Quadratic equation problems are also studied in school curriculum and in universities. They mean equations of the form a*x^2 + b*x + c = 0, where x- variable, a, b, c – constants; a<>0 . The task is to find the roots of the equation.

Geometric meaning of quadratic equation

The graph of a function that is represented by a quadratic equation is a parabola. The solutions (roots) of a quadratic equation are the points of intersection of the parabola with the abscissa (x) axis. It follows that there are three possible cases:
1) the parabola has no points of intersection with the abscissa axis. This means that it is in the upper plane with branches up or the bottom with branches down. In such cases, the quadratic equation has no real roots (it has two complex roots).

2) the parabola has one point of intersection with the Ox axis. Such a point is called the vertex of the parabola, and the quadratic equation at it acquires its minimum or maximum value. In this case, the quadratic equation has one real root (or two identical roots).

3) The last case is more interesting in practice - there are two points of intersection of the parabola with the abscissa axis. This means that there are two real roots of the equation.

Based on the analysis of the coefficients of the powers of the variables, interesting conclusions can be drawn about the placement of the parabola.

1) If the coefficient a is greater than zero, then the parabola’s branches are directed upward; if it is negative, the parabola’s branches are directed downward.

2) If the coefficient b is greater than zero, then the vertex of the parabola lies in the left half-plane, if it takes a negative value, then in the right.

Derivation of the formula for solving a quadratic equation

Let's transfer the constant from the quadratic equation

for the equal sign, we get the expression

Multiply both sides by 4a

To get a complete square on the left, add b^2 on both sides and carry out the transformation

From here we find

Formula for the discriminant and roots of a quadratic equation

The discriminant is the value of the radical expression. If it is positive, then the equation has two real roots, calculated by the formula When the discriminant is zero, the quadratic equation has one solution (two coinciding roots), which can be easily obtained from the above formula for D=0. When the discriminant is negative, the equation has no real roots. However, solutions to the quadratic equation are found in the complex plane, and their value is calculated using the formula

Vieta's theorem

Let's consider two roots of a quadratic equation and construct a quadratic equation on their basis. Vieta's theorem itself easily follows from the notation: if we have a quadratic equation of the form then the sum of its roots is equal to the coefficient p taken from opposite sign, and the product of the roots of the equation is equal to the free term q. The formulaic representation of the above will look like If in a classical equation the constant a is nonzero, then you need to divide the entire equation by it, and then apply Vieta’s theorem.

Factoring quadratic equation schedule

Let the task be set: factor a quadratic equation. To do this, we first solve the equation (find the roots). Next, we substitute the found roots into the expansion formula for the quadratic equation. This will solve the problem.

Quadratic equation problems

Task 1. Find the roots of a quadratic equation

x^2-26x+120=0 .

Solution: Write down the coefficients and substitute them into the discriminant formula

Root of given value is equal to 14, it is easy to find with a calculator, or remember with frequent use, however, for convenience, at the end of the article I will give you a list of squares of numbers that can often be encountered in such problems.
We substitute the found value into the root formula

and we get

Task 2. Solve the equation

2x 2 +x-3=0.

Solution: We have a complete quadratic equation, write out the coefficients and find the discriminant


Using known formulas we find the roots of the quadratic equation

Task 3. Solve the equation

9x 2 -12x+4=0.

Solution: We have a complete quadratic equation. Determining the discriminant

We got a case where the roots coincide. Find the values ​​of the roots using the formula

Task 4. Solve the equation

x^2+x-6=0 .

Solution: In cases where there are small coefficients for x, it is advisable to apply Vieta’s theorem. By its condition we obtain two equations

From the second condition we find that the product must be equal to -6. This means that one of the roots is negative. We have the following possible pair of solutions (-3;2), (3;-2) . Taking into account the first condition, we reject the second pair of solutions.
The roots of the equation are equal

Problem 5. Find the lengths of the sides of a rectangle if its perimeter is 18 cm and its area is 77 cm 2.

Solution: Half the perimeter of a rectangle is equal to the sum of its adjacent sides. Let's denote x as the larger side, then 18-x is its smaller side. The area of ​​the rectangle is equal to the product of these lengths:
x(18-x)=77;
or
x 2 -18x+77=0.
Let's find the discriminant of the equation

Calculating the roots of the equation

If x=11, That 18's=7 , the opposite is also true (if x=7, then 21's=9).

