”, that is, equations of the first degree. In this lesson, we will explore what is a quadratic equation and how to solve it.

What is a quadratic equation

Important!

The degree of an equation is determined by the highest degree to which the unknown stands.

If a maximum degree, in which there is an unknown - " 2", So you have a quadratic equation.

Examples of quadratic equations

• 5x2 - 14x + 17 = 0
• −x 2 + x +

  1

  3

  = 0
• x2 + 0.25x = 0
• x 2 − 8 = 0

Important! The general form of the quadratic equation looks like this:

A x 2 + b x + c = 0

"a", "b" and "c" - given numbers.

• "a" - the first or senior coefficient;
• "b" - the second coefficient;
• "c" is a free member.

To find "a", "b" and "c" You need to compare your equation with the general form of the quadratic equation "ax 2 + bx + c \u003d 0".

Let's practice determining the coefficients "a", "b" and "c" in quadratic equations.

5x2 - 14x + 17 = 0 −7x 2 − 13x + 8 = 0 −x 2 + x +

The equation	Odds
a=5 b = −14 c = 17
a = −7 b = −13 c = 8

1

3

= 0

• a = -1
• b = 1
• c =

  1

  3

x2 + 0.25x = 0

• a = 1
• b = 0.25
• c = 0

x 2 − 8 = 0

• a = 1
• b = 0
• c = −8

How to solve quadratic equations

Unlike linear equations for solutions quadratic equations special formula for finding roots.

Remember!

To solve a quadratic equation you need:

• bring the quadratic equation to the general form "ax 2 + bx + c \u003d 0". That is, only "0" should remain on the right side;
• use the formula for roots:

Let's use an example to figure out how to apply the formula to find the roots of a quadratic equation. Let's solve the quadratic equation.

X 2 - 3x - 4 = 0

The equation "x 2 - 3x - 4 = 0" has already been reduced to the general form "ax 2 + bx + c = 0" and does not require additional simplifications. To solve it, we need only apply formula for finding the roots of a quadratic equation.

Let's define the coefficients "a", "b" and "c" for this equation.

x 1;2 =
x 1;2 =
x 1;2 =
x 1;2 =

With its help, any quadratic equation is solved.

In the formula "x 1; 2 \u003d" the root expression is often replaced
"b 2 − 4ac" to the letter "D" and called discriminant. The concept of a discriminant is discussed in more detail in the lesson "What is a discriminant".

Consider another example of a quadratic equation.

x 2 + 9 + x = 7x

In this form, it is rather difficult to determine the coefficients "a", "b", and "c". Let's first bring the equation to the general form "ax 2 + bx + c \u003d 0".

X 2 + 9 + x = 7x
x 2 + 9 + x − 7x = 0
x2 + 9 - 6x = 0
x 2 − 6x + 9 = 0

Now you can use the formula for the roots.

X 1;2 =
x 1;2 =
x 1;2 =
x 1;2 =
x=

6

2

x=3
Answer: x = 3

There are times when there are no roots in quadratic equations. This situation occurs when a negative number appears in the formula under the root.

With this math program you can solve quadratic equation.

The program not only gives the answer to the problem, but also displays the solution process in two ways:
- using the discriminant
- using the Vieta theorem (if possible).

Moreover, the answer is displayed exact, not approximate.
For example, for the equation \(81x^2-16x-1=0\), the answer is displayed in this form:

$$ x_1 = \frac(8+\sqrt(145))(81), \quad x_2 = \frac(8-\sqrt(145))(81) $$ instead of this: \(x_1 = 0.247; \quad x_2 = -0.05 \)

This program May be useful for high school students general education schools in preparation for control work and exams, when testing knowledge before the exam, parents to control the solution of many problems in mathematics and algebra. Or maybe it's too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get it done as soon as possible? homework math or algebra? In this case, you can also use our programs with a detailed solution.

In this way, you can conduct your own training and/or the training of your younger brothers or sisters, while the level of education in the field of tasks to be solved is increased.

