What is the square root of 64?

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What is the square root of 64?

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Answer 1 · 2016-11-21T05:04:21

$\sqrt{64} = \pm 8$ $sqrt64=+-8$

Answer 2 · 2016-11-21T05:04:54

It's 8 or -8.

Explanation:

8 x 8 = 64.

Answer 3 · 2016-11-21T07:33:47

The principal square root of $64$ $64$ is:

$\sqrt{64} = 8$ $sqrt(64) = 8$

The other (non-principal) square root is:

$- \sqrt{64} = - 8$ $-sqrt(64) = -8$

Explanation:

$64$ $64$ has two square roots, namely $8$ $8$ and $- 8$ $-8$ , since:

$8^{2} = {(- 8)}^{2} = 64$ $8^2 = (-8)^2 = 64$

When we say "the square root", what is usually intended is "the principal square root", which in the case of the Real square root of a positive number is the positive one.

Any non-zero number $n$ $n$ has two square roots. In order to distinguish between them, we call one the "principal" square root, which in the case of $n > 0$ $n > 0$ means the positive one.

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Complex footnote

If $n < 0$ $n < 0$ then it has two Complex non-Real square roots:

$\pm i \sqrt{- n}$ $+-i sqrt(-n)$

In this case we call $i \sqrt{- n}$ $i sqrt(-n)$ the principal square root and $- i \sqrt{- n}$ $-i sqrt(-n)$ the non-principal one.

For example:

$\sqrt{- 64} = 8 i$ $sqrt(-64) = 8i$ is the principal square root.

$- \sqrt{- 64} = - 8 i$ $-sqrt(-64) = -8i$ is the other square root.

Note that $8 i$ $8i$ is not "positive". Unlike Real numbers, Complex numbers are not ordered, but for pure imaginary square roots we choose the one with the positive imaginary part and call it "principal".

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What is the square root of 64?

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