Problem 6. Factor the quadratic equation 10x 2 -11x+3=0.

Solution: Let's calculate the roots of the equation, to do this we find the discriminant

We substitute the found value into the root formula and calculate

We apply the formula for decomposing a quadratic equation by roots

Opening the brackets we obtain an identity.

Quadratic equation with parameter

Example 1. At what parameter values A , does the equation (a-3)x 2 + (3-a)x-1/4=0 have one root?

Solution: By direct substitution of the value a=3 we see that it has no solution. Next, we will use the fact that with a zero discriminant the equation has one root of multiplicity 2. Let's write out the discriminant

Let's simplify it and equate it to zero

We have obtained a quadratic equation with respect to the parameter a, the solution of which can be easily obtained using Vieta’s theorem. The sum of the roots is 7, and their product is 12. By simple search we establish that the numbers 3,4 will be the roots of the equation. Since we already rejected the solution a=3 at the beginning of the calculations, the only correct one will be - a=4. Thus, for a=4 the equation has one root.

Example 2. At what parameter values A , the equation a(a+3)x^2+(2a+6)x-3a-9=0 has more than one root?

Solution: Let's first consider the singular points, they will be the values ​​a=0 and a=-3. When a=0, the equation will be simplified to the form 6x-9=0; x=3/2 and there will be one root. For a= -3 we obtain the identity 0=0.
Let's calculate the discriminant

and find the value of a at which it is positive

From the first condition we get a>3. For the second, we find the discriminant and roots of the equation


Let us determine the intervals where the function takes positive values. By substituting the point a=0 we get 3>0 . So, outside the interval (-3;1/3) the function is negative. Don't forget the point a=0, which should be excluded because the original equation has one root in it.
As a result, we obtain two intervals that satisfy the conditions of the problem

There will be many similar tasks in practice, try to figure out the tasks yourself and do not forget to take into account the conditions that are mutually exclusive. Study well the formulas for solving quadratic equations; they are often needed in calculations in various problems and sciences.

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, back in ancient times, was caused by the need to solve problems related to finding the areas of land plots and earthworks 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 eclipse the glory of another in popular assemblies by 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 complement left side of this equation to the 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 sets out the rules for solving them using particular numerical examples, and then 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.

Derivation of the formula for solving a quadratic equation in general view Viet has it, but Viet recognized only positive roots. Italian mathematicians Tartaglia, Cardano, Bombelli were among the first in the 16th century. They take into account, in addition to the positive, and negative roots. 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 equations with general formulas written using symbols, Viète 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 widely used in solving trigonometric, exponential, logarithmic, irrational and transcendental equations and inequalities. We all know how to solve quadratic equations from school (8th grade) until graduation.

Discriminant is a multi-valued term. In this article we will talk about the discriminant of a polynomial, which allows you to determine whether a given polynomial has valid solutions. The formula for the quadratic polynomial is found in the school course on algebra and analysis. How to find a discriminant? What is needed to solve the equation?

A quadratic polynomial or equation of the second degree is called i * w ^ 2 + j * w + k equals 0, where “i” and “j” are the first and second coefficients, respectively, “k” is a constant, sometimes called the “dismissive term,” and “w” is a variable. Its roots will be all the values ​​of the variable at which it turns into an identity. Such an equality can be rewritten as the product of i, (w - w1) and (w - w2) equal to 0. In this case, it is obvious that if the coefficient “i” does not become zero, then the function on the left side will become zero only if if x takes the value w1 or w2. These values ​​are the result of setting the polynomial equal to zero.

To find the value of a variable at which a quadratic polynomial vanishes, an auxiliary construction is used, built on its coefficients and called a discriminant. This design is calculated according to the formula D equals j * j - 4 * i * k. Why is it used?

1. She says are there any valid results.
2. She helps calculate them.

How does this value show the presence of real roots:

• If it is positive, then two roots can be found in the region of real numbers.
• If the discriminant is zero, then both solutions are the same. We can say that there is only one solution, and it is from the field of real numbers.
• If the discriminant less than zero, then the polynomial has no real roots.