If you are not familiar with the rules for entering a square polynomial, we recommend that you familiarize yourself with them.

Rules for entering a square polynomial

Any Latin letter can act as a variable.
For example: \(x, y, z, a, b, c, o, p, q \) etc.

Numbers can be entered as integers or fractions.
Moreover, fractional numbers can be entered not only in the form of a decimal, but also in the form of an ordinary fraction.

Rules for entering decimal fractions.
In decimal fractions, the fractional part from the integer can be separated by either a dot or a comma.
For example, you can enter decimals so: 2.5x - 3.5x^2

Rules for entering ordinary fractions.
Only a whole number can act as the numerator, denominator and integer part of a fraction.

The denominator cannot be negative.

When you enter numeric fraction The numerator is separated from the denominator by a division sign: /
The integer part is separated from the fraction by an ampersand: &
Input: 3&1/3 - 5&6/5z +1/7z^2
Result: \(3\frac(1)(3) - 5\frac(6)(5) z + \frac(1)(7)z^2 \)

When entering an expression you can use brackets. In this case, when solving a quadratic equation, the introduced expression is first simplified.
For example: 1/2(y-1)(y+1)-(5y-10&1/2)

=0

Decide

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A bit of theory.

Quadratic equation and its roots. Incomplete quadratic equations

Each of the equations
\(-x^2+6x+1,4=0, \quad 8x^2-7x=0, \quad x^2-\frac(4)(9)=0 \)
has the form
\(ax^2+bx+c=0, \)
where x is a variable, a, b and c are numbers.
In the first equation a = -1, b = 6 and c = 1.4, in the second a = 8, b = -7 and c = 0, in the third a = 1, b = 0 and c = 4/9. Such equations are called quadratic equations.

Definition.
quadratic equation an equation of the form ax 2 +bx+c=0 is called, where x is a variable, a, b and c are some numbers, and \(a \neq 0 \).

The numbers a, b and c are the coefficients of the quadratic equation. The number a is called the first coefficient, the number b is the second coefficient and the number c is the intercept.

In each of the equations of the form ax 2 +bx+c=0, where \(a \neq 0 \), the largest power of the variable x is a square. Hence the name: quadratic equation.

Note that a quadratic equation is also called an equation of the second degree, since its left side is a polynomial of the second degree.

A quadratic equation in which the coefficient at x 2 is 1 is called reduced quadratic equation. For example, the given quadratic equations are the equations
\(x^2-11x+30=0, \quad x^2-6x=0, \quad x^2-8=0 \)

If in the quadratic equation ax 2 +bx+c=0 at least one of the coefficients b or c is equal to zero, then such an equation is called incomplete quadratic equation. So, the equations -2x 2 +7=0, 3x 2 -10x=0, -4x 2 =0 are incomplete quadratic equations. In the first of them b=0, in the second c=0, in the third b=0 and c=0.

Incomplete quadratic equations are of three types:
1) ax 2 +c=0, where \(c \neq 0 \);
2) ax 2 +bx=0, where \(b \neq 0 \);
3) ax2=0.

Consider the solution of equations of each of these types.

To solve an incomplete quadratic equation of the form ax 2 +c=0 for \(c \neq 0 \), its free term is transferred to the right side and both parts of the equation are divided by a:
\(x^2 = -\frac(c)(a) \Rightarrow x_(1,2) = \pm \sqrt( -\frac(c)(a)) \)

Since \(c \neq 0 \), then \(-\frac(c)(a) \neq 0 \)

If \(-\frac(c)(a)>0 \), then the equation has two roots.

If \(-\frac(c)(a) To solve an incomplete quadratic equation of the form ax 2 +bx=0 for \(b \neq 0 \) factorize its left side and obtain the equation
\(x(ax+b)=0 \Rightarrow \left\( \begin(array)(l) x=0 \\ ax+b=0 \end(array) \right. \Rightarrow \left\( \begin (array)(l) x=0 \\ x=-\frac(b)(a) \end(array) \right. \)

Hence, an incomplete quadratic equation of the form ax 2 +bx=0 for \(b \neq 0 \) always has two roots.

