detailed analysis and decide, decide, decide!
You can use our tasks (optional) and we, of course, recommend them.
In order to get better at using our tasks, you need to help extend the life of the YouClever textbook you are currently reading.
How? There are two options:
- Unlock all hidden tasks in this article - 299 rub.
- Unlock access to all hidden tasks in all 99 articles of the textbook - 499 rub.
Yes, we have 99 such articles in our textbook and access to all tasks and all hidden texts in them can be opened immediately.
Access to all hidden tasks is provided for the ENTIRE life of the site.
In conclusion...
If you don't like our tasks, find others. Just don't stop at theory.
“Understood” and “I can solve” are completely different skills. You need both.
Find problems and solve them!
Kopyevskaya rural secondary school
10 Ways to Solve Quadratic Equations
Head: Patrikeeva Galina Anatolyevna,
mathematic teacher
village Kopevo, 2007
1. History of the development of quadratic equations
1.1 Quadratic equations in Ancient Babylon
1.2 How Diophantus composed and solved quadratic equations
1.3 Quadratic equations in India
1.4 Quadratic equations by al-Khorezmi
1.5 Quadratic equations in Europe XIII - XVII centuries
1.6 About Vieta's theorem
2. Methods for solving quadratic equations
Conclusion
Literature
1. History of the development of quadratic equations
1.1 Quadratic equations in Ancient Babylon
The need to solve equations not only of the first, but also of the second degree, even in ancient times, was caused by the need to solve problems related to finding the areas of land plots and with excavation work of a military nature, as well as with the development of astronomy and mathematics itself. Quadratic equations could be solved around 2000 BC. e. Babylonians.
Using modern algebraic notation, we can say that in their cuneiform texts there are, in addition to incomplete ones, such, for example, complete quadratic equations:
X 2 + X = ¾; X 2 - X = 14,5
The rule for solving these equations, set out in the Babylonian texts, essentially coincides with the modern one, but it is not known how the Babylonians arrived at this rule. Almost all cuneiform texts found so far provide only problems with solutions laid out in the form of recipes, with no indication as to how they were found.
Despite high level development of algebra in Babylon, the cuneiform texts lack the concept of a negative number and general methods solving quadratic equations.
1.2 How Diophantus composed and solved quadratic equations.
Diophantus' Arithmetic does not contain a systematic presentation of algebra, but it contains a systematic series of problems, accompanied by explanations and solved by constructing equations of various degrees.
When composing equations, Diophantus skillfully selects unknowns to simplify the solution.
Here, for example, is one of his tasks.
Problem 11.“Find two numbers, knowing that their sum is 20 and their product is 96”
Diophantus reasons as follows: from the conditions of the problem it follows that the required numbers are not equal, since if they were equal, then their product would not be equal to 96, but to 100. Thus, one of them will be more than half of their sum, i.e. . 10 + x, the other is less, i.e. 10's. The difference between them 2x .
Hence the equation:
(10 + x)(10 - x) = 96
100 - x 2 = 96
x 2 - 4 = 0 (1)
From here x = 2. One of the required numbers is equal to 12 , other 8 . Solution x = -2 for Diophantus does not exist, since Greek mathematics knew only positive numbers.
If we solve this problem by choosing one of the required numbers as the unknown, then we will come to a solution to the equation
y(20 - y) = 96,
y 2 - 20y + 96 = 0. (2)
It is clear that by choosing the half-difference of the required numbers as the unknown, Diophantus simplifies the solution; he manages to reduce the problem to solving an incomplete quadratic equation (1).
1.3 Quadratic Equations in India
Problems on quadratic equations are found already in the astronomical treatise “Aryabhattiam”, compiled in 499 by the Indian mathematician and astronomer Aryabhatta. Another Indian scientist, Brahmagupta (7th century), outlined general rule solutions of quadratic equations reduced to a single canonical form:
ah 2 + b x = c, a > 0. (1)
In equation (1), the coefficients, except A, can also be negative. Brahmagupta's rule is essentially the same as ours.
