Question

ALGEBRA HELP PLZ WILL FAN & MEDAL <3

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

3. Simplify the radical expression.
square root 150x^3k^4
A) 75xk^2
B) 5xk square root 6xk
C) 5xk^2 square root 6x
D) 5 square root 6x^3k^4

Answer 1

Problem 1:
\(\color{#000000 }{ \displaystyle \sqrt{45} =\sqrt{5\times 9}=\sqrt{5}\times \sqrt{9} }\)

Answer 2

\(\sqrt{9}~=~?\)
(you tell me, and use that to simplify the expression)

Answer 3

\[\sqrt{9} = 3\]

Answer 4

Yes, so

\(\color{#000000 }{ \displaystyle \sqrt{45} =\sqrt{5\times 9}=\sqrt{5}\times \sqrt{9} =\sqrt{5}\times 3~~~\quad or,~~3\sqrt{5} }\)

Answer 5

So, the thing here is to brake down the root into the product of roots, which [at least] one of them is a perfect square - (if possible, of course).

Answer 6

How can you write
\(\color{#000000 }{ \displaystyle 56x^2 }\)

as a product?
(Note that \(x^2\) is a perfect square)

Answer 7

Try it...
if you won;t get it right, it is totally fine.

Answer 8

For example, if I want to find the \(\color{#000000 }{ \displaystyle \sqrt{12x^4} }\)

I can say that
\(\color{#000000 }{ \displaystyle 12x^4=3\times 4 \times x^4 }\)

And notice, I have two perfect squares
\(x^4\) (because \((x^2)^2=x^{2\times 2}=x^4\))
And, \(4\) (because \(2^2=4\))

((And I know that \(\sqrt{17^2}\) for example, will result in just 17...))

So,
\(\color{#000000 }{ \displaystyle \sqrt{12x^4} = }\)
\(\color{#000000 }{ \displaystyle \sqrt{3\times 4 \times x^4} = }\)
\(\color{#000000 }{ \displaystyle \sqrt{3}\times \sqrt{4} \times \sqrt{x^4} = }\)
\(\color{#000000 }{ \displaystyle \sqrt{3}\times 2 \times x^2 = }\)

And when you simplify/rearrange this, you have;
\(\color{#000000 }{ \displaystyle 2x^2\sqrt{3} }\)

Answer 9

In your expression, \(\color{#000000 }{ \displaystyle \sqrt{56x^2} }\), you also have some perfect squares inside the root.

\(\color{#000000 }{ \displaystyle 56x^2=14\times 4\times x^2 }\)
And \(4\) and \(x^2\) are perfect squares (right?)

Answer 10

OK, so write \(\sqrt{56x^2}\) as the product of square-roots, and simplify...

Answer 11

Here, this should be pretty easy to read.
Not a scholarly article or anything of the sort.

https://www.mathsisfun.com/numbers/simplify-square-roots.html