An incomplete quadratic equation of the form ax 2 \u003d 0 is equivalent to the equation x 2 \u003d 0 and therefore has a single root 0.

The formula for the roots of a quadratic equation

Let us now consider how quadratic equations are solved in which both coefficients of the unknowns and the free term are nonzero.

We solve the quadratic equation in general view and as a result we get the formula of the roots. Then this formula can be applied to solve any quadratic equation.

Solve the quadratic equation ax 2 +bx+c=0

Dividing both its parts by a, we obtain the equivalent reduced quadratic equation
\(x^2+\frac(b)(a)x +\frac(c)(a)=0 \)

We transform this equation by highlighting the square of the binomial:
\(x^2+2x \cdot \frac(b)(2a)+\left(\frac(b)(2a)\right)^2- \left(\frac(b)(2a)\right)^ 2 + \frac(c)(a) = 0 \Rightarrow \)

\(x^2+2x \cdot \frac(b)(2a)+\left(\frac(b)(2a)\right)^2 = \left(\frac(b)(2a)\right)^ 2 - \frac(c)(a) \Rightarrow \) \(\left(x+\frac(b)(2a)\right)^2 = \frac(b^2)(4a^2) - \frac( c)(a) \Rightarrow \left(x+\frac(b)(2a)\right)^2 = \frac(b^2-4ac)(4a^2) \Rightarrow \) \(x+\frac(b )(2a) = \pm \sqrt( \frac(b^2-4ac)(4a^2) ) \Rightarrow x = -\frac(b)(2a) + \frac( \pm \sqrt(b^2 -4ac) )(2a) \Rightarrow \) \(x = \frac( -b \pm \sqrt(b^2-4ac) )(2a) \)

The root expression is called discriminant of a quadratic equation ax 2 +bx+c=0 (“discriminant” in Latin - distinguisher). It is denoted by the letter D, i.e.
\(D = b^2-4ac\)

Now, using the notation of the discriminant, we rewrite the formula for the roots of the quadratic equation:
\(x_(1,2) = \frac( -b \pm \sqrt(D) )(2a) \), where \(D= b^2-4ac \)

It's obvious that:
1) If D>0, then the quadratic equation has two roots.
2) If D=0, then the quadratic equation has one root \(x=-\frac(b)(2a)\).
3) If D Thus, depending on the value of the discriminant, the quadratic equation can have two roots (for D > 0), one root (for D = 0) or no roots (for D When solving a quadratic equation using this formula, it is advisable to do the following way:
1) calculate the discriminant and compare it with zero;
2) if the discriminant is positive or equal to zero, then use the root formula, if the discriminant is negative, then write down that there are no roots.

Vieta's theorem

The given quadratic equation ax 2 -7x+10=0 has roots 2 and 5. The sum of the roots is 7, and the product is 10. We see that the sum of the roots is equal to the second coefficient taken from opposite sign, and the product of the roots is equal to the free term. Any reduced quadratic equation that has roots has this property.

The sum of the roots of the given quadratic equation is equal to the second coefficient, taken with the opposite sign, and the product of the roots is equal to the free term.

Those. Vieta's theorem states that the roots x 1 and x 2 of the reduced quadratic equation x 2 +px+q=0 have the property:
\(\left\( \begin(array)(l) x_1+x_2=-p \\ x_1 \cdot x_2=q \end(array) \right. \)

First level

Quadratic equations. Comprehensive guide (2019)

In the term "quadratic equation" the key word is "quadratic". This means that the equation must necessarily contain a variable (the same X) in the square, and at the same time there should not be Xs in the third (or greater) degree.

The solution of many equations is reduced to the solution of quadratic equations.

Let's learn to determine that we have a quadratic equation, and not some other.

Example 1

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 a square!

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, but let's take a closer look. Let's move everything to the left side:

You see, it has shrunk - 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:

1. square;
2. square;
3. not square;
4. not square;
5. not square;
6. square;
7. not square;
8. square.