IN Ancient India Public competitions in solving difficult problems were common. One of the old Indian books says the following about such competitions: “As the sun eclipses the stars with its brilliance, so learned man will eclipse the glory of another people's assemblies, proposing and solving algebraic problems.” Problems were often presented in poetic form.
This is one of the problems of the famous Indian mathematician of the 12th century. Bhaskars.
Problem 13.
“A flock of frisky monkeys, and twelve along the vines...
The authorities, having eaten, had fun. They started jumping, hanging...
There are them in the square, part eight. How many monkeys were there?
I was having fun in the clearing. Tell me, in this pack?
Bhaskara's solution indicates that he knew that the roots of quadratic equations are two-valued (Fig. 3).
The equation corresponding to problem 13 is:
( x /8) 2 + 12 = x
Bhaskara writes under the guise:
x 2 - 64x = -768
and, to complete the left side of this equation to square, adds to both sides 32 2 , then getting:
x 2 - 64x + 32 2 = -768 + 1024,
(x - 32) 2 = 256,
x - 32 = ± 16,
x 1 = 16, x 2 = 48.
1.4 Quadratic equations in al - Khorezmi
In the algebraic treatise of al-Khorezmi, a classification of linear and quadratic equations is given. The author counts 6 types of equations, expressing them as follows:
1) “Squares are equal to roots,” i.e. ax 2 + c = b X.
2) “Squares are equal to numbers”, i.e. ax 2 = c.
3) “The roots are equal to the number,” i.e. ah = s.
4) “Squares and numbers are equal to roots,” i.e. ax 2 + c = b X.
5) “Squares and roots are equal to numbers”, i.e. ah 2 + bx = s.
6) “Roots and numbers are equal to squares,” i.e. bx + c = ax 2 .
For al-Khorezmi, who avoided the use of negative numbers, the terms of each of these equations are addends and not subtractables. In this case, equations that do not have positive solutions are obviously not taken into account. The author sets out methods for solving these equations using the techniques of al-jabr and al-muqabala. His decisions, of course, do not completely coincide with ours. Not to mention that it is purely rhetorical, it should be noted, for example, that when solving an incomplete quadratic equation of the first type
al-Khorezmi, like all mathematicians before the 17th century, does not take into account the zero solution, probably because in specific practical problems it does not matter. When solving complete quadratic equations al-Khorezmi on partial numerical examples lays out the rules for the solution and then the geometric proofs.
Problem 14.“The square and the number 21 are equal to 10 roots. Find the root" (implying the root of the equation x 2 + 21 = 10x).
The author's solution goes something like this: divide the number of roots in half, you get 5, multiply 5 by itself, subtract 21 from the product, what remains is 4. Take the root from 4, you get 2. Subtract 2 from 5, you get 3, this will be the desired root. Or add 2 to 5, which gives 7, this is also a root.
The treatise of al-Khorezmi is the first book that has come down to us, which systematically sets out the classification of quadratic equations and gives formulas for their solution.
1.5 Quadratic equations in Europe XIII - XVII bb
Formulas for solving quadratic equations along the lines of al-Khwarizmi in Europe were first set forth in the Book of Abacus, written in 1202 by the Italian mathematician Leonardo Fibonacci. This voluminous work, which reflects the influence of mathematics, both Islamic countries and Ancient Greece, is distinguished by both completeness and clarity of presentation. The author independently developed some new algebraic examples of solving problems and was the first in Europe to approach the introduction of negative numbers. His book contributed to the spread of algebraic knowledge not only in Italy, but also in Germany, France and other European countries. Many problems from the Book of Abacus were used in almost all European textbooks of the 16th - 17th centuries. and partly XVIII.
The general rule for solving quadratic equations reduced to a single canonical form:
x 2 + bx = c,
for all possible combinations of coefficient signs b , With was formulated in Europe only in 1544 by M. Stiefel.
