Quadratic equations are studied in 8th grade, so there is nothing complicated here. The ability to solve them is absolutely necessary.

A quadratic equation is an equation of the form ax 2 + bx + c = 0, where the coefficients a, b and c are arbitrary numbers, and a ≠ 0.

Before studying specific solution methods, note that all quadratic equations can be divided into three classes:

1. Have no roots;
2. Have exactly one root;
3. Have two various roots.

This is an important difference between quadratic equations and linear ones, where the root always exists and is unique. How to determine how many roots an equation has? There is a wonderful thing for this - discriminant.

Discriminant

Let the quadratic equation ax 2 + bx + c = 0 be given. Then the discriminant is simply the number D = b 2 − 4ac.

You need to know this formula by heart. Where it comes from is not important now. Another thing is important: by the sign of the discriminant you can determine how many roots a quadratic equation has. Namely:

1. If D< 0, корней нет;
2. If D = 0, there is exactly one root;
3. If D > 0, there will be two roots.

Please note: the discriminant indicates the number of roots, and not at all their signs, as for some reason many people believe. Take a look at the examples and you will understand everything yourself:

Task. How many roots do quadratic equations have:

1. x 2 − 8x + 12 = 0;
2. 5x 2 + 3x + 7 = 0;
3. x 2 − 6x + 9 = 0.

Let's write out the coefficients for the first equation and find the discriminant:
a = 1, b = −8, c = 12;
D = (−8) 2 − 4 1 12 = 64 − 48 = 16

So the discriminant is positive, so the equation has two different roots. We analyze the second equation in a similar way:
a = 5; b = 3; c = 7;
D = 3 2 − 4 5 7 = 9 − 140 = −131.

The discriminant is negative, there are no roots. The last equation left is:
a = 1; b = −6; c = 9;
D = (−6) 2 − 4 1 9 = 36 − 36 = 0.

The discriminant is zero - the root will be one.

Please note that coefficients have been written down for each equation. Yes, it’s long, yes, it’s tedious, but you won’t mix up the odds and make stupid mistakes. Choose for yourself: speed or quality.

By the way, if you get the hang of it, after a while you won’t need to write down all the coefficients. You will perform such operations in your head. Most people start doing this somewhere after 50-70 solved equations - in general, not that much.

Roots of a quadratic equation

Now let's move on to the solution itself. If the discriminant D > 0, the roots can be found using the formulas:

Basic root formula quadratic equation

When D = 0, you can use any of these formulas - you will get the same number, which will be the answer. Finally, if D< 0, корней нет — ничего считать не надо.

1. x 2 − 2x − 3 = 0;
2. 15 − 2x − x 2 = 0;
3. x 2 + 12x + 36 = 0.

First equation:
x 2 − 2x − 3 = 0 ⇒ a = 1; b = −2; c = −3;
D = (−2) 2 − 4 1 (−3) = 16.

D > 0 ⇒ the equation has two roots. Let's find them:

Second equation:
15 − 2x − x 2 = 0 ⇒ a = −1; b = −2; c = 15;
D = (−2) 2 − 4 · (−1) · 15 = 64.

D > 0 ⇒ the equation again has two roots. Let's find them

\[\begin(align) & ((x)_(1))=\frac(2+\sqrt(64))(2\cdot \left(-1 \right))=-5; \\ & ((x)_(2))=\frac(2-\sqrt(64))(2\cdot \left(-1 \right))=3. \\ \end(align)\]

Finally, the third equation:
x 2 + 12x + 36 = 0 ⇒ a = 1; b = 12; c = 36;
D = 12 2 − 4 1 36 = 0.

D = 0 ⇒ the equation has one root. Any formula can be used. For example, the first one:

As you can see from the examples, everything is very simple. If you know the formulas and can count, there will be no problems. Most often, errors occur when substituting negative coefficients into the formula. Here again, the technique described above will help: look at the formula literally, write down each step - and very soon you will get rid of errors.

Incomplete quadratic equations

It happens that a quadratic equation is slightly different from what is given in the definition. For example:

1. x 2 + 9x = 0;
2. x 2 − 16 = 0.

