Question

ALGEBRA HELP PLEASE!!!!!! WILL FAN & MEDAL :)

1. Simplify the radical expression.
square root of 45
A) 5 square root 3
B) 3 square root 5
C) 3 square root 15
D) 15

2. Simplify the radical expression.
square root of 56x^2
A) 28x
B) 2x square root 14
C) 2x square root 7
D) 2 square root 14x^2

Answer 1

\(\sqrt{45}\)

factors of 45?

Answer 2

\(\large \sqrt{45}\)
first list the factors of 45

Answer 3

\(\sqrt{9}=\)

Answer 4

good now do you notice a perfect square from either one of those factors?

Answer 5

1,3,5,9,15,45

Answer 6

we can work with 5 and 9

Answer 7

Square root of 9 = 3

Answer 8

so 3 goes out side and 5 stays inside
\(3\sqrt{5}\)

Answer 9

i will let @Mehek14 help you with the first one. seems like @Mehek14 got that question.

Answer 10

so simplified is
\(3\sqrt{5}\)

Answer 11

I actually have to go :(

Answer 12

i will take over now if you may :)

Answer 13

so the second question is almost like the first one.
\(\sqrt{56x^2}\)

this is basically saying \(\sqrt{56 \times x^2}\) which means we need to find the square root of 56 and \(x^2\) individually.

Answer 14

so list the factors of 56 and make sure that one of the factors (of the two) is a perfect square (if there is any)

Answer 15

@calculusxy Factors of 56: 1, 2, 4, 7, 14, 28

Answer 16

but 1 x 2 is not 28

Answer 17

but if we have 4 and 7 (4 x 7 = 28), then we have 4 as the perfect square.

Answer 18

2 is not a perfect square because it leads to 1.414...

Answer 19

we will stick with 4 and 7. do you understand why?

Answer 20

@Answers101

Answer 21

okay

so \(\sqrt{4} = ?\)

Answer 22

good so we will place 2 out of the parenthesis and have 7 inside of the parenthesis (because it is not a perfect square)

we now have to find the square root of \(x^2\). what will that be?

Answer 23

not really because x^2 is the same thing as \(x \times x\) so the square root of x^2 = x. do you get it?

Answer 24

Can you help me with a few more? I can open a new question

Answer 25

okay so that will be placed outside of the square root as well
now we have
\(\large 2x \sqrt{7}\)

Answer 26

i can only help with one more b/c i have to go soon too