√ doesn't make sense, can you please anwser my question(s)?

Question

1) Why does √2²*7³= 2*7*√7?????


2) Using the same method.
√2^5*3²*7³=
doesn't make sense
3*7√2³*√7=21√56 this would be the final anwser, please explain

4) √20√18=2√5*3√2
Why can't i use 4*5 and 9*2???

What is the prime factor of 784?

What is a prime factor?

PLEASE IM SO CONFUSED MATH DOESN'T EVEN MAKES SENSE, THE √ ESPICALLY

Answer 1

The reason is that:
(a) √(a*b) = √(a) * √(b), (the square root of a product is the same as the product of square roots)
(b) √(a^2) = a (square root and squared are inverse functions)

As a result of these, any even exponent can be halved and moved out of the square root. For example:

√(a^11) = a^5 * √(a)

So, to answer your first question:

(1)
√(2^2 * 7^3) =
√(2^2 * 7^2 * 7) =
√(2^2) * √(7^2) * √(7) =
2 * 7 * √7

(2)
√(2^5 * 3^2 * 7^3) =
2^2 * 3 * 7 * √(2*7) =
84 * √14

That should be the final answer. 21√56 is close, but not quite right (there's a factor of two that can be moved out of the square root, and another factor of two missing)

(4)
√20√18 = 2√5*3√2

You *do* use 4*5 and 9*2 when pulling apart the square roots, but 4 is a perfect square and √4 can be simplified as 2, while √9 can be simplified as 3.

(5) Prime factor of 784

Prime factors are factors which are prime numbers (i.e., cannot be factored any further). To break down a number into its prime factorization, divide it up until it can't be divided any further.

784 is even, so it divides by 2.
That leaves 392 which is also divisible by 2.
That leaves 196 which is also divisible by 2.
That leaves 98 which is also divisible by 2.
That leaves 49 which is 7 * 7.

Both 2 and 7 are prime, so that is as far as the number can be broken down.

784 = 2 * 2 * 2 * 2 * 7 * 7
784 = 2^4 * 7^2

Answer 2

Question 1
√(2² x 7³)
= (2² x 7³)^(1/2)
= 2^(2/2) x 7^(3/2)
= 2 x 7 x √7

Question 2
(2^5 x 3² x 7³)^(1/2)
= 2^(5/2) x 3 x 7^(3/2)

Question 3
Not there!?

Question 4
√20 √18
= √(4 x 5) √(9 x 2)
= 2√5 x 3√2
= 6 √5 x √2
= 6 √10

784
= 2 x 392
= 2² x 196
= 2³ x 98
= 2^4 x 49
= 2^4 x 7 x 7
Prime numbers are numbers that can only be divided by themselves and 1.
2 and 7 are prime numbers.

Tip
Roots and indices are not as difficult as you may imagine. Spend a little time learning the basic rules (and there are but a few) and you will manage.
Good luck

Answer 3

1. √(2²*7³) is the same as √(2*2*7*7*7), which is the same as
(√2)*(√2)*(√7)*(√7)*(√7)
Since (√2)*(√2)=2 and (√7)*(√7)=7, we know that
(√2)*(√2)*(√7)*(√7)*(√7)= 2*7*√7.

2. √(2^5*3^2*7^3) = √(2*2*2*2*2*3*3*7*7*7)
=(√2)*(√2)*(√2)*(√2)*(√2)*(√3)*(√3)*(√7)*(√7)*(√7)
=2*2*(√2)*3*7*(√7)
=84*(√2)*(√7)
=84(√14)

21√56 is actually not correct (unless I misread the original problem).

4. You can use √20√18=√(4*5)*√(9*2), and that becomes
√4*√5*√9*√2
=2*√5*3*√2
=6*√10

The prime factorization of 784 is obtained by finding the lowest prime number that goes evenly int 784 and dividing by that. Do this again with the result until you end up with a prime number. The list of the prime numbers that go into 784 is its prime factorization.

The lowest prime number that divides 784 is 2.
2 divides 784 to give 392.
2 divides 392 to give 196.
2 divides 196 to give 98.
2 divides 98 to give 49.
We've run out of twos, so the lowest prime number that divides 49 is 7.
7 divides 49 to give 7.

We're done, and our prime factorization is 784=2*2*2*2*7*7.

The first step in simplifying radicals (that is √) is to take a prime factorization and then group all the prime factors that appear there twice.

Say, √784=√(2*2*2*2*7*7)
=√2*√2*√2*√2*√7*√7*
=2*2*7
=28 (yay! no √ signs left over.)

Answer 4

Square root is like raising something to the 1/2 power.

Look at the equations below, and hopefully they might make the transitive properties make more sense.
Think 2^2=2*2=4
2^3= 2*2*2= 2*(2*2)= (2^1)*(2^2)= 2^(2+1)= 2^3= 8
2^4= 2*2*2*2= (2*2)*(2*2)= (2^2)*(2^2)= 2^(2+2)= 2^4= 16


√(4)=√(2*2)=(2*2)^(1/2)=2

One way to think about it is like this
(2*2)^5 so I'm gonna have 5 times as many 2's mulltiplying
2*2*2*2*2*2*2*2*2*2=1024

(2*2)^3 so I'm gonna have 3 times as many 2's mulltiplying
2*2*2*2*2*2=64

(2*2)^1 so I'm gonn have 1 times as many 2's
2*2=4

(2*2)^(1/2)=√(2*2) so I'm gonna have 1/2 times as many 2's
2=2

Now, for your question
√(2²*7³)
√(2*2*7*7*7) 2 times as many 2's, and 3 times as many 7's
√(2*2)*√(7*7)*√7
2*7*√7

you can't factor out the 7 anymore so you keep it in the square root.

√(2^5*3²*7³)
√(2*2*2*2*2*3*3*7*7*7)
√(2*2*2*2)√2√(3*3)√(7*7)*√7
2*2*√2*3*7*√7
2*2*3*7*√(2*7)
84√14

3*7√2³*√7
3*7√(2*2*2)*√7
3*7√(2*2)*√2*√7
3*7*2*√2*√7
3*7*2*√(2*7)
42*√14

4
√20√18
√(2*2*5)*√(3*3*2)
√(2*2)*√5*√(3*3)*√2
2*√5*3*√2
2*3*√(5*2)
6√10

prime factors of a positive integer are the prime numbers that divide into that integer exactly, without leaving a remainder
784=2*392=2*2*196=2*2*2*98=2*2*2*2*49=2*2*2*2*7*7

So 2,7 are the prime factors of 784.