Quadratic equations. Comprehensive guide (2019)

In the term “quadratic equation,” the key word is “quadratic.” This means that the equation must necessarily contain a variable (that same x) squared, and there should not be xes to the third (or greater) power.

The solution of many equations comes down to solving exactly quadratic equations.

Let's learn to determine that this is a quadratic equation and not some other equation.

Example 1.

Let's get rid of the denominator and multiply each term of the equation by

Let's move everything to left side and arrange the terms in descending order of powers of x

Now we can say with confidence that this equation is quadratic!

Example 2.

Multiply the left and right sides by:

This equation, although it was originally in it, is not quadratic!

Example 3.

Let's multiply everything by:

Scary? The fourth and second degrees... However, if we make a replacement, we will see that we have a simple quadratic equation:

Example 4.

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

See, it's reduced - and now it's a simple linear equation!

Now try to determine for yourself which of the following equations are quadratic and which are not:

Examples:

Answers:

square;
square;
not square;
not square;
not square;
square;
not square;
square.

Mathematicians conventionally divide all quadratic equations into the following types:

Complete quadratic equations- equations in which the coefficients and, as well as the free term c, are not equal to zero (as in the example). In addition, among complete quadratic equations there are given- these are equations in which the coefficient (the equation from example one is not only complete, but also reduced!)

Incomplete quadratic equations- equations in which the coefficient and or the free term c are equal to zero:
They are incomplete because they are missing some element. But the equation must always contain x squared!!! Otherwise, it will no longer be a quadratic equation, but some other equation.

Why did they come up with such a division? It would seem that there is an X squared, and okay. This division is determined by the solution methods. Let's look at each of them in more detail.

Solving incomplete quadratic equations

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

There are types of incomplete quadratic equations:

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

1. i. Since we know how to take the square root, let's express from this equation

The expression can be either negative or positive. A squared number cannot be negative, because when multiplying two negative or two positive numbers, the result will always be a positive number, so: if, then the equation has no solutions.

And if, then we get two roots. There is no need to memorize these formulas. The main thing is that you must know and always remember that it cannot be less.

Let's try to solve some examples.

Example 5:

Solve the equation

Now all that remains is to extract the root from the left and right sides. After all, you remember how to extract roots?

Answer:

Never forget about roots with a negative sign!!!

Example 6:

Solve the equation

Answer:

Example 7:

Solve the equation

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

no roots!

For such equations that have no roots, mathematicians came up with a special icon - (empty set). And the answer can be written like this:

Answer:

Thus, this quadratic equation has two roots. There are no restrictions here, since we did not extract the root.
Example 8:

Solve the equation

Let's take the common factor out of brackets:

Thus,

This equation has two roots.

Answer:

The simplest type of incomplete quadratic equations (although they are all simple, right?). Obviously, this equation always has only one root:

We will dispense with examples here.

Solving complete quadratic equations

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

Solving complete quadratic equations is a little more difficult (just a little) than these.

Remember, Any quadratic equation can be solved using a discriminant! Even incomplete.

The other methods will help you do it faster, but if you have problems with quadratic equations, first master the solution using the discriminant.

1. Solving quadratic equations using a discriminant.

Solving quadratic equations using this method is very simple; the main thing is to remember the sequence of actions and a couple of formulas.

If, then the equation has a root. Special attention take a step. Discriminant () tells us the number of roots of the equation.

If, then the formula in the step will be reduced to. Thus, the equation will only have a root.
If, then we will not be able to extract the root of the discriminant at the step. This indicates that the equation has no roots.

Let's go back to our equations and look at some examples.

Example 9:

Solve the equation

Step 1 we skip.

Step 2.

We find the discriminant:

This means the equation has two roots.

Step 3.

Answer:

Example 10:

Solve the equation

The equation is presented in standard form, so Step 1 we skip.

Step 2.

We find the discriminant:

This means that the equation has one root.

Answer:

Example 11:

Solve the equation

The equation is presented in standard form, so Step 1 we skip.