It is easy to notice that these equations are missing one of the terms. Such quadratic equations are even easier to solve than standard ones: they don’t even require calculating the discriminant. So, let's introduce a new concept:

The equation ax 2 + bx + c = 0 is called an incomplete quadratic equation if b = 0 or c = 0, i.e. the coefficient of the variable x or the free element is equal to zero.

Of course, a very difficult case is possible when both of these coefficients are equal to zero: b = c = 0. In this case, the equation takes the form ax 2 = 0. Obviously, such an equation has a single root: x = 0.

Let's consider the remaining cases. Let b = 0, then we obtain an incomplete quadratic equation of the form ax 2 + c = 0. Let us transform it a little:

Since the arithmetic square root exists only of a non-negative number, the last equality makes sense only for (−c /a) ≥ 0. Conclusion:

1. If in an incomplete quadratic equation of the form ax 2 + c = 0 the inequality (−c /a) ≥ 0 is satisfied, there will be two roots. The formula is given above;
2. If (−c /a)< 0, корней нет.

As you can see, a discriminant was not required—there are no complex calculations at all in incomplete quadratic equations. In fact, it is not even necessary to remember the inequality (−c /a) ≥ 0. It is enough to express the value x 2 and see what is on the other side of the equal sign. If there is a positive number, there will be two roots. If it is negative, there will be no roots at all.

Now let's look at equations of the form ax 2 + bx = 0, in which the free element is equal to zero. Everything is simple here: there will always be two roots. It is enough to factor the polynomial:

Taking the common factor out of brackets

The product is zero when at least one of the factors is zero. This is where the roots come from. In conclusion, let’s look at a few of these equations:

Task. Solve quadratic equations:

1. x 2 − 7x = 0;
2. 5x 2 + 30 = 0;
3. 4x 2 − 9 = 0.

x 2 − 7x = 0 ⇒ x · (x − 7) = 0 ⇒ x 1 = 0; x 2 = −(−7)/1 = 7.

5x 2 + 30 = 0 ⇒ 5x 2 = −30 ⇒ x 2 = −6. There are no roots, because a square cannot be equal to a negative number.

4x 2 − 9 = 0 ⇒ 4x 2 = 9 ⇒ x 2 = 9/4 ⇒ x 1 = 3/2 = 1.5; x 2 = −1.5.

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 is an even number, that is, b = 6 or b = 2k, whence k = 3. Then let’s try to solve the equation using the formulas shown in the diagram of the figure D 1 = 3 2 – 3 · (– 6 ) = 9 + 18 = 27

√(D 1) = √27 = √(9 3) = 3√3

x 1 = (-3 - 3√3)/3 = (3 (-1 - √(3)))/3 = – 1 – √3

x 2 = (-3 + 3√3)/3 = (3 (-1 + √(3)))/3 = – 1 + √3

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

D 2 = 2 2 – 4 (– 2) = 4 + 8 = 12

√(D 2) = √12 = √(4 3) = 2√3

x 1 = (-2 - 2√3)/2 = (2 (-1 - √(3)))/2 = – 1 – √3

x 2 = (-2 + 2√3)/2 = (2 (-1+ √(3)))/2 = – 1 + √3

Answer: –1 – √3; –1 + √3.

As you can see, when solving this equation using different formulas, we received the same answer. Therefore, having thoroughly mastered the formulas shown in the diagram in Figure 1, you will always be able to solve any complete quadratic equation.

website, when copying material in full or in part, a link to the source is required.

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 a discriminant
- using Vieta's theorem (if possible).

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

$$ x_1 = \frac(8+\sqrt(145))(81), \quad x_2 = \frac(8-\sqrt(145))(81) $$ and not like this: \(x_1 = 0.247; \quad x_2 = -0.05\)

This program may be useful for high school students secondary schools in preparation for tests and exams, when testing knowledge before the Unified State Exam, for 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 quickly as possible? homework in mathematics or algebra? In this case, you can also use our programs with detailed solutions.

In this way, you can conduct your own training and/or training of your younger brothers or sisters, while the level of education in the field of solving problems increases.

If you are not familiar with the entry rules quadratic polynomial, we recommend that you familiarize yourself with them.

Rules for entering a quadratic 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 whole or fractional numbers.
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 decimals fraction can be separated from the whole by either a period or a comma.
For example, you can enter decimals like this: 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 parentheses. 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

It was discovered that some scripts necessary to solve this problem were not loaded, and the program may not work.
You may have AdBlock enabled.
In this case, disable it and refresh the page.

