Writing the roots of a quadratic equation. Methods for solving quadratic equations

Entry level

Quadratic equations. The 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 the left side and arrange the terms in descending order of powers of X

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

Example 2.

Multiply the left and right sides by:

This equation, although it was originally in it, is not 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:

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

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

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

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. You need to pay special attention to the 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:

QUADRATE EQUATIONS. MIDDLE 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 are different numbers of roots possible? Let us turn to the geometric meaning of the 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 such pairs of numbers that give in the product, and the difference of which is equal to:

and: their difference is equal - does not fit;

and: - not suitable;

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 a 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 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 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:

1. Vieta's theorem is used only in the quadratic equations given.
2. Using Vieta's theorem, you can find the roots by selection, orally.
3. If the equation is not given or no suitable pair of factors of the free term 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:

It follows: .

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

QUADRATE 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

Just. According to formulas and clear, simple rules. At the first stage

it is necessary to bring the given equation to a standard form, i.e. to the form:

If the equation is already given to you in this form, you do not need to do the first stage. The most important thing is to do it right

determine all the coefficients, A, b And c.

Formula for finding the roots of a quadratic equation.

The expression under the root sign is called discriminant . As you can see, to find X, we

we use only a, b and c. Those. coefficients from quadratic equation. Just carefully set it up

values a, b and c We calculate into this formula. We substitute with their signs!

For example, in the equation:

A =1; b = 3; c = -4.

We substitute the values ​​and write:

The example is almost solved:

This is the answer.

The most common mistakes are confusion with sign values a, b And With. Or rather, with substitution

negative values ​​into the formula for calculating the roots. A detailed recording of the formula comes to the rescue here

with specific numbers. If you have problems with calculations, do it!

Suppose we need to solve the following example:

Here a = -6; b = -5; c = -1

We describe everything in detail, carefully, without missing anything with all the signs and brackets:

Quadratic equations often look slightly different. For example, like this:

Now take note of practical techniques that dramatically reduce the number of errors.

First appointment. Don't be lazy before solving a quadratic equation bring it to standard form.

What does this mean?

Let's say that after all the transformations you get the following equation:

Don't rush to write the root formula! You'll almost certainly get the odds mixed up a, b and c.

Construct the example correctly. First, X squared, then without square, then the free term. Like this:

Get rid of the minus. How? We need to multiply the entire equation by -1. We get:

But now you can safely write down the formula for the roots, calculate the discriminant and finish solving the example.

Decide for yourself. You should now have roots 2 and -1.

Reception second. Check the roots! By Vieta's theorem.

To solve the given quadratic equations, i.e. if the coefficient

x 2 +bx+c=0,

Thenx 1 x 2 =c

x 1 +x 2 =−b

For a complete quadratic equation in which a≠1:

x 2 +bx+c=0,

divide the whole equation by A:

→ →

Where x 1 And x 2 - roots of the equation.

Reception third. If your equation has fractional coefficients, get rid of the fractions! Multiply

equation with a common denominator.

Conclusion. Practical advice:

1. Before solving, we bring the quadratic equation to standard form and build it Right.

2. If there is a negative coefficient in front of the X squared, we eliminate it by multiplying everything

equations by -1.

3. If the coefficients are fractional, we eliminate the fractions by multiplying the entire equation by the corresponding

factor.

4. If x squared is pure, its coefficient is equal to one, the solution can be easily verified by

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:

This program may be useful for high school students in 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 your math or algebra homework done as quickly as possible? 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 rules for entering a 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 as a decimal, but also as an ordinary fraction.

Rules for entering decimal fractions.
In decimal fractions, the fractional part can be separated from the whole part by either a period or a comma.
For example, you can enter decimal fractions 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: /
The whole part is separated from the fraction by the ampersand sign: &
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)

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.