Mathematicians conditionally 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 the complete quadratic equations, there are given 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 free term c are equal to zero:

  They are incomplete because some element is missing from them. But the equation must always contain x squared !!! Otherwise, it will no longer be a quadratic, but some other equation.

Why did they come up with such a division? It would seem that there is an X squared, and okay. Such a division is due to the methods of solution. Let's consider each of them in more detail.

Solving incomplete quadratic equations

First, let's focus on solving incomplete quadratic equations - they are much simpler!

Incomplete quadratic equations are of types:

1. , in this equation the coefficient is equal.
2. , in this equation the free term is equal to.
3. , in this equation the coefficient and the free term are equal.

1. i. Since we know how to take the square root, 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. These formulas do not need to be memorized. The main thing is that you should always know and remember that it cannot be less.

Let's try to solve some examples.

Example 5:

Solve the Equation

Now it remains to extract the root from the left and right parts. After all, do you remember how to extract the roots?

Answer:

Never forget about roots with a negative sign!!!

Example 6:

Solve the Equation

Answer:

Example 7:

Solve the Equation

Ouch! The square of a number cannot be negative, which means that the equation

no roots!

For such equations in which there are 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:

Here we will do without examples.

Solving complete quadratic equations

We remind you that the complete quadratic equation is an equation of the form equation where

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 a root Special attention draw a step. The discriminant () 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'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. Solution of 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.

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. MIDDLE 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.

Why? Because if, the equation will immediately become linear, because will disappear.

In this case, and can be equal to zero. In this stool 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:

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. So:

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:

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 that the 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.

Decision:

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 is it possible different amount roots? Let's turn to geometric sense 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.

Examples:

Solutions:

Answer:

Answer: .

Answer:

This means there are no solutions.

Answer: .

2. Vieta's theorem

Using the Vieta theorem is very easy: 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 to given quadratic equations ().

Let's look at a few examples:

Example #1:

Solve the equation.

Decision:

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 #2:

Decision:

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 #3:

Decision:

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 #4:

Solve the equation.

Decision:

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 #5:

Solve the equation.

Decision:

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:

Solutions for tasks for independent work:

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: ; .

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: ; .

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:

Fine. 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: ; .

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: ; .

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: ; .

Let me summarize:

1. Vieta's theorem is used only in the given quadratic equations.
2. Using the Vieta theorem, you can find the roots by selection, orally.
3. 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 it is possible to represent the equation in the form of an incomplete quadratic equation of the type.

For example:

Example 1:

Solve the equation: .

Decision:

Answer:

Example 2:

Solve the equation: .

Decision:

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) We bring the equation to standard view: ,

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

Some problems in mathematics require the ability to calculate the value of the square root. These problems include solving second-order equations. In this article, we present effective method calculations square roots and use it when working with the formulas of the roots of a quadratic equation.

What is a square root?

In mathematics, this concept corresponds to the symbol √. Historical data says that it began to be used for the first time around the first half of the 16th century in Germany (the first German work on algebra by Christoph Rudolf). Scientists believe that this symbol is a transformed Latin letter r (radix means "root" in Latin).

The root of any number is equal to such a value, the square of which corresponds to the root expression. In the language of mathematics, this definition will look like this: √x = y if y 2 = x.

root of positive number(x > 0) is also a positive number (y > 0), but if you take the root of a negative number (x< 0), то его результатом уже будет комплексное число, включающее мнимую единицу i.

Here are two simple examples:

√9 = 3 because 3 2 = 9; √(-9) = 3i since i 2 = -1.

Heron's iterative formula for finding the values ​​of the square roots

The above examples are very simple, and the calculation of the roots in them is not difficult. Difficulties begin to appear already when finding the values ​​​​of the root for any value that cannot be represented as a square natural number, for example √10, √11, √12, √13, not to mention the fact that in practice it is necessary to find roots for non-integer numbers: for example √(12.15), √(8.5) and so on.