JavaScript is disabled in your browser.
For the solution to appear, you need to enable JavaScript.
Here are instructions on how to enable JavaScript in your browser.

Because There are a lot of people willing to solve the problem, your request has been queued.
In a few seconds the solution will appear below.
Please wait sec...

If you noticed an error in the solution, then you can write about this in the Feedback Form.
Do not forget indicate which task you decide what enter in the fields.

Our games, puzzles, emulators:

A little theory.

Quadratic equation and its roots. Incomplete quadratic equations

Each of the equations
\(-x^2+6x+1.4=0, \quad 8x^2-7x=0, \quad x^2-\frac(4)(9)=0 \)
looks like
\(ax^2+bx+c=0, \)
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 is called an equation of the form ax 2 +bx+c=0, 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 free term.

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 of x 2 is equal to 1 is called given quadratic equation. For example, the quadratic equations given are the equations
\(x^2-11x+30=0, \quad x^2-6x=0, \quad x^2-8=0 \)

If in a 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. Thus, 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.

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

Let's consider solving 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 right side and divide both sides of the equation 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 with \(b \neq 0 \) expand it left side by factors and get the equation
\(x(ax+b)=0 \Rightarrow \left\( \begin(array)(l) x=0 \\ ax+b=0 \end(array) \right. \Rightarrow \left\( \begin (array)(l) x=0 \\ x=-\frac(b)(a) \end(array) \right. \)

This means that 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 =0 is equivalent to the equation x 2 =0 and therefore has a single root 0.

Formula for the roots of a quadratic equation

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

Let's solve the quadratic equation in general view and as a result we get the formula for the roots. This formula can then be used to solve any quadratic equation.

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

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

Let's transform this equation by selecting 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 radical expression is called discriminant of a quadratic equation ax 2 +bx+c=0 (“discriminant” in Latin - discriminator). It is designated by the letter D, i.e.
\(D = b^2-4ac\)

Now, using the discriminant notation, 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, a quadratic equation can have two roots (for D > 0), one root (for D = 0) or have 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 above 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\( \begin(array)(l) x_1+x_2=-p \\ x_1 \cdot x_2=q \end(array) \right. \)

Bibliographic description: Gasanov A. R., Kuramshin A. A., Elkov A. A., Shilnenkov N. V., Ulanov D. D., Shmeleva O. V. Methods for solving quadratic equations // Young scientist. 2016. No. 6.1. P. 17-20..02.2019).



Our project is about ways to solve quadratic equations. Goal of the project: learn to solve quadratic equations in ways not included in the school curriculum. Task: find everything possible ways solving quadratic equations and learning how to use them yourself and introducing these methods to your classmates.

What are “quadratic equations”?

Quadratic equation- equation of the form ax2 + bx + c = 0, Where a, b, c- some numbers ( a ≠ 0), x- unknown.

The numbers a, b, c are called the coefficients of the quadratic equation.

• a is called the first coefficient;
• b is called the second coefficient;
• c - free member.

Who was the first to “invent” quadratic equations?

Some algebraic techniques for solving linear and quadratic equations were known 4000 years ago in Ancient Babylon. The discovery of ancient Babylonian clay tablets, dating from somewhere between 1800 and 1600 BC, provides the earliest evidence of the study of quadratic equations. The same tablets contain methods for solving certain types of quadratic equations.

The need to solve equations not only of the first, but also of the second degree, even in ancient times, was caused by the need to solve problems related to finding the areas of land plots and with excavation work of a military nature, as well as with the development of astronomy and mathematics itself.

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.

Babylonian mathematicians from about the 4th century BC. used the square's complement method to solve equations with positive roots. Around 300 BC Euclid came up with a more general geometric solution method. The first mathematician who found solutions to equations with negative roots in the form of an algebraic formula was an Indian scientist Brahmagupta(India, 7th century AD).

Brahmagupta laid out a general rule for solving quadratic equations reduced to a single canonical form:

ax2 + bx = c, a>0

The coefficients in this equation can also be negative. Brahmagupta's rule is essentially the same as ours.

Public competitions in solving difficult problems were common in India. One of the old Indian books says the following about such competitions: “As the sun eclipses the stars with its brilliance, so learned man will eclipse the glory in people's assemblies, proposing and solving algebraic problems.” Problems were often presented in poetic form.