The derivation of the formula for solving a quadratic equation in general form is available from Viète, but Viète recognized only positive roots. Italian mathematicians Tartaglia, Cardano, Bombelli were among the first in the 16th century. In addition to positive ones, negative roots are also taken into account. Only in the 17th century. Thanks to the work of Girard, Descartes, Newton and other scientists, the method of solving quadratic equations takes on a modern form.
1.6 About Vieta's theorem
The theorem expressing the relationship between the coefficients of a quadratic equation and its roots, named after Vieta, was formulated by him for the first time in 1591 as follows: “If B + D, multiplied by A - A 2 , equals BD, That A equals IN and equal D ».
To understand Vieta, we should remember that A, like any vowel letter, meant the unknown (our X), vowels IN, D- coefficients for the unknown. In the language of modern algebra, the above Vieta formulation means: if there is
(a + b )x - x 2 = ab ,
x 2 - (a + b )x + a b = 0,
x 1 = a, x 2 = b .
Expressing the relationship between the roots and coefficients of the equations general formulas written using symbols, Viet established uniformity in the methods of solving equations. However, the symbolism of Viet is still far from modern look. He did not recognize negative numbers and therefore, when solving equations, he considered only cases where all the roots were positive.
2. Methods for solving quadratic equations
Quadratic equations are the foundation on which the majestic edifice of algebra rests. Quadratic equations are found wide application when solving trigonometric, exponential, logarithmic, irrational and transcendental equations and inequalities. We all know how to solve quadratic equations from school (8th grade) until graduation.
I hope that after studying this article you will learn how to find the roots of a complete quadratic equation.
Using the discriminant, only complete quadratic equations are solved; to solve incomplete quadratic equations, other methods are used, which you will find in the article “Solving incomplete quadratic equations.”
What quadratic equations are called complete? This equations of the form ax 2 + b x + c = 0, where coefficients a, b and c are not equal to zero. So, to solve a complete quadratic equation, we need to calculate the discriminant D.
D = b 2 – 4ac.
Depending on the value of the discriminant, we will write down the answer.
If the discriminant is a negative number (D< 0),то корней нет.
If the discriminant is zero, then x = (-b)/2a. When the discriminant is a positive number (D > 0),
then x 1 = (-b - √D)/2a, and x 2 = (-b + √D)/2a.
For example. Solve the equation x 2– 4x + 4= 0.
D = 4 2 – 4 4 = 0
x = (- (-4))/2 = 2
Answer: 2.
Solve Equation 2 x 2 + x + 3 = 0.
D = 1 2 – 4 2 3 = – 23
Answer: no roots.
Solve Equation 2 x 2 + 5x – 7 = 0.
D = 5 2 – 4 2 (–7) = 81
x 1 = (-5 - √81)/(2 2)= (-5 - 9)/4= – 3.5
x 2 = (-5 + √81)/(2 2) = (-5 + 9)/4=1
Answer: – 3.5; 1.
So let’s imagine the solution of complete quadratic equations using the diagram in Figure 1.
Using these formulas you can solve any complete quadratic equation. You just need to be careful to the equation was written as a polynomial standard view
A x 2 + bx + c, otherwise you may make a mistake. For example, in writing the equation x + 3 + 2x 2 = 0, you can mistakenly decide that
a = 1, b = 3 and c = 2. Then
D = 3 2 – 4 1 2 = 1 and then the equation has two roots. And this is not true. (See solution to example 2 above).
Therefore, if the equation is not written as a polynomial of the standard form, first the complete quadratic equation must be written as a polynomial of the standard form (the monomial with the largest exponent should come first, that is A x 2 , then with less – bx and then a free member With.
When solving the reduced quadratic equation and a quadratic equation with an even coefficient in the second term, you can use other formulas. Let's get acquainted with these formulas. If in a complete quadratic equation the second term has an even coefficient (b = 2k), then you can solve the equation using the formulas shown in the diagram in Figure 2.
