anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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SolomonZelman
  • SolomonZelman
Problem 1: \(\color{#000000 }{ \displaystyle \sqrt{45} =\sqrt{5\times 9}=\sqrt{5}\times \sqrt{9} }\)
SolomonZelman
  • SolomonZelman
\(\sqrt{9}~=~?\) (you tell me, and use that to simplify the expression)
anonymous
  • anonymous
\[\sqrt{9} = 3\]

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SolomonZelman
  • SolomonZelman
Yes, so \(\color{#000000 }{ \displaystyle \sqrt{45} =\sqrt{5\times 9}=\sqrt{5}\times \sqrt{9} =\sqrt{5}\times 3~~~\quad or,~~3\sqrt{5} }\)
SolomonZelman
  • SolomonZelman
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).
SolomonZelman
  • SolomonZelman
How can you write \(\color{#000000 }{ \displaystyle 56x^2 }\) as a product? (Note that \(x^2\) is a perfect square)
SolomonZelman
  • SolomonZelman
Try it... if you won;t get it right, it is totally fine.
SolomonZelman
  • SolomonZelman
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; \(\color{#000000 }{ \displaystyle 2x^2\sqrt{3} }\)
SolomonZelman
  • SolomonZelman
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?)
SolomonZelman
  • SolomonZelman
OK, so write \(\sqrt{56x^2}\) as the product of square-roots, and simplify...
SolomonZelman
  • SolomonZelman
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

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