In an algebraic treatise Al-Khwarizmi 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. ax2 = bx.

2) “Squares are equal to numbers,” i.e. ax2 = c.

3) “The roots are equal to the number,” i.e. ax2 = c.

4) “Squares and numbers are equal to roots,” i.e. ax2 + c = bx.

5) “Squares and roots are equal to the number,” i.e. ax2 + bx = c.

6) “Roots and numbers are equal to squares,” i.e. bx + c == ax2.

For Al-Khwarizmi, 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-mukabal. His decision, of course, does 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 until the 17th century, does not take into account the zero solution, probably because in specific practical it doesn't matter in tasks. When solving complete quadratic equations of Al-Khwarizmi on partial numerical examples lays out the solution rules and then their geometric proofs.

Forms for solving quadratic equations following the model of Al-Khwarizmi in Europe were first set forth in the “Book of the Abacus,” written in 1202. Italian mathematician Leonard Fibonacci. The author independently developed some new algebraic examples of solving problems and was the first in Europe to approach the introduction of negative numbers.

This book contributed to the spread of algebraic knowledge not only in Italy, but also in Germany, France and other European countries. Many problems from this book were used in almost all European textbooks of the 14th-17th centuries. General rule the solution of quadratic equations reduced to a single canonical form x2 + bх = с for all possible combinations of signs and coefficients b, c was formulated in Europe in 1544. M. Stiefel.

The derivation of the formula for solving a quadratic equation in general form is available from Viète, but Viète recognized only positive roots. Italian mathematicians Tartaglia, Cardano, Bombelli among the first in the 16th century. In addition to positive ones, negative roots are also taken into account. Only in the 17th century. thanks to the efforts Girard, Descartes, Newton and other scientists, the method of solving quadratic equations takes a modern form.

Let's look at several ways to solve quadratic equations.

Standard methods for solving quadratic equations from school curriculum:

1. Factoring the left side of the equation.
2. Method for selecting a complete square.
3. Solving quadratic equations using the formula.
4. Graphic solution quadratic equation.
5. Solving equations using Vieta's theorem.

Let us dwell in more detail on the solution of reduced and unreduced quadratic equations using Vieta’s theorem.

Recall that to solve the above quadratic equations, it is enough to find two numbers whose product is equal to the free term, and whose sum is equal to the second coefficient with the opposite sign.

Example.x 2 -5x+6=0

You need to find numbers whose product is 6 and whose sum is 5. These numbers will be 3 and 2.

Answer: x 1 =2, x 2 =3.

But you can also use this method for equations with the first coefficient not equal to one.

Example.3x 2 +2x-5=0

Take the first coefficient and multiply it by the free term: x 2 +2x-15=0

The roots of this equation will be numbers whose product is - 15 and whose sum is - 2. These numbers are 5 and 3. To find the roots original equation, divide the resulting roots by the first coefficient.

Answer: x 1 =-5/3, x 2 =1

6. Solving equations using the "throw" method.

Consider the quadratic equation ax 2 + bx + c = 0, where a≠0.

Multiplying both sides by a, we obtain the equation a 2 x 2 + abx + ac = 0.

Let ax = y, whence x = y/a; then we arrive at the equation y 2 + by + ac = 0, equivalent to the given one. We find its roots for 1 and 2 using Vieta’s theorem.

We finally get x 1 = y 1 /a and x 2 = y 2 /a.

With this method, the coefficient a is multiplied by the free term, as if “thrown” to it, which is why it is called the “throw” method. This method is used when the roots of the equation can be easily found using Vieta's theorem and, most importantly, when the discriminant is an exact square.

Example.2x 2 - 11x + 15 = 0.

Let’s “throw” the coefficient 2 to the free term and make a substitution and get the equation y 2 - 11y + 30 = 0.

According to Vieta's inverse theorem

y 1 = 5, x 1 = 5/2, x 1 = 2.5; y 2 ​​= 6, x 2 = 6/2, x 2 = 3.

Answer: x 1 =2.5; X 2 = 3.

7. Properties of coefficients of a quadratic equation.

Let the quadratic equation ax 2 + bx + c = 0, a ≠ 0 be given.

