Property of square roots. Online calculator

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 can be useful for high school students in general education schools when preparing for tests and exams, when testing knowledge before the Unified State Exam, and 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 in the form of a decimal, but also in the form of an ordinary fraction.

Rules for entering decimal fractions.
In decimal fractions, the fractional part 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)

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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 the 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's obvious that:
1) If D>0, then the quadratic equation has two roots.
2) If D=0, then the quadratic equation has one root \(x=-\frac(b)(2a)\).
3) If D Thus, depending on the value of the discriminant, a quadratic equation can have two roots (for D > 0), one root (for D = 0) or have no roots (for D When solving a quadratic equation using this formula, it is advisable to do the following way:
1) calculate the discriminant and compare it with zero;
2) if the discriminant is positive or equal to zero, then use the root formula; if the discriminant is negative, then write down that there are no roots.

Vieta's theorem

The given quadratic equation ax 2 -7x+10=0 has roots 2 and 5. The sum of the roots is 7, and the product is 10. We see that the sum of the roots is equal to the second coefficient taken 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\( \begin(array)(l) x_1+x_2=-p \\ x_1 \cdot x_2=q \end(array) \right. \)

Lesson 41. Property of square roots.

Make a table containing properties and examples for each property, in the process of explanation.

Lesson objectives:

* Educational:

o study the basic properties of square roots,

o develop the ability to use them to transform expressions containing square roots,

o teach how to calculate the values ​​of square roots.

* Educational:

o cultivate attentiveness, accuracy, perseverance.

* Educational:

o memory development,

o developing skills to overcome difficulties,

o development of skills in working with textbooks and reference materials.

Lesson type: combined.

Forms and methods of work:

* frontal (oral counting),

* individual work with differentiation (cards, teaching material),

* heuristic.

Lesson equipment:

* table,

* cards (4 options),

* didactic material,

* textbook (reference material on the flyleaf of the textbook).

DURING THE CLASSES

I. Organizational moment

Communicate the purpose of the lesson and lesson plan. (Explore the two properties of arithmetic square roots and learn how to use them to transform expressions containing square roots.)

In the last lessons you learned the new concept of square roots. In this regard, the need arose to study the properties of the new concept. So, the purpose of our lesson is studying the properties of the square root, as well as learning to apply these properties in calculations.

II. Verbal counting

III. Repetition

– What is called the arithmetic square root of the number a? ( A non-negative number whose square is equal to a.)

– What property of the arithmetic square root did you use?

– Calculate using the properties of degrees: https://pandia.ru/text/78/318/images/image004_99.gif" width="376" height="240">

Suggestive questions:

– Find the meanings of the expressions lion. and right. parts of equalities.

– Compare the answers. What conclusions can you draw?

Answers: 1. Square root of the sum of numbers not equal the sum of the roots of these

2. Root of the difference between numbers not equal differences of roots of given numbers

3. Root from the work equals the product of the roots of these numbers.

4.Root of a fraction equals root of the numerator divided by the root of

denominator (or the root of the quotient is equal to the quotient of the roots).

– Think about how you can write down the correct equalities in letter form.

(1). and 2).)

– If these equalities are written down in letter form, then what restrictions are imposed on the numbers..gif" width="76" height="21 src=">) - two students write on the board, the rest in notebooks slide 3

– So, we have considered two properties of square meters. roots, they are also called briefly « root of the product and root of the fraction »

–In the textbook, these properties are written in the form of theorems. Please open your textbook on pages 66 and 67 and read these two theorems.

– Apply these properties in your calculations:

V. Consolidation.

Consolidation of new material.

1. Solve problems from textbook No. 14.3(a, b), 14.5(a, b), 14.11(a, b), 14.20(a, b), 14.23(a, b) using the properties of square roots.

2. Using the definition of square root, solve the equations:

VI. Independent educational work

Summarizing.

– What properties have we studied today?

(students formulate properties)

– How to find the root of a work? (The root of the product is equal to the product of the roots.)

– How to find the root of a fraction? (The root of the numbers is divided by the root of the denominator.)

– Can there be a zero in the denominator? (No, you cannot divide by zero.)

– How to find the root of a mixed fraction? (Convert to an improper fraction and apply the arithmetic square root property.)

Giving grades in diaries.

- The lesson is over. Thank you. Goodbye.