A complete quadratic equation is called reduced if the coefficient at x 2 is equal to one and the equation takes the form x 2 + px + q = 0. Such an equation can be given for solution, or it can be obtained by dividing all coefficients of the equation by the coefficient A, standing at x 2 .
Figure 3 shows a diagram for solving the reduced square 
equations. Let's look at an example of the application of the formulas discussed in this article.
Example. Solve the equation
3x 2 + 6x – 6 = 0.
Let's solve this equation using the formulas shown in the diagram in Figure 1.
D = 6 2 – 4 3 (– 6) = 36 + 72 = 108
√D = √108 = √(36 3) = 6√3
x 1 = (-6 - 6√3)/(2 3) = (6 (-1- √(3)))/6 = –1 – √3
x 2 = (-6 + 6√3)/(2 3) = (6 (-1+ √(3)))/6 = –1 + √3
Answer: –1 – √3; –1 + √3
You can notice that the coefficient of x in this equation is an even number, that is, b = 6 or b = 2k, whence k = 3. Then let’s try to solve the equation using the formulas shown in the diagram of the figure D 1 = 3 2 – 3 · (– 6 ) = 9 + 18 = 27
√(D 1) = √27 = √(9 3) = 3√3
x 1 = (-3 - 3√3)/3 = (3 (-1 - √(3)))/3 = – 1 – √3
x 2 = (-3 + 3√3)/3 = (3 (-1 + √(3)))/3 = – 1 + √3
Answer: –1 – √3; –1 + √3. Noticing that all the coefficients in this quadratic equation are divisible by 3 and performing the division, we get the reduced quadratic equation x 2 + 2x – 2 = 0 Solve this equation using the formulas for the reduced quadratic 
equations figure 3.
D 2 = 2 2 – 4 (– 2) = 4 + 8 = 12
√(D 2) = √12 = √(4 3) = 2√3
x 1 = (-2 - 2√3)/2 = (2 (-1 - √(3)))/2 = – 1 – √3
x 2 = (-2 + 2√3)/2 = (2 (-1+ √(3)))/2 = – 1 + √3
Answer: –1 – √3; –1 + √3.
As you can see, when solving this equation using different formulas, we received the same answer. Therefore, having thoroughly mastered the formulas shown in the diagram in Figure 1, you will always be able to solve any complete quadratic equation.
blog.site, when copying material in full or in part, a link to the original source is required.
IN modern society the ability to perform operations with equations containing a variable squared can be useful in many areas of activity and is widely used in practice in scientific and technical developments. Evidence of this can be found in the design of sea and river vessels, aircraft and missiles. Using such calculations, the trajectories of movement of the most different bodies, including space objects. Examples with the solution of quadratic equations are used not only in economic forecasting, in the design and construction of buildings, but also in the most ordinary everyday circumstances. They may be needed on hiking trips, at sporting events, in stores when making purchases and in other very common situations.
Let's break the expression into its component factors
The degree of an equation is determined by the maximum value of the degree of the variable that the expression contains. If it is equal to 2, then such an equation is called quadratic.
If we speak in the language of formulas, then the indicated expressions, no matter how they look, can always be brought to the form when left side expression consists of three terms. Among them: ax 2 (that is, a variable squared with its coefficient), bx (an unknown without a square with its coefficient) and c (a free component, that is, an ordinary number). All this on the right side is equal to 0. In the case when such a polynomial lacks one of its constituent terms, with the exception of ax 2, it is called an incomplete quadratic equation. Examples with the solution of such problems, the values of the variables in which are easy to find, should be considered first.
If the expression looks like it has two terms on the right side, more precisely ax 2 and bx, the easiest way to find x is by putting the variable out of brackets. Now our equation will look like this: x(ax+b). Next, it becomes obvious that either x=0, or the problem comes down to finding a variable from the following expression: ax+b=0. This is dictated by one of the properties of multiplication. The rule states that the product of two factors results in 0 only if one of them is zero.