Step 2.

We find the discriminant:

This means we will not be able to extract the root of the discriminant. There are no roots of the equation.

Now we know how to correctly write down such answers.

Answer: no roots

2. Solving quadratic equations using Vieta's theorem.

If you remember, there is a type of equation that is called reduced (when the coefficient a is equal to):

Such equations are very easy to solve using Vieta’s theorem:

Sum of roots given quadratic equation is equal, and the product of the roots is equal.

Example 12:

Solve the equation

This equation can be solved using Vieta's theorem because .

The sum of the roots of the equation is equal, i.e. we get the first equation:

And the product is equal to:

Let's compose and solve the system:

And. The amount is equal to;
And. The amount is equal to;
And. The amount is equal.

and are the solution to the system:

Answer: ; .

Example 13:

Solve the equation

Answer:

Example 14:

Solve the equation

The equation is given, which means:

Answer:

QUADRATIC EQUATIONS. AVERAGE LEVEL

What is a quadratic equation?

In other words, a quadratic equation is an equation of the form, where - the unknown, - some numbers, and.

The number is called the highest or first coefficient quadratic equation, - second coefficient, A - free member.

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

In this case, and can be equal to zero. In this chair equation is called incomplete. If all the terms are in place, that is, the equation is complete.

Solutions to various types of quadratic equations

Methods for solving incomplete quadratic equations:

First, let's look at methods for solving incomplete quadratic equations - they are simpler.

We can distinguish the following types of equations:

I., in this equation the coefficient and the free term are equal.

II. , in this equation the coefficient is equal.

III. , in this equation the free term is equal to.

Now let's look at the solution to each of these subtypes.

Obviously, this equation always has only one root:

A squared number cannot be negative, because when you multiply two negative or two positive numbers, the result will always be a positive number. That's why:

if, then the equation has no solutions;

if we have two roots

There is no need to memorize these formulas. The main thing to remember is that it cannot be less.

Examples:

Solutions:

Answer:

Never forget about roots with a negative sign!

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

no roots.

To briefly write down that a problem has no solutions, we use the empty set icon.

Answer:

So, this equation has two roots: and.

Answer:

Let's take the common factor out of brackets:

The product is equal to zero if at least one of the factors is equal to zero. This means that the equation has a solution when:

So, this quadratic equation has two roots: and.

Example:

Solve the equation.

Solution:

Let's factor the left side of the equation and find the roots:

Answer:

Methods for solving complete quadratic equations:

1. Discriminant

Solving quadratic equations this way is easy, the main thing is to remember the sequence of actions and a couple of formulas. Remember, any quadratic equation can be solved using a discriminant! Even incomplete.

Did you notice the root from the discriminant in the formula for roots? But the discriminant can be negative. What to do? We need to pay special attention to step 2. The discriminant tells us the number of roots of the equation.

If, then the equation has roots:
If, then the equation has the same roots, and in fact, one root:
Such roots are called double roots.
If, then the root of the discriminant is not extracted. This indicates that the equation has no roots.

Why is it possible different quantities roots? Let's turn to geometric sense quadratic equation. The graph of the function is a parabola:

In a special case, which is a quadratic equation, . This means that the roots of a quadratic equation are the points of intersection with the abscissa axis (axis). A parabola may not intersect the axis at all, or may intersect it at one (when the vertex of the parabola lies on the axis) or two points.

In addition, the coefficient is responsible for the direction of the branches of the parabola. If, then the branches of the parabola are directed upward, and if, then downward.

Examples:

Solutions:

Answer:

Answer: .

Answer:

This means there are no solutions.

Answer: .

2. Vieta's theorem

It is very easy to use Vieta's theorem: you just need to choose a pair of numbers whose product is equal to the free term of the equation, and the sum is equal to the second coefficient taken with the opposite sign.

It is important to remember that Vieta's theorem can only be applied in reduced quadratic equations ().

Let's look at a few examples:

Example #1:

Solve the equation.