1. If a+ b + c = 0 (i.e. the sum of the coefficients of the equation is zero), then x 1 = 1.

2. If a - b + c = 0, or b = a + c, then x 1 = - 1.

Example.345x 2 - 137x - 208 = 0.

Since a + b + c = 0 (345 - 137 - 208 = 0), then x 1 = 1, x 2 = -208/345.

Answer: x 1 =1; X 2 = -208/345 .

Example.132x 2 + 247x + 115 = 0

Because a-b+c = 0 (132 - 247 +115=0), then x 1 = - 1, x 2 = - 115/132

Answer: x 1 = - 1; X 2 =- 115/132

There are other properties of the coefficients of a quadratic equation. but their use is more complex.

8. Solving quadratic equations using a nomogram.

Fig 1. Nomogram

This is an old and currently forgotten method of solving quadratic equations, placed on p. 83 of the collection: Bradis V.M. Four-digit math tables. - M., Education, 1990.

Table XXII. Nomogram for solving the equation z 2 + pz + q = 0. This nomogram allows, without solving a quadratic equation, to determine the roots of the equation from its coefficients.

The curvilinear scale of the nomogram is built according to the formulas (Fig. 1):

Believing OS = p, ED = q, OE = a(all in cm), from Fig. 1 similarities of triangles SAN And CDF we get the proportion

which, after substitutions and simplifications, yields the equation z 2 + pz + q = 0, and the letter z means the mark of any point on a curved scale.

Rice. 2 Solving quadratic equations using a nomogram

Examples.

1) For the equation z 2 - 9z + 8 = 0 the nomogram gives the roots z 1 = 8.0 and z 2 = 1.0

Answer:8.0; 1.0.

2) Using a nomogram, we solve the equation

2z 2 - 9z + 2 = 0.

Divide the coefficients of this equation by 2, we get the equation z 2 - 4.5z + 1 = 0.

The nomogram gives roots z 1 = 4 and z 2 = 0.5.

Answer: 4; 0.5.

9. Geometric method for solving quadratic equations.

Example.X 2 + 10x = 39.

In the original, this problem is formulated as follows: “The square and ten roots are equal to 39.”

Consider a square with side x, rectangles are constructed on its sides so that the other side of each of them is 2.5, therefore the area of ​​each is 2.5x. The resulting figure is then completed to a new square ABCD, adding four squares in the corners. equal square, the side of each of them is 2.5, and the area is 6.25

Rice. 3 Graphical method for solving the equation x 2 + 10x = 39

The area S of square ABCD can be represented as the sum of the areas of: the original square x 2, four rectangles (4∙2.5x = 10x) and four additional squares (6.25∙4 = 25), i.e. S = x 2 + 10x = 25. Replacing x 2 + 10x with the number 39, we get that S = 39 + 25 = 64, which means that the side of the square is ABCD, i.e. segment AB = 8. For the required side x of the original square we obtain

10. Solving equations using Bezout's theorem.

Bezout's theorem. The remainder of dividing the polynomial P(x) by the binomial x - α is equal to P(α) (that is, the value of P(x) at x = α).

If the number α is the root of the polynomial P(x), then this polynomial is divisible by x -α without a remainder.

Example.x²-4x+3=0

Р(x)= x²-4x+3, α: ±1,±3, α =1, 1-4+3=0. Divide P(x) by (x-1): (x²-4x+3)/(x-1)=x-3

x²-4x+3=(x-1)(x-3), (x-1)(x-3)=0

x-1=0; x=1, or x-3=0, x=3; Answer: x1 =2, x2 =3.

Conclusion: The ability to quickly and rationally solve quadratic equations is simply necessary for solving more complex equations, for example, fractional rational equations, equations of higher degrees, biquadratic equations, and in high school trigonometric, exponential and logarithmic equations. Having studied all the found methods for solving quadratic equations, we can advise our classmates, in addition to the standard methods, to solve by the transfer method (6) and solve equations using the property of coefficients (7), since they are more accessible to understanding.

