simplifying radicals

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hmwin

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Hi,
I'm trying to make sure I'm following rules for simplifying radicals. Please tell me if this is correct, or where I went wrong.

simplify the square root of 40b^5

I am assuming that you turn 40 into 4 X 10 with 4 being a perfect square...

so... 2 square root of 10b^5

Then... b^2 + b^3 with b^2 being a perfect square...

so... 2b square root 10b^3 being the final answer...
 


Hi:

The expression b^3 also contains a squared factor, so the expression sqrt(10b^3) can also be simplified.


\(\displaystyle \sqrt{40b^5}\)

\(\displaystyle \sqrt{4 \cdot 10 \cdot b^2 \cdot b^2 \cdot b}\)

\(\displaystyle \sqrt{4} \cdot \sqrt{10} \cdot \sqrt{b^2} \cdot \sqrt{b^2} \cdot \sqrt{b}\)

\(\displaystyle 2 \cdot \sqrt{10} \cdot b \cdot b \cdot \sqrt{b}\)

\(\displaystyle 2 \cdot b \cdot b \cdot \sqrt{10} \cdot \sqrt{b}\)

\(\displaystyle 2 \cdot b \cdot b \cdot \sqrt{10 \cdot b}\)

\(\displaystyle 2b^2 \sqrt{10b}\)

 
Would it stand to reason, then, that you would never have more than "b" inside the square root?
 
hmwin said:
Would it stand to reason, then, that you would never have more than one factor of "b" inside the square root after simplifying ?
Click to expand...


I altered your question to match what I think you're actually trying to ask.

The answer is "YES".

For the following examples, assume that b represents a non-negative Real number.

\(\displaystyle \sqrt{b^2} = b\)

\(\displaystyle \sqrt{b^3} = b \sqrt{b}\)

\(\displaystyle \sqrt{b^4} = b^2\)

\(\displaystyle \sqrt{b^5} = b^2 \sqrt{b}\)

\(\displaystyle \sqrt{b^6} = b^3\)

\(\displaystyle \sqrt{b^7} = b^3 \sqrt{b}\)

\(\displaystyle \sqrt{b^8} = b^4\)

\(\displaystyle \sqrt{b^9} = b^4 \sqrt{b}\)

\(\displaystyle \sqrt{b^{10}} = b^5\)


Et cetera!

If the symbol b represents a negative Real number, then the expression sqrt(40b^5) is not defined.
 
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