In all the above cases, a special method for calculating the square root should be used. Currently, several such methods are known: for example, expansion in a Taylor series, division by a column, and some others. Of all known methods Perhaps the simplest and most effective is to use Heron's iterative formula, which is also known as the Babylonian method for determining square roots (there is evidence that the ancient Babylonians used it in their practical calculations).

Let it be necessary to determine the value of √x. The formula for finding the square root has next view:

a n+1 = 1/2(a n +x/a n), where lim n->∞ (a n) => x.

Let's decipher this mathematical notation. To calculate √x, you should take some number a 0 (it can be arbitrary, however, to quickly get the result, you should choose it so that (a 0) 2 is as close as possible to x. Then substitute it into the specified formula for calculating the square root and get a new the number a 1, which will already be closer to the desired value.After that, it is necessary to substitute a 1 into the expression and get a 2. This procedure should be repeated until the required accuracy is obtained.

An example of applying Heron's iterative formula

The algorithm described above for obtaining the square root of some given number may sound rather complicated and confusing for many, but in reality everything turns out to be much simpler, since this formula converges very quickly (especially if a good number a 0 is chosen).

Let's give a simple example: it is necessary to calculate √11. We choose a 0 \u003d 3, since 3 2 \u003d 9, which is closer to 11 than 4 2 \u003d 16. Substituting into the formula, we get:

a 1 \u003d 1/2 (3 + 11/3) \u003d 3.333333;

a 2 \u003d 1/2 (3.33333 + 11 / 3.33333) \u003d 3.316668;

a 3 \u003d 1/2 (3.316668 + 11 / 3.316668) \u003d 3.31662.

There is no point in continuing the calculations, since we have found that a 2 and a 3 begin to differ only in the 5th decimal place. Thus, it was enough to apply the formula only 2 times to calculate √11 with an accuracy of 0.0001.

At present, calculators and computers are widely used to calculate the roots, however, it is useful to remember the marked formula in order to be able to manually calculate their exact value.

Second order equations

Understanding what a square root is and the ability to calculate it is used when solving quadratic equations. These equations are equalities with one unknown, the general form of which is shown in the figure below.

Here c, b and a are some numbers, and a must not be equal to zero, and the values ​​of c and b can be completely arbitrary, including being equal to zero.

Any values ​​of x that satisfy the equality indicated in the figure are called its roots (this concept should not be confused with the square root √). Since the equation under consideration has the 2nd order (x 2), then there cannot be more roots for it than two numbers. We will consider later in the article how to find these roots.

Finding the roots of a quadratic equation (formula)

This method of solving the type of equalities under consideration is also called universal, or the method through the discriminant. It can be applied to any quadratic equations. The formula for the discriminant and roots of the quadratic equation is as follows:

It can be seen from it that the roots depend on the value of each of the three coefficients of the equation. Moreover, the calculation of x 1 differs from the calculation of x 2 only by the sign in front of the square root. The radical expression, which is equal to b 2 - 4ac, is nothing more than the discriminant of the considered equality. The discriminant in the formula for the roots of a quadratic equation plays an important role because it determines the number and type of solutions. So, if it is zero, then there will be only one solution, if it is positive, then the equation has two real roots, and finally, a negative discriminant leads to two complex roots x 1 and x 2.

Vieta's theorem or some properties of the roots of second-order equations

At the end of the 16th century, one of the founders of modern algebra, a Frenchman, studying second-order equations, was able to obtain the properties of its roots. Mathematically, they can be written like this:

x 1 + x 2 = -b / a and x 1 * x 2 = c / a.

Both equalities can easily be obtained by everyone; for this, it is only necessary to perform the appropriate mathematical operations with the roots obtained through a formula with a discriminant.

The combination of these two expressions can rightfully be called the second formula of the roots of a quadratic equation, which makes it possible to guess its solutions without using the discriminant. Here it should be noted that although both expressions are always valid, it is convenient to use them to solve an equation only if it can be factored.

The task of consolidating the acquired knowledge

We will decide math problem, in which we will demonstrate all the techniques discussed in the article. The conditions of the problem are as follows: you need to find two numbers for which the product is -13, and the sum is 4.

