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What is the square root of 64y16

User Phanf
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2 Answers

3 votes

Answer:

Explanation:

The idea here is to "match" the exponent on the radicands (the number/variables under the radical sign) to the index (the little number that sits in the "arm" of the radical sign). Your problem looks like this:


\sqrt[2]{64y^(16)}

Our index is a 2. If we could rewrite both the 64 and the y^16 with bases to the power of 2 (that's why I say to "match" the exponent to the index), we could pull out the base. For example,


\sqrt[2]{x^2}=x because the power is a 2 and so is the index, so we pull out the base of x.

Our rewrite would look like this:


\sqrt[2]{8^2(y^8)^2} (remember that power to power on a base means you multiply the exponents so 8 * 2 = 16).

The power on the 8 is a 2 which matches our index of 2 so we will pull out the 8; the power on the y^8 is a 2 which also matches our index of 2 so we will pull out the y^8. The simplification of this is


8y^8

User Rick Wayne
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7.9k points
4 votes
The answer to the square root of 64y16 is 8y8.
User Nabeel Ahmed
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7.3k points