Example
x=0 or 8x - 3 = 0
As a result, we get two roots of the equation: 0 and 0.375.
Equations of this kind can describe the movement of bodies under the influence of gravity, which began to move from a certain point taken as the origin of coordinates. Here mathematical notation takes the following form: y = v 0 t + gt 2 /2. By substituting the necessary values, equating the right side to 0 and finding possible unknowns, you can find out the time that passes from the moment the body rises to the moment it falls, as well as many other quantities. But we'll talk about this later.
Factoring an Expression
The rule described above makes it possible to solve these problems in more difficult cases. Let's look at examples of solving quadratic equations of this type.
X 2 - 33x + 200 = 0
This quadratic trinomial is complete. First, let's transform the expression and factor it. There are two of them: (x-8) and (x-25) = 0. As a result, we have two roots 8 and 25.
Examples with solving quadratic equations in grade 9 allow this method to find a variable in expressions not only of the second, but even of the third and fourth orders.
For example: 2x 3 + 2x 2 - 18x - 18 = 0. When factoring the right side into factors with a variable, there are three of them, that is, (x+1), (x-3) and (x+3).
As a result, it becomes obvious that this equation has three roots: -3; -1; 3.
Square Root
Another case of an incomplete second-order equation is an expression represented in the language of letters in such a way that right part is constructed from the components ax 2 and c. Here, to obtain the value of the variable, the free term is transferred to right side, and after that the square root is taken from both sides of the equality. It should be noted that in in this case There are usually two roots of the equation. The only exceptions can be equalities that do not contain a term with at all, where the variable is equal to zero, as well as variants of expressions when the right side turns out to be negative. In the latter case, there are no solutions at all, since the above actions cannot be performed with roots. Examples of solutions to quadratic equations of this type should be considered.
In this case, the roots of the equation will be the numbers -4 and 4.
Calculation of land area
The need for this kind of calculations appeared in ancient times, because the development of mathematics in those distant times was largely determined by the need to determine with the greatest accuracy the areas and perimeters of land plots.
We should also consider examples of solving quadratic equations based on problems of this kind.
So, let's say there is a rectangular plot of land, the length of which is 16 meters greater than the width. You should find the length, width and perimeter of the site if you know that its area is 612 m2.
To get started, let's first create the necessary equation. Let us denote by x the width of the area, then its length will be (x+16). From what has been written it follows that the area is determined by the expression x(x+16), which, according to the conditions of our problem, is 612. This means that x(x+16) = 612.
Solving complete quadratic equations, and this expression is exactly that, cannot be done in the same way. Why? Although the left side still contains two factors, their product does not equal 0 at all, so different methods are used here.
Discriminant
First of all, let's make the necessary transformations, then appearance of this expression will look like this: x 2 + 16x - 612 = 0. This means that we have received an expression in a form corresponding to the previously specified standard, where a=1, b=16, c=-612.
This could be an example of solving quadratic equations using a discriminant. Here necessary calculations are produced according to the scheme: D = b 2 - 4ac. This auxiliary quantity not only makes it possible to find the required quantities in a second-order equation, it determines the quantity possible options. If D>0, there are two of them; for D=0 there is one root. In case D<0, никаких шансов для решения у уравнения вообще не имеется.
About roots and their formula
In our case, the discriminant is equal to: 256 - 4(-612) = 2704. This suggests that our problem has an answer. If you know k, the solution of quadratic equations must be continued using the formula below. It allows you to calculate the roots.
This means that in the presented case: x 1 =18, x 2 =-34. The second option in this dilemma cannot be a solution, because the dimensions of the land plot cannot be measured in negative quantities, which means x (that is, the width of the plot) is 18 m. From here we calculate the length: 18+16=34, and the perimeter 2(34+ 18)=104(m2).
Examples and tasks
We continue our study of quadratic equations. Examples and detailed solutions of several of them will be given below.
