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youngbuck
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Can I use a common root here? 17 Dec 2014, 10:38
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If b ≠ 0, [(8b^8)/(4b^4)^1/2]

I know there is a simpler solution but can I just square the top and bottom so that im left with: 8b^16/4b^4
Then take common roots so that im left with 2^48/2^8

How come this doesnt work?

Thanks



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bluefly
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Can I use a common root here? 17 Dec 2014, 21:54
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youngbuck
If b ≠ 0, [(8b^8)/(4b^4)^1/2]

I know there is a simpler solution but can I just square the top and bottom so that im left with: 8b^16/4b^4
Then take common roots so that im left with 2^48/2^8

How come this doesnt work?

Thanks
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The simplification should be as follows:-

\(\frac{(8b^8)}{\sqrt{(4b^4)}}\)

=\(\frac{8b^8}{2b^2}\)

=\(4b^6\)
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KarishmaB
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Can I use a common root here? 18 Dec 2014, 04:32
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youngbuck
If b ≠ 0, [(8b^8)/(4b^4)^1/2]

I know there is a simpler solution but can I just square the top and bottom so that im left with: 8b^16/4b^4
Then take common roots so that im left with 2^48/2^8

How come this doesnt work?

Thanks
Show more

When you square an expression, it changes.
For example, if you square 6, it becomes 6^2 which is 36.
Similarly, if you square this expression, it will change.

Say \(x = \frac{(8b^8)}{(4b^4)^{\frac{1}{2}}}\)
Square to get \(x^2 = \frac{8^2 * b^{16}}{4b^4}\)

Simplify to get \(x^2 = 16b^{16 - 4} = 16b^{12}\)

Now you need to take the square root back: \(x = (16b^{12})^{\frac{1}{2}} = 4b^6\)

You need the value of x, not of x^2.

When you have an equation, then you can square both sides and the equation still holds.

x = 6
x^2 = 36



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!