This condition immediately reminds of Vieta's theorem, using the formulas for the sum of square roots and their product, we write:

x 1 + x 2 \u003d -b / a \u003d 4;

x 1 * x 2 \u003d c / a \u003d -13.

Assuming a = 1, then b = -4 and c = -13. These coefficients allow us to compose a second-order equation:

x 2 - 4x - 13 = 0.

We use the formula with the discriminant, we get the following roots:

x 1.2 = (4 ± √D)/2, D = 16 - 4 * 1 * (-13) = 68.

That is, the task was reduced to finding the number √68. Note that 68 = 4 * 17, then, using the square root property, we get: √68 = 2√17.

Now we use the considered square root formula: a 0 \u003d 4, then:

a 1 \u003d 1/2 (4 + 17/4) \u003d 4.125;

a 2 \u003d 1/2 (4.125 + 17 / 4.125) \u003d 4.1231.

There is no need to calculate a 3 because the found values ​​differ by only 0.02. Thus, √68 = 8.246. Substituting it into the formula for x 1,2, we get:

x 1 \u003d (4 + 8.246) / 2 \u003d 6.123 and x 2 \u003d (4 - 8.246) / 2 \u003d -2.123.

As you can see, the sum of the numbers found is really equal to 4, but if you find their product, then it will be equal to -12.999, which satisfies the condition of the problem with an accuracy of 0.001.

Kopyevskaya rural secondary school

10 Ways to Solve Quadratic Equations

Head: Patrikeeva Galina Anatolyevna,

mathematic teacher

s.Kopyevo, 2007

1. History of the development of quadratic equations

1.1 Quadratic equations in ancient Babylon

1.2 How Diophantus compiled and solved quadratic equations

1.3 Quadratic equations in India

1.4 Quadratic equations in al-Khwarizmi

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 in ancient times was caused by the need to solve problems related to finding the areas of land and earthworks military nature, as well as with the development of astronomy and mathematics itself. Quadratic equations were able to solve about 2000 BC. e. Babylonians.

Using modern algebraic notation, we can say that in their cuneiform texts, in addition to incomplete ones, there are such, for example, complete quadratic equations:

X 2 + X = ¾; X 2 - X = 14,5

The rule for solving these equations, stated in the Babylonian texts, coincides essentially with the modern one, but it is not known how the Babylonians came to this rule. Almost all the cuneiform texts found so far give only problems with solutions stated in the form of recipes, with no indication of how they were found.

In spite of high level development of algebra in Babylon, in cuneiform texts there is no concept of a negative number and common methods solutions of quadratic equations.

1.2 How Diophantus compiled and solved quadratic equations.

Diophantus' Arithmetic does not contain a systematic exposition of algebra, but it contains a systematic series of problems, accompanied by explanations and solved by formulating equations of various degrees.

When compiling equations, Diophantus skillfully chooses unknowns to simplify the solution.

Here, for example, is one of his tasks.

Task 11."Find two numbers knowing that their sum is 20 and their product is 96"

Diophantus argues as follows: it follows from the condition of the problem that the desired numbers are not equal, since if they were equal, then their product would be equal not to 96, but to 100. Thus, one of them will be more than half of their sum, i.e. . 10+x, the other is smaller, 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 desired numbers is 12 , other 8 . Decision x = -2 for Diophantus does not exist, since Greek mathematics knew only positive numbers.

If we solve this problem by choosing one of the desired numbers as the unknown, then we will come to the solution of the equation

y(20 - y) = 96,

y 2 - 20y + 96 = 0. (2)

It is clear that Diophantus simplifies the solution by choosing the half-difference of the desired numbers as the unknown; he manages to reduce the problem to solving an incomplete quadratic equation (1).

1.3 Quadratic equations in India

Problems for quadratic equations are already found in the astronomical tract "Aryabhattam", compiled in 499 by the Indian mathematician and astronomer Aryabhatta. Another Indian scholar, Brahmagupta (7th century), expounded 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 for a, can also be negative. Brahmagupta's rule essentially coincides with ours.

