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You are watching: What is the square root of 64?

sometimes a an easy question favor what is the square root of 64 has solution that have the right to confuse a few. In this case, we will certainly dispel a pair of myths.

The key objective in this tutorial is to find out a couple of things about square roots and radicals, so that you space able to answer questions around it without hesitation.


an initial thing is first. Let united state spell the end the definition of the square root:

The square root of a provided number is the optimistic number (or zero) so that once squared results in that provided number .

that is it. So, provided a number \(x\), that square root is a number \(b\) so the \(b \ge 0\) and also


by looking at the above expression, we deserve to see that if \(b\) is walk to it is in the square root of \(x\), then \(x = b^2\), and also since a square number can not be negative, \(x\) have the right to only be non-negative (if we desire to be able to find its square root).

Conclusion : We have the right to only compute square roots of non-negative worths \(x\). Or stated differently, the domain of the role \(\sqrt x\) is \(<0,+\infty)\).

therefore then, comment our early stage question: What is the Square root of 64?

based upon what we defined, we require to discover a non-negative value \(b\) so the \(b^2 = 64\). Any kind of number conference those properties involved mind?

Well, yes, what if we tried v \(b = 8\)? Ok, for this reason \(b = 8\) is non-negative, and also \(b^2 = 8^2 = 64\).

so then, we have discovered the square root of 64, which is 8, because 8 is non-negative, and \(8^2 = 64\). We write this as:

\< \sqrt64 = 8 \>

The Myth about the Square Root duty

currently we go to the topic that urged this tutorial... The above meaning given of the square root enables us to discard the usual statement the "the square root of 64 is plus or minus 8" , i beg your pardon is wrong. Indeed

\<\sqrt64 =\not \pm 8\>

Now, we have the right to understand why together myth tote on. Indeed, both 8 and also -8 have actually the property that \(8^2 = 64\) and also \((-8)^2 = 64\). So then, why is -8 not the square root of 64?

since by definition, we claimed that the square root demands to be that non-negative number that has the property that as soon as squared they same the given number. And -8 stops working the problem of gift non-negative.

The Graph that the Square Root duty

Look in ~ the graph that the square root function below:


as you deserve to see, that duty only takes non-negative values, and also it actually passes the vertical heat test, so that is a function.

so in the end, the meaning of the square root as the non-negative \(b\) so the \(b^2 = x\) renders the square root a function.

If without doubt we had actually that \(\sqrt64 = \pm 8\), then \(\sqrt x\) would not be a function, would certainly be a relationship instead, since the vertical heat at \(x = 64\) would cross the graph twice (at 8 and -8).

What around Other Radical Functions?

There space other varieties of radical functions. For example, the cubic source \(\sqrt<3> x\). In this case, there is no should make a preeminence for what radical to select from, since the cubic root of a provided number \(x\) is the number \(b\) so the \(b^3 = x\).

Cubic source

because that the cubic source case, over there is no have to make distinctions due to the fact that for a given \(x\) there will be just ONE number \(b\) such that \(b^3 = x\).

For instance

\<\sqrt<3>64 = 4\>

simply because \(4^3 = 64\). Or

\<\sqrt<3>-64 = -4\>

simply because \((-4)^3 = -64\). This is, there is no ambiguity choose in the situation of the square root.

Quartic root

for the quartic source case, the is comparable to the square root. Us will have actually that \(\sqrt<4> x = b\) if \(b \ge 0\) and also \(b^4 = x\).

For example

\<\sqrt<4>16 = 2\>

since \(2^4 = 16\) and \(2 \ge 0\). But

\<\sqrt<4>16 =\not -2\>

because although \((-2)^4 = -16\), we have that \(-2

much more About the calculate of The Square root

One point we make the emphasis on was the the square root function \(\sqrt x\) requirements to take it a non-negative dispute \(x\) if we want to be able to compute the square root.

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we cheated over there a small bit, due to the fact that we walk not write the complete sentence: The square root role \(\sqrt x\) needs to take a non-negative argument \(x\) if we want to be able to compute the square root in the genuine LINE.