Advice 1: How to multiply degrees

In mathematics there is such a thing as "degree". The degree is the product of several equal factors. Have much have the base equal to one of these factors. There is also a measure of the degree to which built one of these factors. For example, 23 = 2 * 2 * 2 = 16, where 2 is the base and 3 is the rate. There are various simplifications in the multiplication of degrees between them. It is proposed this instruction.
The rules for the multiplication of degrees
Instruction
1
Define degree what kind of are multiplied. If the members of such works have the same base classes, and the exponents vary, for example, 22 * 23 , the result will be the base of power with the same base members works by degrees raised to the exponent equal to the sum of all the multiplied degrees.

That is

22 * 23 = 22⁺3 = 2⁵ = 32
2
If the members of a work degrees have different base classes, and the exponents are the same, for example, 23 * 53 , then the result will be the product of reason these degrees raised to the exponent, equal this same exponent.

That is

23 * 53 = (2*5)3 = 103 = 1000
3
If you multiply the degrees are equal, for example, 53 * 53 the result will be a power with base equal to the same grounds of degrees, raised to the exponent, equal to the amount of degrees multiplied by the number of these same degrees.

That is

53 * 53 = (53)2 = 53*2 = 5⁶ = 15625

Or another example with the same result:

52 * 52 * 52 = (52)3 = 52*3 = 5⁶ = 15625

Advice 2 : How to solve roots

Solve roots, or irrational equations, is taught in 8th grade. As a rule, the basic method for finding the solution in this case is the method of squaring.
how to solve roots
Instruction
1
Irrational equations should be rational in order to find the answer, solving it the traditional way. However, in addition to squaring then add one more action: discard the extraneous root. This concept is connected with the irrationality of the roots, i.e. the solution to an equation, substitution of which leads to meaninglessness, for example, the root of a negative number.
2
Consider a simple example: √(2•x + 1) = 3. Lift both sides of the square:2•x + 1 = 9 → x = 4.
3
It turns out that x=4 is the root and simultaneously the normal equations 2•x + 1 = 9 and the source of the irrational √(2•x + 1) = 3. Unfortunately, this is not always easy. Sometimes the method of squaring leads to absurdity, for example:√(2•x - 5) = √(4•x - 7)
4
It would seem, you just need to build both parts in the second degree and all, a solution is found. In reality, however, we get the following:2•x – 5 = 4•x – 7 → -2•x = -2 → x=1.The substitute found the root of the original equation:√(-3) = √(-3).x=1 and is called an extraneous root rational equations that has no other roots.
5
Now a more complicated example: √(2•x2 + 5•x - 2) = x – 6 ↑22•x2 + 5•x – 2 = x2 – 12•x + 36x2 + 17•x – 38 = 0
6
Solve ordinary quadratic equation:D = 289 + 152 = 441x1 = (-17 + 21)/2 = 2; x2 = (-17 - 21)/2 = -19.
7
Substitute x1 and x2 into the original equation to avoid extraneous roots:√(2•22 + 5•2 - 2) = 2 – 6 → √16 = -4;√(2•(-19)2 - 5•19 - 2) = -19 – 6 → √625 = -25.This decision is wrong, therefore, the equation, like the previous one, has no roots.
8
Example with variable change.Sometimes, the simple construction of both parts of the equation in the square does not release from the roots. In this case, you can use the method of substitution:√(x2 + 1) + √(x2 + 4) = 3 [y2 = x2 + 1]y + √(y2 + 3) = 3 → √(y2 + 3) = 3 – y ^ a 2
9
y2 + 3 = 9 – 6•and + y26•y = 6 → y=1.x2 + 1 = 1 → x=0.
10
Check the result:√(02 + 1) + √(02 + 4) = 1 + 2 = 3 – equality are met, then the root x=0 is a valid solution of irrational equations.
Is the advice useful?
Search