Literature:

1. Bradis V.M. Four-digit math tables. - M., Education, 1990.
2. Algebra 8th grade: textbook for 8th grade. general education institutions Makarychev Yu. N., Mindyuk N. G., Neshkov K. I., Suvorova S. B. ed. S. A. Telyakovsky 15th ed., revised. - M.: Education, 2015
3. https://ru.wikipedia.org/wiki/%D0%9A%D0%B2%D0%B0%D0%B4%D1%80%D0%B0%D1%82%D0%BD%D0%BE%D0 %B5_%D1%83%D1%80%D0%B0%D0%B2%D0%BD%D0%B5%D0%BD%D0%B8%D0%B5
4. Glazer G.I. History of mathematics at school. Manual for teachers. / Ed. V.N. Younger. - M.: Education, 1964.

This topic may seem difficult at first due to many not so simple formulas. Not only do the quadratic equations themselves have long notations, but the roots are also found through the discriminant. In total, three new formulas are obtained. Not very easy to remember. This is possible only after solving such equations frequently. Then all the formulas will be remembered by themselves.

General view of a quadratic equation

Here we propose their explicit recording, when the largest degree is written first, and then in descending order. There are often situations when the terms are inconsistent. Then it is better to rewrite the equation in descending order of the degree of the variable.

Let us introduce some 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 designated number one.

When an equation is given, it is not clear how many roots there 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 will have no roots at all.

And until the decision is finalized, it is difficult to understand which option will appear in a particular case.

Types of recordings of quadratic equations

There may be different entries in tasks. They won't always look like general formula quadratic equation. Sometimes it will be missing some terms. What was written above is 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 terms with 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 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 - three.

Discriminant and dependence of the number of roots on its value

You need to know this number 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 number four.

After substituting the coefficient values ​​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. At negative number the roots of the quadratic equation will be missing. If it is equal to zero, there will be only one answer.

How to solve a complete quadratic equation?

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

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

Formula number five. From the same record it is clear that if the discriminant is equal to zero, then both roots will take the same values.

If solving 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 to solve an incomplete quadratic equation?

Everything is much simpler here. There is not even a need for additional formulas. And those that have already been written down for the discriminant and the unknown will not be needed.

Let's first consider incomplete equation at number two. In this equality, it is necessary to take the unknown quantity out of brackets and solve the linear equation, which will remain in brackets. The answer will have two roots. The first one is necessarily equal to zero, because there is a multiplier consisting of the variable itself. The second one will be obtained by solving a linear equation.

Incomplete equation number three is solved by moving the number from the left side of the equality to the right. Then you need to divide by the coefficient facing the unknown. All that remains is to extract the square root and remember to write it down twice with opposite signs.

Below are some steps that will 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 can cause poor grades when studying the extensive topic “Quadratic Equations (8th Grade).” Subsequently, these actions will not need to be performed constantly. Because a stable skill will appear.

• 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 a degree, and last - just a number.

• If a minus appears before the coefficient “a”, it can complicate the work for a beginner studying 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.

• It is recommended to get rid of fractions in the same way. 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 = 0;

15 − 2x − x 2 = 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 = 0. It is incomplete, therefore it is solved as described for formula number two.

After taking it out of brackets, it turns out: x (x - 7) = 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: 5x 2 + 30 = 0. Again incomplete. Only it is solved as described for the third formula.

After moving 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 the numbers: x 1 = √6, x 2 = - √6.

The third equation: 15 − 2х − x 2 = 0. Here and further, solving quadratic equations will begin with their rewriting in standard view: − x 2 − 2x + 15 = 0. Now it’s time to use the second useful advice and multiply everything by minus one. It turns out x 2 + 2x - 15 = 0. Using the fourth formula, you need to calculate the discriminant: D = 2 2 - 4 * (- 15) = 4 + 60 = 64. It is a positive number. From what is said above, it turns out that the equation has two roots. They need to be calculated using the fifth formula. It turns out that x = (-2 ± √64) / 2 = (-2 ± 8) / 2. Then x 1 = 3, x 2 = - 5.

The fourth equation x 2 + 8 + 3x = 0 is transformed into this: x 2 + 3x + 8 = 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 = -12/ (2 * 1) = -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 similar terms, first opening the brackets. In place of the first there will be the following expression: x 2 + 2x + 1. After the equality, this entry will appear: x 2 + 3x + 2. After similar terms are counted, the equation will take the form: x 2 - x = 0. It has become incomplete . Something similar to this has already been discussed a little higher. The roots of this will be the numbers 0 and 1.