The sum and product of the roots of the equation are equal. How to find the sum of the roots of an equation

can be found using multiplication. For example: 5+5+5+5+5+5=5x6. Such an expression is said to be that the sum of equal terms is folded into a product. And vice versa, if we read this equality from right to left, we find that we have expanded the sum of equal terms. Similarly, you can collapse the product of several equal factors 5x5x5x5x5x5=5 6.

That is, instead of multiplying six identical multipliers 5x5x5x5x5x5 write 5 6 and say “five to the sixth power.”

The expression 5 6 is a power of a number, where:

5 - degree base;

6 - exponent.

Actions by which the product of equal factors is reduced to a power are called raising to a power.

IN general view degree with base "a" and exponent "n" is written like this

Raising the number a to the power n means finding the product of n factors, each of which is equal to a

If the base of the degree “a” is equal to 1, then the value of the degree for any natural number n will be equal to 1. For example, 1 5 =1, 1 256 =1

If you raise the number “a” to first degree, then we get the number a itself: a 1 = a

If you raise any number to zero degree, then as a result of calculations we get one. a 0 = 1

The second and third powers of a number are considered special. They came up with names for them: the second degree is called square the number, third - cube this number.

Any number can be raised to a power - positive, negative or zero. In this case, the following rules do not apply:

When finding the power of a positive number, the result is a positive number.

When calculating zero in natural degree we get zero.

x m · x n = x m + n

for example: 7 1.7 7 - 0.9 = 7 1.7+(- 0.9) = 7 1.7 - 0.9 = 7 0.8

To divide powers with the same bases We do not change the base, but subtract the exponents:

x m / x n = x m - n , Where, m > n,

for example: 13 3.8 / 13 -0.2 = 13 (3.8 -0.2) = 13 3.6

When calculating raising a power to a power We do not change the base, but multiply the exponents by each other.

(at m ) n = y m n

for example: (2 3) 2 = 2 3 2 = 2 6

(X · y) n = x n · y m ,

for example:(2 3) 3 = 2 n 3 m,

When performing calculations according to raising a fraction to a power we raise the numerator and denominator of the fraction to a given power

(x/y)n = x n / y n

for example: (2 / 5) 3 = (2 / 5) · (2 ​​/ 5) · (2 ​​/ 5) = 2 3 / 5 3.

The sequence of calculations when working with expressions containing a degree.

When performing calculations of expressions without parentheses, but containing powers, first of all, they perform exponentiation, then multiplication and division, and only then addition and subtraction operations.

If you need to calculate an expression containing brackets, then first do the calculations in the brackets in the order indicated above, and then the remaining actions in the same order from left to right.

Very widely in practical calculations, ready-made tables of powers are used to simplify calculations.

Continuing the conversation about the power of a number, it is logical to figure out how to find the value of the power. This process is called exponentiation. In this article we will study how exponentiation is performed, while we will touch on all possible exponents - natural, integer, rational and irrational. And according to tradition, we will consider in detail solutions to examples of raising numbers to various powers.

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What does "exponentiation" mean?

Let's start by explaining what is called exponentiation. Here is the relevant definition.

Definition.

Exponentiation- this is finding the value of the power of a number.

Thus, finding the value of the power of a number a with exponent r and raising the number a to the power r are the same thing. For example, if the task is “calculate the value of the power (0.5) 5,” then it can be reformulated as follows: “Raise the number 0.5 to the power 5.”

Now you can go directly to the rules by which exponentiation is performed.

Raising a number to a natural power

In practice, equality based on is usually applied in the form . That is, when raising a number a to a fractional power m/n, first the nth root of the number a is taken, after which the resulting result is raised to an integer power m.

Let's look at solutions to examples of raising to a fractional power.

Example.

Calculate the value of the degree.

Solution.

We will show two solutions.

First way. By definition of a degree with a fractional exponent. We calculate the value of the degree under the root sign, and then extract cube root: .

Second way. By the definition of a degree with a fractional exponent and based on the properties of the roots, the following equalities are true: . Now we extract the root , finally, we raise it to an integer power .

Obviously, the obtained results of raising to a fractional power coincide.

Answer:

Note that the fractional exponent can be written as decimal or mixed number, in these cases it should be replaced by the corresponding ordinary fraction, and then raised to a power.

Example.

Calculate (44.89) 2.5.

Solution.

Let us write the exponent in the form common fraction(if necessary, see the article): . Now we perform the raising to a fractional power:

Answer:

(44,89) 2,5 =13 501,25107 .

It should also be said that raising numbers to rational powers is a rather labor-intensive process (especially when the numerator and denominator fractional indicator the degrees are sufficient big numbers), which is usually carried out using computer technology.