VII. Homework: No. 14.2, 14.6, c, d), 14.23 (a, b) theory on pp. 66-69.

Determining the sum of the roots of an equation is one of the necessary steps when solving quadratic equations (equations of the form ax² + bx + c = 0, where the exponents a, b and c are arbitrary numbers, and a ? 0) with the support of Vieta’s theorem.

Instructions

1. Write the quadratic equation as ax² + bx + c = 0 Example: Initial equation: 12 + x²= 8x Correctly written equation: x² – 8x + 12 = 0

2. Apply Vieta’s theorem, according to which the sum of the roots of the equation will be equal to the number “b”, taken with the opposite sign, and their product will be equal to the number “c”. Example: In the equation under consideration, b = -8, c = 12, respectively: x1 + x2 =8×1∗x2=12

3. Find out whether the roots of the equations are correct or negative numbers. If both the product and the sum of the roots are positive numbers, all of the roots are a valid number. If the product of the roots is regular and the sum of the roots is a negative number, then both roots are negative. If the product of roots is negative, then one root has a “+” sign, and the other has a “-” sign. In this case, you need to use an additional rule: “If the sum of the roots is a positive number, the larger root in modulus is also positive, and if the sum of the roots is a negative number is a root with a larger modulus - negative.” Example: In the equation under consideration, both the sum and the product are correct numbers: 8 and 12, which means both roots are positive numbers.

4. Solve the resulting system of equations by selecting the roots. It will be more convenient to start the selection with factors, and then, to check, substitute any pair of factors into the second equation and check whether the sum of these roots corresponds to the solution. Example: x1∗x2=12 Suitable pairs of roots will be, respectively: 12 and 1, 6 and 2, 4 and 3Check the resulting pairs using the equation x1+x2=8. Pairs 12 + 1 ≠ 86 + 2 = 84 + 3 ≠ 8 Accordingly, the roots of the equation are the numbers 6 and 8.

An equation is an equality of the form f(x,y,…)=g(x,y,..), where f and g are functions of one or more variables. To discover the root of an equation means to discover a set of arguments in which this equality is satisfied.

You will need

• Knowledge of mathematical review.

Instructions

1. It is possible that you have an equation of the form: x+2=x/5. First, let’s move all the components of this equality from the right side to the left, changing the sign of the component to the opposite one. There will be a zero on the right side of this equation, that is, we get the following: x+2-x/5 = 0.

2. Let us present similar terms. We get the following: 4x/5 + 2 = 0.

3. Next, from the resulting reduced equation we will find the unknown term, in this case it is x. The resulting value of the unknown variable will be the solution to the initial equation. In this case, we get the following: x = -2.5.

Video on the topic

Note!
As a result of the solution, extra roots may appear. They will not be a solution to the initial equation, even if you solved everything positively. Be sure to check all the solutions you receive.

Helpful advice
Always check the obtained values ​​for the unknown. This can be done simply by substituting the resulting value into the initial equation. If the equality is correct, then the solution is correct.

Vieta's theorem establishes a direct connection between the roots (x1 and x2) and exponents (b and c, d) of an equation of the type bx2+cx+d=0. With the help of this theorem, it is possible, without determining the meaning of the roots, to calculate their sum, boldly speaking, in the mind. There is nothing difficult about this, the main thing is to know some rules.

You will need

• - calculator;
• – paper for notes.

Instructions

1. Bring the quadratic equation under study to a standard form, so that all exponents are in descending order, that is, first the highest degree is x2, and at the end the zero degree is x0. The equation will take the form: b*x2 + c*x1 + d*x0 = b*x2 + c*x + d = 0.

2. Check the non-negativity of the discriminant. This check is needed to make sure that the equation has roots. D (discriminant) takes the form: D = c2 – 4*b*d. There are several options here. D – discriminant – correct, which means that the equation has two roots. D is equal to zero, from this it follows that there is a root, but it is dual, that is, x1 = x2. D is negative, for a school algebra course this condition means that there are no roots, for higher mathematics there are roots, but they are complex.

3. Determine the sum of the roots of the equation. Using Vieta's theorem, this is easy to do: b*x2+c*x+d = 0. The sum of the roots of the equation is directly proportional to “–c” and inversely proportional to the exponent “b”. Namely, x1+x2 = -c/b. Determine the product of roots according to the formulation - the product of the roots of an equation is directly proportional to “d” and inversely proportional to the indicator “b”: x1*x2 = d/b.