Calculation options for securing material

For the sum (7 * w^2; 3 * w; 1) equal to 0 We calculate D using the formula 3 * 3 - 4 * 7 * 1 = 9 - 28, we get -19. A discriminant value below zero indicates that there are no results on the actual line.

If we consider 2 * w^2 - 3 * w + 1 equivalent to 0, then D is calculated as (-3) squared minus the product of numbers (4; 2; 1) and equals 9 - 8, that is, 1. Positive value says there are two results on the real line.

If we take the sum (w ^ 2; 2 * w; 1) and equate it to 0, D is calculated as two squared minus the product of the numbers (4; 1; 1). This expression will simplify to 4 - 4 and go to zero. It turns out that the results are the same. If you look closely at this formula, it will become clear that this is a “complete square”. This means that the equality can be rewritten in the form (w + 1) ^ 2 = 0. It became obvious that the result in this problem is “-1”. In a situation where D is equal to 0, the left side of the equality can always be collapsed using the “square of the sum” formula.

Using a discriminant in calculating roots

This auxiliary construction not only shows the number of real solutions, but also helps to find them. General formula The calculation for the second degree equation is:

w = (-j +/- d) / (2 * i), where d is the discriminant to the power of 1/2.

Let's say the discriminant is below zero, then d is imaginary and the results are imaginary.

D is zero, then d equal to D to the power of 1/2 is also zero. Solution: -j / (2 * i). Again considering 1 * w ^ 2 + 2 * w + 1 = 0, we find results equivalent to -2 / (2 * 1) = -1.

Suppose D > 0, then d is a real number, and the answer here breaks down into two parts: w1 = (-j + d) / (2 * i) and w2 = (-j - d) / (2 * i) . Both results will be valid. Let's look at 2 * w ^ 2 - 3 * w + 1 = 0. Here the discriminant and d are ones. It turns out that w1 is equal to (3 + 1) divided by (2 * 2) or 1, and w2 is equal to (3 - 1) divided by 2 * 2 or 1/2.

The result of equating a quadratic expression to zero is calculated according to the algorithm:

1. Determining the number of valid solutions.
2. Calculation d = D^(1/2).
3. Finding the result according to the formula (-j +/- d) / (2 * i).
4. Substituting the obtained result into the original equality for verification.

Some special cases

Depending on the coefficients, the solution may be somewhat simplified. Obviously, if the coefficient of a variable to the second power is zero, then a linear equality is obtained. When the coefficient of a variable to the first power is zero, then two options are possible:

1. the polynomial is expanded into a difference of squares when the free term is negative;
2. for a positive constant, no real solutions can be found.

If the free term is zero, then the roots will be (0; -j)

But there are other special cases that simplify finding a solution.

Reduced second degree equation

The given is called such a quadratic trinomial, where the coefficient of the leading term is one. For this situation, Vieta’s theorem is applicable, which states that the sum of the roots is equal to the coefficient of the variable to the first power, multiplied by -1, and the product corresponds to the constant “k”.

Therefore, w1 + w2 equals -j and w1 * w2 equals k if the first coefficient is one. To verify the correctness of this representation, you can express w2 = -j - w1 from the first formula and substitute it into the second equality w1 * (-j - w1) = k. The result is the original equality w1 ^ 2 + j * w1 + k = 0.

It is important to note, that i * w ^ 2 + j * w + k = 0 can be achieved by dividing by “i”. The result will be: w^2 + j1 * w + k1 = 0, where j1 is equal to j/i and k1 is equal to k/i.

Let's look at the already solved 2 * w^2 - 3 * w + 1 = 0 with the results w1 = 1 and w2 = 1/2. We need to divide it in half, as a result w ^ 2 - 3/2 * w + 1/2 = 0. Let's check that the conditions of the theorem are true for the results found: 1 + 1/2 = 3/2 and 1*1/2 = 1 /2.

Even second factor

If the factor of a variable to the first power (j) is divisible by 2, then it will be possible to simplify the formula and look for a solution through a quarter of the discriminant D/4 = (j / 2) ^ 2 - i * k. it turns out w = (-j +/- d/2) / i, where d/2 = D/4 to the power of 1/2.

