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 can 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 entering a numerical fraction, the numerator is separated from the denominator by a division sign: /
whole part 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 form and as a result we obtain 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 with the 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\( \begin(array)(l) x_1+x_2=-p \\ x_1 \cdot x_2=q \end(array) \right. \)

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 calculating 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.

The root of a 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 root values ​​for any value that cannot be represented as a square of a 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 of the above cases, apply special method calculating the square root. At present, 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 most simple and effective is the use of 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 is as follows:

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 indicated 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

For many, the algorithm for obtaining the square root of a given number may sound rather complicated and confusing, 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, general form 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 solve a mathematical 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.

This topic may seem complicated at first due to the many not-so-simple formulas. Not only do the quadratic equations themselves have long entries, but the roots are also found through the discriminant. There are three new formulas in total. Not very easy to remember. This is possible only after the frequent solution of such equations. Then all the formulas will be remembered by themselves.

General view of the quadratic equation

Here their explicit notation is proposed, when the largest degree is written first, and then - in descending order. Often there are situations when the terms stand apart. Then it is better to rewrite the equation in descending order of the degree of the variable.

Let us introduce notation. They are presented in the table below.

If we accept these notations, all quadratic equations are reduced to the following notation.

Moreover, the coefficient a ≠ 0. Let this formula be denoted by number one.

When the equation is given, it is not clear how many roots will be in the answer. Because one of three options is always possible:

• the solution will have two roots;
• the answer will be one number;
• The equation has no roots at all.

And while the decision is not brought to the end, it is difficult to understand which of the options will fall out in a particular case.

Types of records of quadratic equations

Tasks may have different entries. They will not always look like the general formula of a quadratic equation. Sometimes it will lack some terms. What was written above is the complete equation. If you remove the second or third term in it, you get something else. These records are also called quadratic equations, only incomplete.

Moreover, only the terms for which the coefficients "b" and "c" can disappear. The number "a" cannot be equal to zero under any circumstances. Because in this case the formula turns into a linear equation. The formulas for the incomplete form of the equations will be as follows:

So, there are only two types, in addition to complete ones, there are also incomplete quadratic equations. Let the first formula be number two, and the second number three.

The discriminant and the dependence of the number of roots on its value

This number must be known in order to calculate the roots of the equation. It can always be calculated, no matter what the formula of the quadratic equation is. In order to calculate the discriminant, you need to use the equality written below, which will have the number four.

After substituting the values ​​of the coefficients into this formula, you can get numbers with different signs. If the answer is yes, then the answer to the equation will be two different roots. With a negative number, the roots of the quadratic equation will be absent. If it is equal to zero, the answer will be one.

How is a complete quadratic equation solved?

In fact, consideration of this issue has already begun. Because first you need to find the discriminant. After it is clarified that there are roots of the quadratic equation, and their number is known, you need to use the formulas for the variables. If there are two roots, then you need to apply such a formula.

Since it contains the “±” sign, there will be two values. The expression under the square root sign is the discriminant. Therefore, the formula can be rewritten in a different way.

Formula five. From the same record it can be seen that if the discriminant is zero, then both roots will take the same values.

If the solution of quadratic equations has not yet been worked out, then it is better to write down the values ​​of all coefficients before applying the discriminant and variable formulas. Later this moment will not cause difficulties. But at the very beginning there is confusion.

How is an incomplete quadratic equation solved?

Everything is much simpler here. Even there is no need for additional formulas. And you won't need those that have already been written for the discriminant and the unknown.

First, consider the incomplete equation number two. In this equality, it is supposed to take the unknown value out of the bracket and solve the linear equation, which will remain in the brackets. The answer will have two roots. The first one is necessarily equal to zero, because there is a factor consisting of the variable itself. The second is obtained by solving a linear equation.

The incomplete equation at number three is solved by transferring the number from the left side of the equation to the right. Then you need to divide by the coefficient in front of the unknown. It remains only to extract the square root and do not forget to write it down twice with opposite signs.