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.
Don't 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 \(c \neq 0 \), move its free term to the 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 \(b \neq 0 \) factor its left side and obtain the equation
\(x(ax+b)=0 \Rightarrow \left\( \begin(array)(l) x=0 \\ ax+b=0 \end(array) \right. \Rightarrow \left\( \begin (array)(l) x=0 \\ x=-\frac(b)(a) \end(array) \right.

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 coefficients of the unknowns and the free term are nonzero.

Let us solve the quadratic equation in general form and as a result we obtain 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:
\(x^2+2x \cdot \frac(b)(2a)+\left(\frac(b)(2a)\right)^2- \left(\frac(b)(2a)\right)^ 2 + \frac(c)(a) = 0 \Rightarrow \)

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 is 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 with the opposite sign, and the product of the roots is equal to the free term. Any reduced quadratic equation that has roots has this property.

The sum of the roots of the 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\( \begin(array)(l) x_1+x_2=-p \\ x_1 \cdot x_2=q \end(array) \right. \)

Quadratic equations. Discriminant. Solution, examples.

Attention!
There are additional
materials in Special Section 555.
For those who are very "not very..."
And for those who “very much…”)

Types of quadratic equations

What is a quadratic equation? What does it look like? In term quadratic equation the keyword is "square". This means that in the equation Necessarily there must be an x ​​squared. In addition to it, the equation may (or may not!) contain just X (to the first power) and just a number (free member). And there should be no X's to a power greater than two.

In mathematical terms, a quadratic equation is an equation of the form:

Here a, b and c- some numbers. b and c- absolutely any, but A– anything other than zero. For example:

Here A =1; b = 3; c = -4

Here A =2; b = -0,5; c = 2,2

Here A =-3; b = 6; c = -18

Well, you understand...

In these quadratic equations on the left there is full set members. X squared with a coefficient A, x to the first power with coefficient b And free member s.

Such quadratic equations are called full.

What if b= 0, what do we get? We have X will be lost to the first power. This happens when multiplied by zero.) It turns out, for example:

5x 2 -25 = 0,

2x 2 -6x=0,

-x 2 +4x=0

Etc. And if both coefficients b And c are equal to zero, then it’s even simpler:

2x 2 =0,

-0.3x 2 =0

Such equations where something is missing are called incomplete quadratic equations. Which is quite logical.) Please note that x squared is present in all equations.

By the way, why A can't be equal to zero? And you substitute instead A zero.) Our X squared will disappear! The equation will become linear. And the solution is completely different...

That's all the main types of quadratic equations. Complete and incomplete.

Solving quadratic equations.

Solving complete quadratic equations.

Quadratic equations are easy to solve. According to formulas and clear, simple rules. At the first stage, it is necessary to bring the given equation to a standard form, i.e. to the form:

If the equation is already given to you in this form, you do not need to do the first stage.) The main thing is to correctly determine all the coefficients, A, b And c.

The formula for finding the roots of a quadratic equation looks like this:

The expression under the root sign is called discriminant. But more about him below. As you can see, to find X, we use only a, b and c. Those. coefficients from a quadratic equation. Just carefully substitute the values a, b and c We calculate into this formula. Let's substitute with your own signs! For example, in the equation:

A =1; b = 3; c= -4. Here we write it down:

The example is almost solved:

This is the answer.

It's very simple. And what, you think it’s impossible to make a mistake? Well, yes, how...

The most common mistakes are confusion with sign values a, b and c. Or rather, not with their signs (where to get confused?), but with the substitution of negative values ​​into the formula for calculating the roots. What helps here is a detailed recording of the formula with specific numbers. If there are problems with calculations, do that!

Suppose we need to solve the following example:

Here a = -6; b = -5; c = -1

Let's say you know that you rarely get answers the first time.

Well, don't be lazy. It will take about 30 seconds to write an extra line. And the number of errors will decrease sharply. So we write in detail, with all the brackets and signs:

It seems incredibly difficult to write out so carefully. But it only seems so. Give it a try. Well, or choose. What's better, fast or right? Besides, I will make you happy. After a while, there will be no need to write everything down so carefully. It will work out right on its own. Especially if you use practical techniques that are described below. This evil example with a bunch of minuses can be solved easily and without errors!