In ancient India, public competitions in solving difficult problems were common. In one of the old Indian books, the following is said about such competitions: “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.” Tasks were often dressed in poetic form.

Here is one of the problems of the famous Indian mathematician of the XII century. Bhaskara.

Task 13.

“A frisky flock of monkeys And twelve in vines ...

Having eaten power, had fun. They began to jump, hanging ...

Part eight of them in a square How many monkeys were there,

Having fun in the meadow. You tell me, in this flock?

Bhaskara's solution indicates that he knew about the two-valuedness of the roots of quadratic equations (Fig. 3).

The equation corresponding to problem 13 is:

( x /8) 2 + 12 = x

Bhaskara writes under the guise of:

x 2 - 64x = -768

and, to complete the left side of this equation to a square, he adds to both sides 32 2 , getting then:

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

Al-Khorezmi's algebraic treatise gives a classification of linear and quadratic equations. The author lists 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 number", i.e. ax 2 = s.

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 the number", i.e. ah 2+ bx = s.

6) "Roots and numbers are equal to squares", i.e. bx + c \u003d ax 2.

For al-Khorezmi, who avoided the use of negative numbers, the terms of each of these equations are summands, not subtrahends. In this case, equations that do not have positive solutions are obviously not taken into account. The author outlines the methods for solving these equations, using the methods of al-jabr and al-muqabala. His decisions, of course, do not completely coincide with ours. Not to mention the fact 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 it does not matter in specific practical problems. When solving complete quadratic equations, al-Khorezmi sets out the rules for solving, and then geometric proofs, using particular numerical examples.

Task 14.“The square and the number 21 are equal to 10 roots. Find the root" (assuming 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, 4 remains. Take the root of 4, you get 2. Subtract 2 from 5, you get 3, this will be the desired root. Or add 2 to 5, which will give 7, this is also a root.

Treatise al - Khorezmi is the first book that has come down to us, in which the classification of quadratic equations is systematically stated and formulas for their solution are given.

1.5 Quadratic equations in Europe XIII - XVII centuries

Formulas for solving quadratic equations on the model of al - Khorezmi in Europe were first set forth in the "Book of the Abacus", written in 1202 by the Italian mathematician Leonardo Fibonacci. This voluminous work, which reflects the influence of mathematics, both the countries of Islam and Ancient Greece, differs in both completeness and clarity of presentation. The author has independently developed some new algebraic examples problem solving 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 tasks from the "Book of the Abacus" passed into 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 = with,

for all possible combinations of signs of the coefficients b , with was formulated in Europe only in 1544 by M. Stiefel.

Vieta has a general derivation of the formula for solving a quadratic equation, but Vieta recognized only positive roots. The Italian mathematicians Tartaglia, Cardano, Bombelli were among the first in the 16th century. Take into account, in addition to positive, and negative roots. Only in the XVII century. Thanks to the work of Girard, Descartes, Newton and other scientists, the way to solve quadratic equations takes on a modern look.

1.6 About Vieta's theorem

The theorem expressing the relationship between the coefficients of a quadratic equation and its roots, bearing the name of Vieta, was formulated by him for the first time in 1591 as follows: “If B + D multiplied by A - A 2 , equals BD, then A equals AT and equal D ».

To understand Vieta, one must remember that BUT, like any vowel, meant for him the unknown (our X), the vowels AT, D- coefficients for the unknown. In the language of modern algebra, Vieta's formulation above means: if

(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 equations by general formulas written using symbols, Viet established uniformity in the methods of solving equations. However, the symbolism of Vieta is still far from modern look. He did not recognize negative numbers, and therefore, when solving equations, he considered only cases where all roots are positive.

2. Methods for solving quadratic equations

Quadratic equations are the foundation on which the majestic edifice of algebra rests. Quadratic equations find wide application when solving trigonometric, exponential, logarithmic, irrational and transcendental equations and inequalities. We all know how to solve quadratic equations from school (grade 8) until graduation.