The following are some actions that help you learn how to solve all kinds of equalities that turn into quadratic equations. They will help the student to avoid mistakes due to inattention. These shortcomings are the cause of poor grades when studying the extensive topic "Quadric Equations (Grade 8)". Subsequently, these actions will not need to be constantly performed. Because there will be a stable habit.

• First you need to write the equation in standard form. That is, first the term with the largest degree of the variable, and then - without the degree and the last - just a number.

• If a minus appears before the coefficient "a", then it can complicate the work for a beginner to study quadratic equations. It's better to get rid of it. For this purpose, all equality must be multiplied by "-1". This means that all terms will change sign to the opposite.

• In the same way, it is recommended to get rid of fractions. Simply multiply the equation by the appropriate factor so that the denominators cancel out.

Examples

It is required to solve the following quadratic equations:

x 2 - 7x \u003d 0;

15 - 2x - x 2 \u003d 0;

x 2 + 8 + 3x = 0;

12x + x 2 + 36 = 0;

(x+1) 2 + x + 1 = (x+1)(x+2).

The first equation: x 2 - 7x \u003d 0. It is incomplete, therefore it is solved as described for formula number two.

After bracketing, it turns out: x (x - 7) \u003d 0.

The first root takes the value: x 1 = 0. The second will be found from linear equation: x - 7 = 0. It is easy to see that x 2 = 7.

Second equation: 5x2 + 30 = 0. Again incomplete. Only it is solved as described for the third formula.

After transferring 30 to the right side of the equation: 5x 2 = 30. Now you need to divide by 5. It turns out: x 2 = 6. The answers will be numbers: x 1 = √6, x 2 = - √6.

Third equation: 15 - 2x - x 2 \u003d 0. Here and below, the solution of quadratic equations will begin by rewriting them into a standard form: - x 2 - 2x + 15 \u003d 0. Now it's time to use the second useful advice and multiply everything by minus one. It turns out x 2 + 2x - 15 \u003d 0. According to the fourth formula, you need to calculate the discriminant: D \u003d 2 2 - 4 * (- 15) \u003d 4 + 60 \u003d 64. It is a positive number. From what was said above, it turns out that the equation has two roots. They need to be calculated according to the fifth formula. According to it, it turns out that x \u003d (-2 ± √64) / 2 \u003d (-2 ± 8) / 2. Then x 1 \u003d 3, x 2 \u003d - 5.

The fourth equation x 2 + 8 + 3x \u003d 0 is converted to this: x 2 + 3x + 8 \u003d 0. Its discriminant is equal to this value: -23. Since this number is negative, the answer to this task will be the following entry: "There are no roots."

The fifth equation 12x + x 2 + 36 = 0 should be rewritten as follows: x 2 + 12x + 36 = 0. After applying the formula for the discriminant, the number zero is obtained. This means that it will have one root, namely: x \u003d -12 / (2 * 1) \u003d -6.

The sixth equation (x + 1) 2 + x + 1 = (x + 1) (x + 2) requires transformations, which consist in the fact that you need to bring like terms, before opening the brackets. In place of the first one there will be such an expression: x 2 + 2x + 1. After equality, this entry will appear: x 2 + 3x + 2. After similar terms are counted, the equation will take the form: x 2 - x \u003d 0. It has become incomplete . Similar to it has already been considered a little higher. The roots of this will be the numbers 0 and 1.

I hope that after studying this article, you will learn how to find the roots of a complete quadratic equation.

With the help of 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? it equations of the form ax 2 + b x + c = 0, where the coefficients a, b and c are not equal to zero. So, to solve the complete quadratic equation, you need to calculate the discriminant D.

D \u003d b 2 - 4ac.

Depending on what value the discriminant has, 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 \u003d (-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 \u003d 4 2 - 4 4 \u003d 0

x = (- (-4))/2 = 2

Answer: 2.

Solve Equation 2 x 2 + x + 3 = 0.

