# Square root and power of a negative number

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Hello everyone,

When you have square root of (-2)^10    which is the result?

Because if I use the fraction notation results (-2)^(10/2)  = (-2)^5           .... a negative number.

On the other hand, if the power is calculated first the result is positive.

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Posted (edited)
5 hours ago, neonwarrior said:

When you have square root of (-2)^10    which is the result?

You mean to compare these two different calculations?

$\sqrt{-2}^{10}$
and
$\sqrt{-2^{10}}$

The order is important and well defined. Adding parentheses explicitly to show:

$(\sqrt{-2})^{10}$
and
$\sqrt{(-2^{10})}$

Since the calculations are different there are two results:

$\sqrt{-2}^{10}=-32$
$\sqrt{-2^{10}}=32$

Edited by Ghideon
math tags
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7 hours ago, neonwarrior said:

Hello everyone,

When you have square root of (-2)^10    which is the result?

Because if I use the fraction notation results (-2)^(10/2)  = (-2)^5           .... a negative number.

On the other hand, if the power is calculated first the result is positive.

This is improper use of brackets.

and of the square root function.

If you are going to say  "the square root of"

this is incomplete by itself.
It must be the square root of something. That something is called 'the argument'
The result of applying the function to the argument is called the result of the function.
But it must be the whole of argument.
So we often put the whole of the argument in brackets.

In this case the something or argument we want to take the square root of is stated to be minus 2 to the power 10,

So we put all of that in brackets

$\left( {{{\left( { - 2} \right)}^{10}}} \right)$

and take the square root

$\sqrt {\left( {{{\left( { - 2} \right)}^{10}}} \right)}$

Now the question is unambigous and you can work on it using the PEMDAS or BODMAS rules you should have been taught.

Note that the rules about brackets are often not included in the PEMDAS statements.

Brackets are evaluated from the inside out.  You work out the innermost pair of brackets first, then the next and so on.

So only Ghideon's second answer to your question is correct as you have posed it.

Here is a more complicate SAT question.

$\frac{{126 - {5^0}}}{{{5^3}}} - {\left( {\sqrt {\frac{{\frac{{32}}{4}*6 - 3\left( {2 - 3*\left( { - 1} \right)} \right) + 4 \div 2 - 116}}{{ - 2 + {5^2} - 32}}} } \right)^3} + \sqrt[3]{{ - 8}}*27 - 12{\left( {1 - 2} \right)^3}$

This is worked out for you in fine detail here

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Posted (edited)

Alternatively if you want to use brackets to write the square root as the power 1/2 then you must use

${\left( {{{\left( { - 2} \right)}^{10}}} \right)^{\frac{1}{2}}} = {\left( {1024} \right)^{\frac{1}{2}}} = 32$

Again you must work from the inside out, working out the value of the innermost bracket before applying the out one.

Edited by studiot
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Every number (except 0)  has two square roots.  Using the convention (ignore negative root) leads to these kinds of problems.

l

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1 hour ago, mathematic said:

Every number (except 0)  has two square roots.  Using the convention (ignore negative root) leads to these kinds of problems.

Yes you are nearly right, good catch.  +1

-32 is also a square root of the OP wanted to ask, but didn't pose correctly.

neonwarrior did however identify his/her problem of taking the square root of (-2)5 = -32

However within the real number system, negative numbers do not have even one single square root.

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My statement is about all (including complex) numbers.

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2 hours ago, mathematic said:

My statement is about all (including complex) numbers.

What level are we discussing this topic at ?

I am suprised at this argument because I think complex numbers are above the level of this question.

However I could ask what is the square root of infinity in the extended field of the reals or the square root of 5 in the integers.

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