Question

Please help me! I am behind and really need to catch up! (Algebra 2)

Answer 1

Okay, this one is fun...

Answer 2

That one looks exciting lol

Answer 3

By "fun" you mean hard? -.-

Answer 4

First, you need to break up the numbers under the radicals, so you can simplify. Do you have any idea how to do that?

Answer 5

It shouldn't be that hard...

Answer 6

Not really. :/ I learned it a while back and just don't remember how to.

Answer 7

Factor each number into prime numbers. For example

75 = 5*15 = 5*3*5 = 3*5*5
so \[\sqrt{75} = \sqrt{3*5*5} = \sqrt{3}*\sqrt{5*5} = \sqrt{3} * 5 = 5\sqrt{3}\]

Answer 8

each time you have a pair of identical factors under the square root sign, you can remove both of them and replace by one of them outside of the square root sign

Answer 9

Wait, why did you do all of those different numbers multiplied by each other? O.o

Answer 10

I was showing you the steps in simplifying \(\sqrt{75}\)

Answer 11

So you have to figure out random numbers that equal 75 when multiplied?

Answer 12

no, they aren't random numbers, they are the factors...

Answer 13

OHHHHH Numbers that go INTO 75?

Answer 14

75 is odd (ends in 1,3,5,7,9) so it isn't divisible by 2. It ends in 5, so it is divisible by 5, so I divided it by 5. could have also seen that it is divisible by 3 (sum of digits = 12 which is divisible by 3, so the number is also divisible by 3).

your other option is to try dividing out all the perfect squares less than the number:
1*1 = 1
2*2 = 4
3*3 = 9
4*4 = 16
5*5 = 25
6*6 = 36
7*7 =49
8*8 = 64

only 5*5 = 25 divides 75 evenly, and the quotient is 3, so 75 = 3*5*5

Answer 15

Oh okay. So I have to do this will ALL of the other numbers under the radical?

Answer 16

yeah, that should take a minute each, tops...

Answer 17

Okayy

Answer 18

80 is 4√5

48 is 4√3

20 is 3√2

Is this correct? @whpalmer4

Answer 19

\[\sqrt{20} = \sqrt{2*2*5} = \]
others are correct