1) 15x 2 + 20x + 5 = 12x 2 + 27x + 1
Let’s move everything to the left side of the equality, make a transformation, that is, we’ll get the type of equation that is usually called standard, and equate it to zero.
15x 2 + 20x + 5 - 12x 2 - 27x - 1 = 0
Adding similar ones, we determine the discriminant: D = 49 - 48 = 1. This means our equation will have two roots. Let's calculate them according to the above formula, which means that the first of them will be equal to 4/3, and the second to 1.
2) Now let's solve mysteries of a different kind.
Let's find out if there are any roots here x 2 - 4x + 5 = 1? To obtain a comprehensive answer, let’s reduce the polynomial to the corresponding usual form and calculate the discriminant. In the above example, it is not necessary to solve the quadratic equation, because this is not the essence of the problem at all. In this case, D = 16 - 20 = -4, which means there really are no roots.
Vieta's theorem
It is convenient to solve quadratic equations using the above formulas and the discriminant, when the square root is taken from the value of the latter. But this does not always happen. However, there are many ways to obtain the values of variables in this case. Example: solving quadratic equations using Vieta's theorem. She is named after who lived in the 16th century in France and made a brilliant career thanks to his mathematical talent and connections at court. His portrait can be seen in the article.
The pattern that the famous Frenchman noticed was as follows. He proved that the roots of the equation add up numerically to -p=b/a, and their product corresponds to q=c/a.
Now let's look at specific tasks.
3x 2 + 21x - 54 = 0
For simplicity, let's transform the expression:
x 2 + 7x - 18 = 0
Let's use Vieta's theorem, this will give us the following: the sum of the roots is -7, and their product is -18. From here we get that the roots of the equation are the numbers -9 and 2. After checking, we will make sure that these variable values really fit into the expression.
Parabola graph and equation
The concepts of quadratic function and quadratic equations are closely related. Examples of this have already been given earlier. Now let's look at some mathematical riddles in a little more detail. Any equation of the described type can be represented visually. Such a relationship, drawn as a graph, is called a parabola. Its various types are presented in the figure below.
Any parabola has a vertex, that is, a point from which its branches emerge. If a>0, they go high to infinity, and when a<0, они рисуются вниз. Простейшим примером подобной зависимости является функция y = x 2 . В данном случае в уравнении x 2 =0 неизвестное может принимать только одно значение, то есть х=0, а значит существует только один корень. Это неудивительно, ведь здесь D=0, потому что a=1, b=0, c=0. Выходит формула корней (точнее одного корня) квадратного уравнения запишется так: x = -b/2a.
Visual representations of functions help solve any equations, including quadratic ones. This method is called graphical. And the value of the x variable is the abscissa coordinate at the points where the graph line intersects with 0x. The coordinates of the vertex can be found using the formula just given x 0 = -b/2a. And by substituting the resulting value into the original equation of the function, you can find out y 0, that is, the second coordinate of the vertex of the parabola, which belongs to the ordinate axis.
The intersection of the branches of a parabola with the abscissa axis
There are a lot of examples of solving quadratic equations, but there are also general patterns. Let's look at them. It is clear that the intersection of the graph with the 0x axis for a>0 is possible only if 0 takes negative values. And for a<0 координата у 0 должна быть положительна. Для указанных вариантов D>0. Otherwise D<0. А когда D=0, вершина параболы расположена непосредственно на оси 0х.
From the graph of the parabola you can also determine the roots. The opposite is also true. That is, if it is not easy to obtain a visual representation of a quadratic function, you can equate the right side of the expression to 0 and solve the resulting equation. And knowing the points of intersection with the 0x axis, it is easier to construct a graph.
From the history
Using equations containing a squared variable, in the old days they not only made mathematical calculations and determined the areas of geometric figures. The ancients needed such calculations for grand discoveries in the fields of physics and astronomy, as well as for making astrological forecasts.