Solution:

This equation can be solved using Vieta's theorem because . Other coefficients: ; .

The sum of the roots of the equation is:

And the product is equal to:

Let's select pairs of numbers whose product is equal and check whether their sum is equal:

And. The amount is equal to;
And. The amount is equal to;
And. The amount is equal.

and are the solution to the system:

Thus, and are the roots of our equation.

Answer: ; .

Example #2:

Solution:

Let's select pairs of numbers that give in the product, and then check whether their sum is equal:

and: they give in total.

and: they give in total. To obtain, it is enough to simply change the signs of the supposed roots: and, after all, the product.

Answer:

Example #3:

Solution:

The free term of the equation is negative, and therefore the product of the roots is a negative number. This is only possible if one of the roots is negative and the other is positive. Therefore the sum of the roots is equal to differences of their modules.

Let us select pairs of numbers that give in the product, and whose difference is equal to:

and: their difference is equal - does not fit;

and: - not suitable;

and: - suitable. All that remains is to remember that one of the roots is negative. Since their sum must be equal, the root with the smaller modulus must be negative: . We check:

Answer:

Example #4:

Solve the equation.

Solution:

The equation is given, which means:

The free term is negative, and therefore the product of the roots is negative. And this is only possible when one root of the equation is negative and the other is positive.

Let's select pairs of numbers whose product is equal, and then determine which roots should have a negative sign:

Obviously, only the roots and are suitable for the first condition:

Answer:

Example #5:

Solve the equation.

Solution:

The equation is given, which means:

The sum of the roots is negative, which means that, according to at least, one of the roots is negative. But since their product is positive, it means both roots have a minus sign.

Let us select pairs of numbers whose product is equal to:

Obviously, the roots are the numbers and.

Answer:

Agree, it’s very convenient to come up with roots orally, instead of counting this nasty discriminant. Try to use Vieta's theorem as often as possible.

But Vieta’s theorem is needed in order to facilitate and speed up finding the roots. In order for you to benefit from using it, you must bring the actions to automaticity. And for this, solve five more examples. But don't cheat: you can't use a discriminant! Only Vieta's theorem:

Solutions to tasks for independent work:

Task 1. ((x)^(2))-8x+12=0

According to Vieta's theorem:

As usual, we start the selection with the piece:

Not suitable because the amount;

: the amount is just what you need.

Answer: ; .

Task 2.

And again our favorite Vieta theorem: the sum must be equal, and the product must be equal.

But since it must be not, but, we change the signs of the roots: and (in total).

Answer: ; .

Task 3.

Hmm... Where is that?

You need to move all the terms into one part:

The sum of the roots is equal to the product.

Okay, stop! The equation is not given. But Vieta's theorem is applicable only in the given equations. So first you need to give an equation. If you can’t lead, give up this idea and solve it in another way (for example, through a discriminant). Let me remind you that to give a quadratic equation means to make the leading coefficient equal:

Great. Then the sum of the roots is equal to and the product.

Here it’s as easy as shelling pears to choose: after all, it’s a prime number (sorry for the tautology).

Answer: ; .

Task 4.

The free member is negative. What's special about this? And the fact is that the roots will have different signs. And now, during the selection, we check not the sum of the roots, but the difference in their modules: this difference is equal, but a product.

So, the roots are equal to and, but one of them is minus. Vieta's theorem tells us that the sum of the roots is equal to the second coefficient with the opposite sign, that is. This means that the smaller root will have a minus: and, since.

Answer: ; .

Task 5.

What should you do first? That's right, give the equation:

Again: we select the factors of the number, and their difference should be equal to:

The roots are equal to and, but one of them is minus. Which? Their sum should be equal, which means that the minus will have a larger root.

Answer: ; .

Let me summarize:

Vieta's theorem is used only in the quadratic equations given.
Using Vieta's theorem, you can find the roots by selection, orally.
If the equation is not given or no suitable pair of factors of the free term is found, then there are no whole roots, and you need to solve it in another way (for example, through a discriminant).

