Question

Can somene teach me about how to do square roots like changing it to a radicand and a number? Im really confuse... can someone give me a lesson on it?

Answer 1

can you give an example problem?

Answer 2

square root of 45....can u explain how to do it?

Answer 3

i see... my method is to express 45 as products of prime numbers..

do you understand what that meanS?

Answer 4

yes i know what that means. do you use something called a prime factorization tree? or explain your way of doing it(:

Answer 5

well the tree can be useful...is that your teacher's method?

Answer 6

yeah. but if you have a different way that is easier... could you show me howto do it?

Answer 7

well then i think it'll be the same

\[45 = 3 \times 3 \times 5\] \[\sqrt{45} = \sqrt{(3)(3)(5)}\]

now..since 3 occurred twice, we can pull it out of the square root giving us \(3\sqrt 5\)

Answer 8

so when it occurs twice we pull it out? what if there are two numbers that occure twice? maybe you could think of another example to show that(:

Answer 9

\[\sqrt{48}\]
\[48 = 2 \times 2 \times 2 \times \times 2 \times 3\]
\[\sqrt{48} = \sqrt{(2)(2)(2)(2)(3)}\]

2 occurred twice so pull it out

\[2\sqrt{\cancel{(2)(2)}(2)(2)(3)}\]

another 2 occurred twice soo
\[2(2)\sqrt{\cancel{(2)(2)(2)(2)}(3)} = 4\sqrt 3\]

Answer 10

so if the number occurs more than twice then we multiply them together like how you multiplied 2 and 2 to get 4?

Answer 11

yep

Answer 12

can you find the \(\sqrt{24}\)?

Answer 13

\[4\sqrt{3}\] i think?

Answer 14

but that's the answer to sqrt 48 :P i knew you'd get confused

\[\sqrt{24} = \sqrt{(2)(2)(2)(3)}\]

2 occurred twice so pull it out

\[2\sqrt{\cancel{(2)(2)}(2)(3)}\]

note that no other number appeared TWICE so \(2\sqrt 6\) is the final answer

Answer 15

oh ok. thanks for the clarification. give me one more then ill b good i think

Answer 16

nevermind i g2g. i got it