Note!
If you receive a negative discriminant, this does not mean that there are no roots. This means that the roots of the equation are the so-called complex roots. Vieta's theorem is also applicable in this case, but its form will be slightly changed: [-c+(-i)*(-c2 + 4*b*d)0.5]/ = x1,2

Helpful advice
If you are faced not with a quadratic equation, but with a cubic or equation of degree n: b0*xn + b1*xn-1 +…..+ bn = 0, then to calculate the sum or product of the roots of the equation, you can also correctly use Vieta’s theorem :1. –b1/b0 = x1 + x2 + x3 +….+ xn,2. b2/b0 = x1*x2+….+xn-1*xn,3. (-1)n * (bn/b0) = x1*x2*x3*….*xn.

If, when substituting a number into an equation, the correct equality is obtained, such a number is called a root. Roots can be regular, negative or zero. Among each set of roots of the equation, maximum and minimum are distinguished.

Instructions

1. Find all the roots of the equation, choose the negative one among them, if there is one. Let's say we are given a quadratic equation 2x?-3x+1=0. Apply the formula for finding the roots of a quadratic equation: x(1,2)=/2=/2=/2, then x1=2, x2=1. It is easy to notice that there are no negative ones among them.

2. You can also find the roots of a quadratic equation using Vieta's theorem. According to this theorem, x1+x1=-b, x1?x2=c, where b and c are the exponents of the equation x?+bx+c=0, respectively. By applying this theorem, it is possible not to calculate the discriminant b?-4ac, which in some cases can significantly simplify the problem.

3. If in a quadratic equation the exponent at x is even, you can use not the main, but an abbreviated formula to find the roots. If the basic formula looks like x(1,2)=[-b±?(b?-4ac)]/2a, then in abbreviated form it is written as follows: x(1,2)=[-b/2±?( b?/4-ac)]/a. If there is no dummy term in a quadratic equation, it is quite easy to move x out of brackets. And occasionally the left side folds into a complete square: x?+2x+1=(x+1)?.

4. There are types of equations that give not just one number, but a whole bunch of solutions. Let's say trigonometric equations. So, the result to the equation 2sin?(2x)+5sin(2x)-3=0 will be x=?/4+?k, where k is an integer. That is, when substituting any integer value of the parameter k, the argument x will satisfy the given equation.

5. In trigonometry problems, you may need to find all negative roots or the highest of the negative ones. To solve such problems, logical reasoning or the method of mathematical induction is used. Plug some integer values ​​for k into the expression x=?/4+?k and observe how the argument works. By the way, the largest negative root in the previous equation will be x=-3?/4 with k=1.

Video on the topic

Note!
In this example, we considered a version of a quadratic equation in which a=1. In order to solve a complete quadratic equation using the same method, where a&ne 1, you need to create an auxiliary equation, bringing “a” to unity.

Helpful advice
Use this method of solving equations to quickly discover the roots. It will also help if you need to solve an equation in your head without taking notes.

I. Vieta's theorem for the reduced quadratic equation.

Sum of roots of the reduced quadratic equation x 2 +px+q=0 is equal to the second coefficient taken with the opposite sign, and the product of the roots is equal to the free term:

x 1 + x 2 = -p; x 1 ∙x 2 =q.

Find the roots of the given quadratic equation using Vieta's theorem.

Example 1) x 2 -x-30=0. This is the reduced quadratic equation ( x 2 +px+q=0), second coefficient p=-1, and the free member q=-30. First, let's make sure that this equation has roots, and that the roots (if any) will be expressed in integers. To do this, it is enough that the discriminant be a perfect square of an integer.

Finding the discriminant D=b 2 — 4ac=(-1) 2 -4∙1∙(-30)=1+120=121= 11 2 .

Now, according to Vieta’s theorem, the sum of the roots must be equal to the second coefficient taken with the opposite sign, i.e. ( -p), and the product is equal to the free term, i.e. ( q). Then:

x 1 +x 2 =1; x 1 ∙x 2 =-30. We need to choose two numbers such that their product is equal to -30 , and the amount is unit. These are numbers -5 And 6 . Answer: -5; 6.

Example 2) x 2 +6x+8=0. We have the reduced quadratic equation with the second coefficient p=6 and free member q=8. Let's make sure that there are integer roots. Let's find the discriminant D 1 D 1=3 2 -1∙8=9-8=1=1 2 . The discriminant D 1 is the perfect square of the number 1 , which means that the roots of this equation are integers. Let us select the roots using Vieta’s theorem: the sum of the roots is equal to –р=-6, and the product of the roots is equal to q=8. These are numbers -4 And -2 .