3. Method for selecting a complete square

If all terms containing the unknown are represented in the form of terms from abbreviated multiplication formulas - the square of the sum or difference - then after replacing variables, the equation can be presented in the form of an incomplete quadratic equation of the type.

For example:

Example 1:

Solve the equation: .

Solution:

Answer:

Example 2:

Solve the equation: .

Solution:

Answer:

IN general view the transformation will look like this:

This implies: .

Doesn't remind you of anything? This is a discriminatory thing! That's exactly how we got the discriminant formula.

QUADRATIC EQUATIONS. BRIEFLY ABOUT THE MAIN THINGS

Quadratic equation- this is an equation of the form, where - the unknown, - the coefficients of the quadratic equation, - the free term.

Complete quadratic equation- an equation in which the coefficients are not equal to zero.

Reduced quadratic equation- an equation in which the coefficient, that is: .

Incomplete quadratic equation- an equation in which the coefficient and or the free term c are equal to zero:

if the coefficient, the equation looks like: ,
if there is a free term, the equation has the form: ,
if and, the equation looks like: .

1. Algorithm for solving incomplete quadratic equations

1.1. An incomplete quadratic equation of the form, where, :

1) Let's express the unknown: ,

2) Check the sign of the expression:

if, then the equation has no solutions,
if, then the equation has two roots.

1.2. An incomplete quadratic equation of the form, where, :

1) Let’s take the common factor out of brackets: ,

2) The product is equal to zero if at least one of the factors is equal to zero. Therefore, the equation has two roots:

1.3. An incomplete quadratic equation of the form, where:

This equation always has only one root: .

2. Algorithm for solving complete quadratic equations of the form where

2.1. Solution using discriminant

1) Let's bring the equation to standard form: ,

2) Let's calculate the discriminant using the formula: , which indicates the number of roots of the equation:

3) Find the roots of the equation:

if, then the equation has roots, which are found by the formula:
if, then the equation has a root, which is found by the formula:
if, then the equation has no roots.

2.2. Solution using Vieta's theorem

The sum of the roots of the reduced quadratic equation (equation of the form where) is equal, and the product of the roots is equal, i.e. , A.

2.3. Solution by the method of selecting a complete square

Quadratic equation problems are studied both in the 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 with the 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

By known formulas finding the roots of a 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 the formulas for solving quadratic equations well; they are often needed when calculating in different tasks and sciences.

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 in ancient times was caused by the need to solve problems related to finding areas land plots and with earthworks 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 his glory in public assemblies by 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 consumption negative numbers, the terms of each of these equations are addends, 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, Al-Khwarizmi sets out the rules for solving them using particular numerical examples, 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 solving problems and was the first in Europe to introduce 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:

Factoring the left side of the equation.
Method for selecting a complete square.
Solving quadratic equations using the formula.
Graphic solution quadratic equation.
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 equal to - 15, and whose sum is equal to - 2. These numbers are 5 and 3. To find the roots of the 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 supplemented to a new square ABCD, building four equal squares in the corners, 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 to solve more complex equations, for example, fractional rational equations, equations 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:

Bradis V.M. Four-digit math tables. - M., Education, 1990.
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
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
Glazer G.I. History of mathematics at school. Manual for teachers. / Ed. V.N. Younger. - M.: Education, 1964.

It is known that it is a particular version of the equality ax 2 + bx + c = 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, b ≠ o or vice versa. We almost remembered the definition of a quadratic equation.