To conclude this point, let us dwell on raising the number zero to a fractional power. Fractional degree zero of the form we gave the following meaning: when we have , and at zero to the m/n power is not defined. So, zero in the fraction positive degree equal to zero, for example, . And zero in a fractional negative power does not make sense, for example, the expressions 0 -4.3 do not make sense.

Raising to an irrational power

Sometimes it becomes necessary to find out the value of the power of a number with an irrational exponent. At the same time, in practical purposes Usually it is enough to obtain the value of the degree up to a certain sign. Let us immediately note that this value in practice is calculated using electronic computer technology, since raising it to ir rational degree manually requires large quantity cumbersome calculations. But still we will describe in general outline the essence of the action.

To obtain an approximate value of the power of a number a with an irrational exponent, some decimal approximation of the exponent is taken and the value of the power is calculated. This value is an approximate value of the power of the number a with an irrational exponent. The more accurate the decimal approximation of a number is taken initially, the more accurate the value of the degree will be obtained in the end.

As an example, let's calculate the approximate value of the power of 2 1.174367... . Let's take the following decimal approximation irrational indicator: . Now we raise 2 to the rational power 1.17 (we described the essence of this process in the previous paragraph), we get 2 1.17 ≈2.250116. Thus, 2 1,174367... ≈2 1,17 ≈2,250116 . If we take a more accurate decimal approximation of the irrational exponent, for example, then we obtain a more accurate value of the original exponent: 2 1,174367... ≈2 1,1743 ≈2,256833 .

Bibliography.

• Vilenkin N.Ya., Zhokhov V.I., Chesnokov A.S., Shvartsburd S.I. Mathematics textbook for 5th grade. educational institutions.
• Makarychev Yu.N., Mindyuk N.G., Neshkov K.I., Suvorova S.B. Algebra: textbook for 7th grade. educational institutions.
• Makarychev Yu.N., Mindyuk N.G., Neshkov K.I., Suvorova S.B. Algebra: textbook for 8th grade. educational institutions.
• Makarychev Yu.N., Mindyuk N.G., Neshkov K.I., Suvorova S.B. Algebra: textbook for 9th grade. educational institutions.
• Kolmogorov A.N., Abramov A.M., Dudnitsyn Yu.P. and others. Algebra and the beginnings of analysis: Textbook for grades 10 - 11 of general education institutions.
• Gusev V.A., Mordkovich A.G. Mathematics (a manual for those entering technical schools).

Lesson and presentation on the topic: "Exponent with a negative exponent. Definition and examples of problem solving"

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Determination of degree with a negative exponent

Guys, we are good at raising numbers to powers.
For example: $2^4=2*2*2*2=16$  $((-3))^3=(-3)*(-3)*(-3)=27$.

We know well that any number to the zero power is equal to one. $a^0=1$, $a≠0$.
The question arises, what happens if you raise a number to a negative power? For example, what will the number $2^(-2)$ be equal to?
The first mathematicians who asked this question decided that it was not worth reinventing the wheel, and it was good that all the properties of degrees remained the same. That is, when multiplying powers with the same basis, the exponents add up.
Let's consider this case: $2^3*2^(-3)=2^(3-3)=2^0=1$.
We found that the product of such numbers should give one. The unit in the product is obtained by multiplying the reciprocal numbers, that is, $2^(-3)=\frac(1)(2^3)$.

Such reasoning led to the following definition.
Definition. If $n$ – natural number and $a≠0$, then the equality holds: $a^(-n)=\frac(1)(a^n)$.

An important identity that is often used is: $(\frac(a)(b))^(-n)=(\frac(b)(a))^n$.
In particular, $(\frac(1)(a))^(-n)=a^n$.

Examples of solutions

Example 1.
Calculate: $2^(-3)+(\frac(2)(5))^(-2)-8^(-1)$.

Solution.
Let's consider each term separately.
1. $2^(-3)=\frac(1)(2^3)=\frac(1)(2*2*2)=\frac(1)(8)$.
2. $(\frac(2)(5))^(-2)=(\frac(5)(2))^2=\frac(5^2)(2^2)=\frac(25) (4)$.
3. $8^(-1)=\frac(1)(8)$.
It remains to perform addition and subtraction operations: $\frac(1)(8)+\frac(25)(4)-\frac(1)(8)=\frac(25)(4)=6\frac(1) (4)$.
Answer: $6\frac(1)(4)$.

Example 2.
Represent a given number as a power prime number$\frac(1)(729)$.