As modern scientists suggest, the inhabitants of Babylon were among the first to solve quadratic equations. This happened four centuries before our era. Of course, their calculations were radically different from those currently accepted and turned out to be much more primitive. For example, Mesopotamian mathematicians had no idea about the existence of negative numbers. They were also unfamiliar with other subtleties that any modern schoolchild knows.
Perhaps even earlier than the scientists of Babylon, the sage from India Baudhayama began solving quadratic equations. This happened about eight centuries before the era of Christ. True, the second-order equations, the methods for solving which he gave, were the simplest. Besides him, Chinese mathematicians were also interested in similar questions in the old days. In Europe, quadratic equations began to be solved only at the beginning of the 13th century, but later they were used in their works by such great scientists as Newton, Descartes and many others.
Consider the quadratic equation:
(1)
.
Roots of a quadratic equation(1) are determined by the formulas:
;
.
These formulas can be combined like this:
.
When the roots of a quadratic equation are known, then a polynomial of the second degree can be represented as a product of factors (factored):
.
Next we assume that are real numbers.
Let's consider discriminant of a quadratic equation:
.
If the discriminant is positive, then the quadratic equation (1) has two different real roots:
;
.
Then the factorization of the quadratic trinomial has the form:
.
If the discriminant is equal to zero, then the quadratic equation (1) has two multiple (equal) real roots:
.
Factorization:
.
If the discriminant is negative, then the quadratic equation (1) has two complex conjugate roots:
;
.
Here is the imaginary unit, ;
and are the real and imaginary parts of the roots:
;
.
Then
.
Graphic interpretation
If you plot the function
,
which is a parabola, then the points of intersection of the graph with the axis will be the roots of the equation
.
At , the graph intersects the x-axis (axis) at two points.
When , the graph touches the x-axis at one point.
When , the graph does not cross the x-axis.
Below are examples of such graphs.
Useful formulas related to quadratic equation
(f.1) ;
(f.2) ;
(f.3) .
Derivation of the formula for the roots of a quadratic equation
We carry out transformations and apply formulas (f.1) and (f.3):
,
Where
;
.
So, we got the formula for a polynomial of the second degree in the form:
.
This shows that the equation
performed at
And .
That is, and are the roots of the quadratic equation
.
Examples of determining the roots of a quadratic equation
Example 1
(1.1)
.
Solution
.
Comparing with our equation (1.1), we find the values of the coefficients:
.
We find the discriminant:
.
Since the discriminant is positive, the equation has two real roots:
;
;
.
From here we obtain the factorization of the quadratic trinomial:
.
Graph of the function y = 2 x 2 + 7 x + 3 intersects the x-axis at two points.
Let's plot the function
.
The graph of this function is a parabola. It crosses the abscissa axis (axis) at two points:
And .
These points are the roots of the original equation (1.1).
Answer
;
;
.
Example 2
Find the roots of a quadratic equation:
(2.1)
.
Solution
Let's write the quadratic equation in general form:
.
Comparing with the original equation (2.1), we find the values of the coefficients:
.
We find the discriminant:
.
Since the discriminant is zero, the equation has two multiple (equal) roots:
;
.
Then the factorization of the trinomial has the form:
.
Graph of the function y = x 2 - 4 x + 4 touches the x-axis at one point.
Let's plot the function
.
The graph of this function is a parabola. It touches the x-axis (axis) at one point:
.
This point is the root of the original equation (2.1). Because this root is factored twice:
,
then such a root is usually called a multiple. That is, they believe that there are two equal roots:
.
Answer
;
.
Example 3
Find the roots of a quadratic equation:
(3.1)
.
Solution
Let's write the quadratic equation in general form:
(1)
.
Let's rewrite the original equation (3.1):
.
Comparing with (1), we find the values of the coefficients:
.
We find the discriminant:
.
The discriminant is negative, . Therefore there are no real roots.
You can find complex roots:
;
;
Let's plot the function
.
The graph of this function is a parabola. It does not intersect the x-axis (axis). Therefore there are no real roots.
Answer
There are no real roots. Complex roots:
;
;
.