But, often, quadratic equations look slightly different. For example, like this:

Did you recognize it?) Yes! This incomplete quadratic equations.

Solving incomplete quadratic equations.

They can also be solved using a general formula. You just need to understand correctly what they are equal to here. a, b and c.

Have you figured it out? In the first example a = 1; b = -4; A c? It's not there at all! Well yes, that's right. In mathematics this means that c = 0 ! That's it. Substitute zero into the formula instead c, and we will succeed. Same with the second example. Only we don’t have zero here With, A b !

But incomplete quadratic equations can be solved much more simply. Without any formulas. Let's consider the first incomplete equation. What can you do on the left side? You can take X out of brackets! Let's take it out.

So what of this? And the fact that the product equals zero if and only if any of the factors equals zero! Don't believe me? Okay, then come up with two non-zero numbers that, when multiplied, will give zero!
Doesn't work? That's it...
Therefore, we can confidently write: x 1 = 0, x 2 = 4.

All. These will be the roots of our equation. Both are suitable. When substituting any of them into the original equation, we get the correct identity 0 = 0. As you can see, the solution is much simpler than using the general formula. Let me note, by the way, which X will be the first and which will be the second - absolutely indifferent. It is convenient to write in order, x 1- what is smaller and x 2- that which is greater.

The second equation can also be solved simply. Move 9 to the right side. We get:

All that remains is to extract the root from 9, and that’s it. It will turn out:

Also two roots . x 1 = -3, x 2 = 3.

This is how all incomplete quadratic equations are solved. Either by placing X out of brackets, or by simply moving the number to the right and then extracting the root.
It is extremely difficult to confuse these techniques. Simply because in the first case you will have to extract the root of X, which is somehow incomprehensible, and in the second case there is nothing to take out of brackets...

Discriminant. Discriminant formula.

Magic word discriminant ! Rarely a high school student has not heard this word! The phrase “we solve through a discriminant” inspires confidence and reassurance. Because there is no need to expect tricks from the discriminant! It is simple and trouble-free to use.) I remind you of the most general formula for solving any quadratic equations:

The expression under the root sign is called a discriminant. Typically the discriminant is denoted by the letter D. Discriminant formula:

D = b 2 - 4ac

And what is so remarkable about this expression? Why did it deserve a special name? What the meaning of the discriminant? After all -b, or 2a in this formula they don’t specifically call it anything... Letters and letters.

Here's the thing. When solving a quadratic equation using this formula, it is possible only three cases.

1. The discriminant is positive. This means the root can be extracted from it. Whether the root is extracted well or poorly is another question. What is important is what is extracted in principle. Then your quadratic equation has two roots. Two different solutions.

2. The discriminant is zero. Then you will have one solution. Since adding or subtracting zero in the numerator does not change anything. Strictly speaking, this is not one root, but two identical. But, in a simplified version, it is customary to talk about one solution.

3. The discriminant is negative. The square root of a negative number cannot be taken. Oh well. This means there are no solutions.

Honestly speaking, when simple solution quadratic equations, the concept of a discriminant is not particularly required. We substitute the values ​​of the coefficients into the formula and count. Everything happens there by itself, two roots, one, and none. However, when solving more complex tasks, without knowledge meaning and formula of the discriminant can't get by. Especially in equations with parameters. Such equations are aerobatics for the State Examination and the Unified State Examination!)

So, how to solve quadratic equations through the discriminant you remembered. Or you learned, which is also not bad.) You know how to correctly determine a, b and c. Do you know how? attentively substitute them into the root formula and attentively count the result. You understand that the key word here is attentively?

Now take note of practical techniques that dramatically reduce the number of errors. The same ones that are due to inattention... For which it later becomes painful and offensive...

