Question

Will medal and fan whoever helps :)

Answer 1

8 / 2 = ? for the sqrt(y^8)

Answer 2

8/2=4

Answer 3

Okay so you pull out y^8 and get y^4, how about 20?
What are some factors of 20 that are perfect squares?

Answer 4

I have no clue, sorry this is so confusing for me

Answer 5

4 * 5 = 20 right?
4 is a perfect square (2*2)
Therefore, you can pull out 4 and make it 2.
Then you leave the five in the sqrt
(2*y^4)*sqrt5

Answer 6

Okay?

Answer 7

\[\sqrt{20y^{2}}=\sqrt{20} \times \sqrt{y^{8}}\]

to simplify a square root of a number you look for the highest square number that goes into that number

so square root of 20

highest square number that goes into 20 is 4
it goes into 20 5 times so you get

\[\sqrt{4} \times \sqrt{5} \times \sqrt{y^{8}}\]

you know what square root of 4 is 2 but we dont know the square root of 5 so we have 2sqrt5

now we simplify sqrt y^8

\[\sqrt{a^{n}}=(a^{n})^{\frac{ 1 }{ 2 }}=a^{n \times \frac{ 1 }{ 2 }}= a^{\frac{ n }{ 2 }}\]

Answer 8

so we just get
\[2y^{4}\sqrt{5}\]

Answer 9

Oh, that makes sense lol. thanks so much