The second degree trinomial is zero. Its first coefficient a ≠ o, b and c can take any values. The value of the variable x will then be when substitution turns it into a correct numerical equality. Let's focus on real roots, although the equations can also be solutions. It is customary to call an equation complete in which none of the coefficients is equal to o, a ≠ o, b ≠ o, c ≠ o.
Let's solve an example. 2x 2 -9x-5 = oh, we find
D = 81+40 = 121,
D is positive, which means 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

Using the discriminant, you can solve any equation on the left side of which there is a known quadratic trinomial for a ≠ o. In our example. 2x 2 -9x-5 = 0 (ax 2 +in+s = o)

Let's consider what incomplete equations of the second degree are

ax 2 +in = o. The free term, the coefficient c at x 0, is equal to zero here, in ≠ o.
How to solve an incomplete quadratic equation of this type? Let's take x out of brackets. Let's remember when the product of two factors is equal to zero.
x(ax+b) = o, this can be when x = o or when ax+b = o.
Having solved the 2nd we have x = -в/а.
As a result, we have roots x 1 = 0, according to calculations x 2 = -b/a.
Now the coefficient of x is equal to o, and c is not equal (≠) o.
x 2 +c = o. Let's move c to the right side of the equality, we get x 2 = -с. 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.
The last option: b = c = o, that is, ax 2 = o. Naturally, such a simple equation has one root, x = o.

Special cases

We looked at how to solve an incomplete quadratic equation, and now let’s take any types.

In a complete 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 = k 2 - ac, the roots are calculated as x 1,2 = (-k±√(D/4))/a for D › o.
x = -k/a at D = o.
There are no roots for D ‹ o.
There are given quadratic equations, when the coefficient of x squared is equal to 1, they are usually written x 2 + рх + q = o. All the above formulas apply to them, but the calculations are somewhat simpler.
Example, x 2 -4x-9 = 0. Calculate D: 2 2 +9, D = 13.
x 1 = 2+√13, x 2 = 2-√13.
In addition, it is easy to apply to those given. It says that the sum of the roots of the equation is equal to -p, the second coefficient with a minus (meaning opposite sign), and the product of these same roots will be equal to q, the free term. See how easy it would be to determine the roots of this equation verbally. For unreduced coefficients (for all coefficients not equal to zero), this theorem is applicable as follows: the sum x 1 + x 2 is equal to -b/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 (easy to prove), the first one is necessarily equal to -1, and the second -c/a, if it exists. You can check how to solve an incomplete quadratic equation yourself. As easy as pie. The coefficients may be in certain relationships with each other

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

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

5x (x - 4) = 0

5 x = 0 or x - 4 = 0

x = ± √ 25/4

Having learned to solve equations of the first degree, of course, you want to work with others, in particular, with equations of the second degree, which are otherwise called quadratic.

Quadratic equations are equations like ax² + bx + c = 0, where the variable is x, the numbers are a, b, c, where a is not equal to zero.

If in a quadratic equation one or the other coefficient (c or b) is equal to zero, then this equation will be classified as an incomplete quadratic equation.

How to solve an incomplete quadratic equation if students have so far only been able to solve equations of the first degree? Consider incomplete quadratic equations different types and simple ways to solve them.

a) If coefficient c is equal to 0, and coefficient b is not equal to zero, then ax ² + bx + 0 = 0 is reduced to an equation of the form ax ² + bx = 0.

To solve such an equation, you need to know the formula for solving an incomplete quadratic equation, which consists in factoring the left side of it and later using the condition that the product is equal to zero.

For example, 5x² - 20x = 0. We factor the left side of the equation, while performing the usual mathematical operation: taking the common factor out of brackets

5x (x - 4) = 0

We use the condition that the products are equal to zero.

5 x = 0 or x - 4 = 0

The answer will be: the first root is 0; the second root is 4.

b) If b = 0, and the free term is not equal to zero, then the equation ax ² + 0x + c = 0 is reduced to an equation of the form ax ² + c = 0. The equations are solved in two ways: a) by factoring the polynomial of the equation on the left side ; b) using the properties of arithmetic square root. Such an equation can be solved using one of the methods, for example:

x = ± √ 25/4

x = ± 5/2. The answer will be: the first root is 5/2; the second root is equal to - 5/2.

c) If b is equal to 0 and c is equal to 0, then ax ² + 0 + 0 = 0 is reduced to an equation of the form ax ² = 0. In such an equation x will be equal to 0.

As you can see, incomplete quadratic equations can have no more than two roots.