In fact: -4-2=-6=-р; -4∙(-2)=8=q. Answer: -4; -2.

Example 3) x 2 +2x-4=0. In this reduced quadratic equation, the second coefficient p=2, and the free member q=-4. Let's find the discriminant D 1, since the second coefficient is an even number. D 1=1 2 -1∙(-4)=1+4=5. The discriminant is not a perfect square of the number, so we do conclusion: The roots of this equation are not integers and cannot be found using Vieta’s theorem. This means that we solve this equation, as usual, using the formulas (in this case, using the formulas). We get:

Example 4). Write a quadratic equation using its roots if x 1 =-7, x 2 =4.

Solution. The required equation will be written in the form: x 2 +px+q=0, and, based on Vieta’s theorem –p=x 1 +x 2=-7+4=-3 → p=3; q=x 1 ∙x 2=-7∙4=-28 . Then the equation will take the form: x 2 +3x-28=0.

Example 5). Write a quadratic equation using its roots if:

II. Vieta's theorem for a complete quadratic equation ax 2 +bx+c=0.

The sum of the roots is minus b, divided by A, the product of the roots is equal to With, divided by A:

x 1 + x 2 = -b/a; x 1 ∙x 2 =c/a.

Example 6). Find the sum of the roots of a quadratic equation 2x 2 -7x-11=0.

Solution.

We make sure that this equation will have roots. To do this, it is enough to create an expression for the discriminant, and, without calculating it, just make sure that the discriminant is greater than zero. D=7 2 -4∙2∙(-11)>0 . Now let's use theorem Vieta for complete quadratic equations.

x 1 +x 2 =-b:a=- (-7):2=3,5.

Example 7). Find the product of the roots of a quadratic equation 3x 2 +8x-21=0.

Solution.

Let's find the discriminant D 1, since the second coefficient ( 8 ) is an even number. D 1=4 2 -3∙(-21)=16+63=79>0 . The quadratic equation has 2 root, according to Vieta’s theorem, the product of roots x 1 ∙x 2 =c:a=-21:3=-7.

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Determining the sum of the roots of an equation is one of the necessary steps in solving quadratic equations (equations of the form ax²- + bx + c = 0, where the coefficients a, b and c are arbitrary numbers, and a 0) using Vieta’s theorem.

Instructions

Write the quadratic equation as ax²- + bx + c = 0

Example:
Original equation: 12 + x²-= 8x
Correctly written equation: x²- - 8x + 12 = 0

Apply Vieta's theorem, according to which the sum of the roots of the equation will be equal to the number "b" taken with the opposite sign, and their product will be equal to the number "c".

Example:
In the equation under consideration, b=-8, c=12, respectively:
x1+x2=8
x1&lowast-x2=12

Find out whether the roots of equations are positive or negative numbers. If both the product and the sum of the roots are positive numbers, each of the roots is a positive number. If the product of the roots is positive and the sum of the roots is a negative number, then both are negative. If the product of roots is negative, then one root has a “+” sign, and the other has a “-” sign. In this case, you must use an additional rule: “If the sum of the roots is a positive number, the larger root is also positive, and if the sum of the roots is - a negative number is a root with a larger modulus - negative."

Example:
In the equation under consideration, both the sum and the product are positive numbers: 8 and 12, which means both roots are positive numbers.

Solve the resulting system of equations by selecting the roots. It will be more convenient to start the selection with factors, and then, to check, substitute each pair of factors into the second equation and check whether the sum of these roots corresponds to the solution.

Example:
x1&lowast-x2=12
Suitable pairs of roots will be, respectively: 12 and 1, 6 and 2, 4 and 3

Check the resulting pairs using the equation x1+x2=8. Couples
12 + 1 &ne- 8
6 + 2 = 8
4 + 3 &ne- 8

Accordingly, the roots of the equation are the numbers 6 and 8.

note

In this example, we considered a version of a quadratic equation in which a=1. In order to solve the complete quadratic equation in the same way, where a&ne 1, it is necessary to create an auxiliary equation, bringing “a” to unity.

Helpful advice

Use this method of solving equations to quickly find roots. It will also help if you need to solve an equation in your head without resorting to notes.

Everything interesting

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