First appointment . Don’t be lazy before solving a quadratic equation and bring it to standard form. What does this mean?
Let's say that after all the transformations you get the following equation:

Don't rush to write the root formula! You'll almost certainly get the odds mixed up a, b and c. Construct the example correctly. First, X squared, then without square, then the free term. Like this:

And again, don’t rush! A minus in front of an X squared can really upset you. It's easy to forget... Get rid of the minus. How? Yes, as taught in the previous topic! We need to multiply the entire equation by -1. We get:

But now you can safely write down the formula for the roots, calculate the discriminant and finish solving the example. Decide for yourself. You should now have roots 2 and -1.

Reception second. Check the roots! According to Vieta's theorem. Don't be scared, I'll explain everything! Checking last equation. Those. the one we used to write down the root formula. If (as in this example) the coefficient a = 1, checking the roots is easy. It is enough to multiply them. The result should be a free member, i.e. in our case -2. Please note, not 2, but -2! Free member with your sign . If it doesn’t work out, it means you’ve already screwed up somewhere. Look for the error.

If it works, you need to add the roots. Last and final check. The coefficient should be b With opposite familiar. In our case -1+2 = +1. A coefficient b, which is before the X, is equal to -1. So, everything is correct!
It’s a pity that this is so simple only for examples where x squared is pure, with a coefficient a = 1. But at least check in such equations! There will be fewer and fewer errors.

Reception third . If your equation has fractional coefficients, get rid of the fractions! Multiply the equation by a common denominator as described in the lesson "How to solve equations? Identity transformations." When working with fractions, errors keep creeping in for some reason...

By the way, I promised to simplify the evil example with a bunch of minuses. Please! Here he is.

In order not to get confused by the minuses, we multiply the equation by -1. We get:

That's it! Solving is a pleasure!

So, let's summarize the topic.

Practical tips:

1. Before solving, we bring the quadratic equation to standard form and build it Right.

2. If there is a negative coefficient in front of the X squared, we eliminate it by multiplying the entire equation by -1.

3. If the coefficients are fractional, we eliminate the fractions by multiplying the entire equation by the corresponding factor.

4. If x squared is pure, its coefficient is equal to one, the solution can be easily verified using Vieta’s theorem. Do it!

Now we can decide.)

Solve equations:

8x 2 - 6x + 1 = 0

x 2 + 3x + 8 = 0

x 2 - 4x + 4 = 0

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

Answers (in disarray):

x 1 = 0
x 2 = 5

x 1.2 =2

x 1 = 2
x 2 = -0.5

x - any number

x 1 = -3
x 2 = 3

no solutions

x 1 = 0.25
x 2 = 0.5

Does everything fit? Great! Quadratic equations are not your headache. The first three worked, but the rest didn’t? Then the problem is not with quadratic equations. The problem is in identical transformations of equations. Take a look at the link, it's helpful.

Doesn't quite work out? Or does it not work out at all? Then Section 555 will help you. All these examples are broken down there. Shown main errors in the solution. Of course, we also talk about the use of identical transformations in solving various equations. Helps a lot!

If you like this site...

By the way, I have a couple more interesting sites for you.)

You can practice solving examples and find out your level. Testing with instant verification. Let's learn - with interest!)

You can get acquainted with functions and derivatives.

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. Project goal: learn to solve quadratic equations in ways not included in the school curriculum. Task: find all possible ways to solve quadratic equations and learn how to use them yourself and introduce 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 the high level of development of algebra in Babylon, the cuneiform texts do not contain 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 outshines the stars with its brilliance, so a learned man will outshine 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 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, Al-Khwarizmi sets out the rules of solution 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 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 Vieth, but Vieth 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 the 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. Graphical solution of a 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 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 you can easily find the roots of the equation 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 the replacement, we get the equation y 2 - 11y + 30 = 0.

According to converse of the theorem Vieta

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 essential for solving more complex equations, such as fractional rational equations, higher power equations, 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.