Solution.
Obviously, $\frac(1)(729)=729^(-1)$.
But 729 is not a prime number ending in 9. It can be assumed that this number is a power of three. Consistently divide 729 by 3.
1) $\frac(729)(3)=243$;
2) $\frac(243)(3)=81$;
3) $\frac(81)(3)=27$;
4) $\frac(27)(3)=9$;
5) $\frac(9)(3)=3$;
6) $\frac(3)(3)=1$.
Six operations were performed and that means: $729=3^6$.
For our task:
$729^{-1}=(3^6)^{-1}=3^{-6}$.
Answer: $3^(-6)$.

Example 3. Express the expression as a power: $\frac(a^6*(a^(-5))^2)((a^(-3)*a^8)^(-1))$.
Solution. The first action is always performed inside parentheses, then multiplication $\frac(a^6*(a^(-5))^2)((a^(-3)*a^8)^(-1))=\frac (a^6*a^(-10))((a^5)^(-1))=\frac(a^((-4)))(a^((-5)))=a^ (-4-(-5))=a^(-4+5)=a$.
Answer: $a$.

Example 4. Prove the identity:
$(\frac(y^2 (xy^(-1)-1)^2)(x(1+x^(-1)y)^2)*\frac(y^2(x^(-2 )+y^(-2)))(x(xy^(-1)+x^(-1)y)):\frac(1-x^(-1) y)(xy^(-1 )+1)=\frac(x-y)(x+y)$.

Solution.
On the left side, we consider each factor in brackets separately.
1. $\frac(y^2(xy^(-1)-1)^2)(x(1+x^(-1)y)^2)=\frac(y^2(\frac(x )(y)-1)^2)(x(1+\frac(y)(x))^2) =\frac(y^2(\frac(x^2)(y^2)-2\ frac(x)(y)+1))(x(1+2\frac(y)(x)+\frac(y^2)(x^2)))=\frac(x^2-2xy+ y^2)(x+2y+\frac(y^2)(x))=\frac(x^2-2xy+y^2)(\frac(x^2+2xy+y^2)(x) )=\frac(x(x^2-2xy+y^2))((x^2+2xy+y^2))$.
2. $\frac(y^2(x^(-2)+y^(-2)))(x(xy^(-1)+x^(-1)y))=\frac(y^ 2(\frac(1)(x^2)+\frac(1)(y^2)))(x(\frac(x)(y)+\frac(y)(x))) =\frac (\frac(y^2)(x^2)+1)(\frac(x^2)(y)+y)=\frac(\frac(y^2+x^2)(x^2) )((\frac(x^2+y^2)(y)))=\frac(y^2+x^2)(x^2) *\frac(y)(x^2+y^2 )=\frac(y)(x^2)$.
3. $\frac(x(x^2-2xy+y^2))((x^2+2xy+y^2))*\frac(y)(x^2)=\frac(y(x ^2-2xy+y^2))(x(x^2+2xy+y^2))=\frac(y(x-y)^2)(x(x+y)^2)$.
4. Let's move on to the fraction we are dividing by.
$\frac(1-x^(-1)y)(xy^(-1)+1)=\frac(1-\frac(y)(x))(\frac(x)(y)+1 )=\frac(\frac(x-y)(x))(\frac(x+y)(y))=\frac(x-y)(x)*\frac(y)(x+y)=\frac( y(x-y))(x(x+y))$.
5. Let's do the division.
$\frac(y(x-y)^2)(x(x+y)^2):\frac(y(x-y))(x(x+y))=\frac(y(x-y)^2)( x(x+y)^2)*\frac(x(x+y))(y(x-y))=\frac(x-y)(x+y)$.
We obtained the correct identity, which was what we needed to prove.

At the end of the lesson, we will once again write down the rules for working with powers, here the exponent is an integer.
$a^s*a^t=a^(s+t)$.
$\frac(a^s)(a^t)=a^(s-t)$.
$(a^s)^t=a^(st)$.
$(ab)^s=a^s*b^s$.
$(\frac(a)(b))^s=\frac(a^s)(b^s)$.

Problems to solve independently

1. Calculate: $3^(-2)+(\frac(3)(4))^(-3)+9^(-1)$.
2. Represent the given number as a power of a prime number $\frac(1)(16384)$.
3. Express the expression as a power:
$\frac(b^(-8)*(b^3)^(-4))((b^2*b^(-7))^3)$.
4. Prove the identity:
$(\frac(b^(-m)-c^(-m))(b^(-m)+c^(-m))+\frac(b^(-m)+c^(-m ))(c^(-m)-b^(-m)))=\frac(4)(b^m c^(-m)-b^(-m)c^m) $.