D \u003d 1 2 - 4 2 3 \u003d - 23

Answer: no roots.

Solve Equation 2 x 2 + 5x - 7 = 0.

D \u003d 5 2 - 4 2 (-7) \u003d 81

x 1 \u003d (-5 - √81) / (2 2) \u003d (-5 - 9) / 4 \u003d - 3.5

x 2 \u003d (-5 + √81) / (2 2) \u003d (-5 + 9) / 4 \u003d 1

Answer: - 3.5; one.

So let's imagine the solution of complete quadratic equations by the scheme in Figure 1.

These formulas can be used to solve any complete quadratic equation. You just need to be careful to the equation was written as a polynomial of standard form

a x 2 + bx + c, otherwise you can 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 \u003d 3 2 - 4 1 2 \u003d 1 and then the equation has two roots. And this is not true. (See example 2 solution 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 (in the first place there should be a monomial with the largest exponent, that is a x 2 , then with less – bx, and then the free term With.

When solving the above quadratic equation and the quadratic equation with an even coefficient for the second term, other formulas can also be used. Let's get acquainted with these formulas. If in the full quadratic equation with the second term the coefficient is even (b = 2k), then the equation can be solved using the formulas shown in the diagram of Figure 2.

A complete quadratic equation is called reduced if the coefficient at x 2 equals unity and the equation takes the form x 2 + px + q = 0. Such an equation can be given to solve, or is obtained by dividing all the coefficients of the equation by the coefficient a standing at x 2 .

Figure 3 shows a diagram of the solution of the reduced square
equations. Consider the 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 Figure 1.

D \u003d 6 2 - 4 3 (- 6) \u003d 36 + 72 \u003d 108

√D = √108 = √(36 3) = 6√3

x 1 \u003d (-6 - 6 √ 3) / (2 3) \u003d (6 (-1- √ (3))) / 6 \u003d -1 - √ 3

x 2 \u003d (-6 + 6 √ 3) / (2 3) \u003d (6 (-1 + √ (3))) / 6 \u003d -1 + √ 3

Answer: -1 - √3; –1 + √3

You can see that the coefficient at x in this equation even number, that is, b \u003d 6 or b \u003d 2k, whence k \u003d 3. Then let's try to solve the equation using the formulas shown in the diagram of the figure D 1 \u003d 3 2 - 3 (- 6) \u003d 9 + 18 \u003d 27

√(D 1) = √27 = √(9 3) = 3√3

x 1 \u003d (-3 - 3√3) / 3 \u003d (3 (-1 - √ (3))) / 3 \u003d - 1 - √3

x 2 \u003d (-3 + 3√3) / 3 \u003d (3 (-1 + √ (3))) / 3 \u003d - 1 + √3

Answer: -1 - √3; –1 + √3. Noticing that all the coefficients in this quadratic equation are divisible by 3 and dividing, we get the reduced quadratic equation x 2 + 2x - 2 = 0 We solve this equation using the formulas for the reduced quadratic
equations figure 3.

D 2 \u003d 2 2 - 4 (- 2) \u003d 4 + 8 \u003d 12

√(D 2) = √12 = √(4 3) = 2√3

x 1 \u003d (-2 - 2√3) / 2 \u003d (2 (-1 - √ (3))) / 2 \u003d - 1 - √3

x 2 \u003d (-2 + 2 √ 3) / 2 \u003d (2 (-1 + √ (3))) / 2 \u003d - 1 + √ 3

Answer: -1 - √3; –1 + √3.

As you can see, when solving this equation using different formulas, we got the same answer. Therefore, having well mastered the formulas shown in the diagram of Figure 1, you can always solve any complete quadratic equation.

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It is known that it is a particular version of the equality ax 2 + in + c \u003d o, where a, b and c are real coefficients for unknown x, and where a ≠ o, and b and c will be zeros - simultaneously or separately. For example, c = o, v ≠ o or vice versa. We almost remembered the definition of a quadratic equation.