If i = 1, and the coefficient j is even, then the solution will be the product of -1 and half the coefficient of the variable w, plus/minus the root of the square of this half minus the constant “k”. Formula: w = -j/2 +/- (j^2/4 - k)^1/2.

Higher discriminant order

The discriminant of the second degree trinomial discussed above is the most commonly used special case. In the general case, the discriminant of a polynomial is multiplied squares of the differences of the roots of this polynomial. Therefore, a discriminant equal to zero indicates the presence of at least two multiple solutions.

Consider i * w^3 + j * w^2 + k * w + m = 0.

D = j^2 * k^2 - 4 * i * k^3 - 4 * i^3 * k - 27 * i^2 * m^2 + 18 * i * j * k * m.

Suppose the discriminant exceeds zero. This means that there are three roots in the region of real numbers. At zero there are multiple solutions. If D< 0, то два корня комплексно-сопряженные, которые дают отрицательное значение при возведении в квадрат, а также один корень — вещественный.

Video

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For example, for the trinomial \(3x^2+2x-7\), the discriminant will be equal to \(2^2-4\cdot3\cdot(-7)=4+84=88\). And for the trinomial \(x^2-5x+11\), it will be equal to \((-5)^2-4\cdot1\cdot11=25-44=-19\).

The discriminant is denoted by the letter \(D\) and is often used in solving. Also, by the value of the discriminant, you can understand what the graph approximately looks like (see below).

Discriminant and roots of a quadratic equation

The discriminant value shows the number of quadratic equations:
- if \(D\) is positive, the equation will have two roots;
- if \(D\) is equal to zero – there is only one root;
- if \(D\) is negative, there are no roots.

This does not need to be taught, it is not difficult to come to such a conclusion, simply knowing that from the discriminant (that is, \(\sqrt(D)\) is included in the formula for calculating the roots of a quadratic equation: \(x_(1)=\)\( \frac(-b+\sqrt(D))(2a)\) and \(x_(2)=\)\(\frac(-b-\sqrt(D))(2a)\) Let's look at each case more details.

If the discriminant is positive

In this case, the root of it is some positive number, which means \(x_(1)\) and \(x_(2)\) will have different meanings, because in the first formula \(\sqrt(D)\) is added , and in the second it is subtracted. And we have two different roots.

Example : Find the roots of the equation \(x^2+2x-3=0\)
Solution :

Answer : \(x_(1)=1\); \(x_(2)=-3\)

If the discriminant is zero

How many roots will there be if the discriminant is zero? Let's reason.

The root formulas look like this: \(x_(1)=\)\(\frac(-b+\sqrt(D))(2a)\) and \(x_(2)=\)\(\frac(-b- \sqrt(D))(2a)\) . And if the discriminant is zero, then its root is also zero. Then it turns out:

\(x_(1)=\)\(\frac(-b+\sqrt(D))(2a)\) \(=\)\(\frac(-b+\sqrt(0))(2a)\) \(=\)\(\frac(-b+0)(2a)\) \(=\)\(\frac(-b)(2a)\)

\(x_(2)=\)\(\frac(-b-\sqrt(D))(2a)\) \(=\)\(\frac(-b-\sqrt(0))(2a) \) \(=\)\(\frac(-b-0)(2a)\) \(=\)\(\frac(-b)(2a)\)

That is, the values ​​of the roots of the equation will be the same, because adding or subtracting zero does not change anything.

Example : Find the roots of the equation \(x^2-4x+4=0\)
Solution :

\(x^2-4x+4=0\)		We write out the coefficients:
\(a=1;\) \(b=-4;\) \(c=4;\)		We calculate the discriminant using the formula \(D=b^2-4ac\)
\(D=(-4)^2-4\cdot1\cdot4=\) \(=16-16=0\)		Finding the roots of the equation
\(x_(1)=\) \(\frac(-(-4)+\sqrt(0))(2\cdot1)\)\(=\)\(\frac(4)(2)\) \(=2\) \(x_(2)=\) \(\frac(-(-4)-\sqrt(0))(2\cdot1)\)\(=\)\(\frac(4)(2)\) \(=2\)		We got two identical roots, so there is no point in writing them separately - we write them as one.

Answer : \(x=2\)

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 of the standard form

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 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.

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