A trinomial of the second degree is equal to zero. Its first coefficient a ≠ o, b and c can take on any values. The value of the variable x will then be when, when substituting, it will turn it into the correct numerical equality. Let us dwell on real roots, although the solutions of the equation can also be complete. It is customary to call an equation in which none of the coefficients is equal to o, a ≠ o, b ≠ o, c ≠ o.
Let's solve an example. 2x2 -9x-5 = oh, we find
D \u003d 81 + 40 \u003d 121,
D is positive, so there are roots, x 1 = (9+√121): 4 = 5, and the second x 2 = (9-√121): 4 = -o.5. Checking will help make sure they are correct.

Here is a step by step solution to the quadratic equation

Through the discriminant, you can solve any equation, on the left side of which there is a known square trinomial with a ≠ o. In our example. 2x 2 -9x-5 \u003d 0 (ax 2 + in + c \u003d o)

Consider what are incomplete equations of the second degree

1. ax 2 + in = o. The free term, the coefficient c at x 0, is zero here, in ≠ o.
   How to solve an incomplete quadratic equation of this kind? Let's take x out of brackets. Remember when the product of two factors is zero.
   x(ax+b) = o, this can be when x = o or when ax+b = o.
   Solving the 2nd, we have x = -v/a.
   As a result, we have roots x 1 \u003d 0, according to calculations x 2 \u003d -b / a.
2. Now the coefficient of x is o, but c is not equal to (≠) o.
   x 2 + c \u003d o. We transfer c to the right side of the equality, we get x 2 \u003d -c. This equation only has real roots when -c is a positive number (c ‹ o),
   x 1 is then equal to √(-c), respectively, x 2 is -√(-c). Otherwise, the equation has no roots at all.
3. The last option: b \u003d c \u003d o, that is, ax 2 \u003d o. Naturally, such a simple equation has one root, x = o.

Special cases

We have considered how to solve an incomplete quadratic equation, and now we will take any kind.

• In the full quadratic equation, the second coefficient of x is an even number.
  Let k = o,5b. We have formulas for calculating the discriminant and roots.
  D / 4 \u003d k 2 - ac, the roots are calculated as follows x 1,2 \u003d (-k ± √ (D / 4)) / a for D › o.
  x = -k/a at D = o.
  There are no roots for D ‹ o.
• There are reduced quadratic equations, when the coefficient of x squared is 1, they are usually written x 2 + px + q \u003d o. All of the above formulas apply to them, but the calculations are somewhat simpler.
  Example, x 2 -4x-9 \u003d 0. We calculate D: 2 2 +9, D \u003d 13.
  x 1 = 2+√13, x 2 = 2-√13.
• In addition, it is easily applied to the given ones. It says that the sum of the roots of the equation is equal to -p, the second coefficient with a minus (meaning the opposite sign), and the product of these same roots will be equal to q, the free term. Check out how easy it would be to verbally determine the roots of this equation. For non-reduced (for all coefficients that are not equal to zero), this theorem is applicable as follows: the sum x 1 + x 2 is equal to -v / a, the product x 1 x 2 is equal to c / a.

The sum of the free term c and the first coefficient a is equal to the coefficient b. In this situation, the equation has at least one root (it is easy to prove), the first one is necessarily equal to -1, and the second - c / a, if it exists. How to solve an incomplete quadratic equation, you can check it yourself. Easy peasy. Coefficients can be in some ratios among themselves

• x 2 + x \u003d o, 7x 2 -7 \u003d o.
• The sum of all coefficients is o.
  The roots of such an equation are 1 and c / a. Example, 2x 2 -15x + 13 = o.
  x 1 \u003d 1, x 2 \u003d 13/2.

There are a number of other ways to solve different equations of the second degree. Here, for example, is a method for extracting a full square from a given polynomial. There are several graphic ways. When you often deal with such examples, you will learn to “click” them like seeds, because all methods come to mind automatically.