The results of raising things into power strictly depends on the writer's interpretation; whereas \(\frac{6}{2}=3\), raising to \(\frac{6}{2}\) is not the same as raising to \(3\). First, let's focus on \(\frac{6}{2}\).

This is because there's a difference between \(\sqrt{(-2)^6}\) and \((\sqrt{-2})^6\).

(i) In one, you're ordering the operation raise this value into the sixth power and square root it

(ii) In another, you're ordering the operation square root this value and raise it into the sixth power.

Here's how they differ: In (i), you will be taking a square root of a single value, and will thus obtain two distinct roots (unless it's 0). In (ii), you will be finding the two distinct square roots, and then raising each of them individually into the sixth power. If your answer was in a complex form, or were being raised into an odd power, you would actually get two distinct answers upon raising into that power; but either way, the results shall differ.

Now, raising to \(3\) is another story. When you choose to simplify \(\frac{6}{2}\) to \(3\), you're killing off roots. Take a look at the following example:

\[x^2=x\].

Now, you may be tempted to cancel the \(x\) on both sides and cheer \(x=1\)!, but then you'd be killing a root:

\[x^2-x=0\]

\[x(x-1)=0\]

\[\boxed{x=0,1}\]

(iii) Here, you're ordering the operation raise this value into the third power

In conclusion, it's just a matter of notation. This is one of those confusing wings of mathematics that teachers and professors leave out in high schools and colleges. And this is why generally mathematicians refer to "principal roots." But all of this jibber-jabr isn't for nothing. In real-world applications, depending on what you're trying to calculate, you may want to take the square root of a value to the sixth power to observe symmetry, or just cube it because your model is simplistic. It's all up to you.

\(\sqrt{a} \times \sqrt{b}=\sqrt{ab}\) if and only if, at least one of a and b is positive

If both a and b are negative, then:

\(\sqrt{a} \times \sqrt{b}=\sqrt{|a|}.i \times \sqrt{|b|}.i = i^{2}. \sqrt{|a|} \times \sqrt{|b|}=-\sqrt{|a|.|b|}\)
–
Sandeep Bhardwaj
·
2 years, 10 months ago

Log in to reply

@Sandeep Bhardwaj
–
Hm that's interesting! But I'm afraid the operation you're applying when splitting the squareroots is \((\sqrt{-2})^6\), so you're changing the operation in a way. In \(\sqrt{(-2)^6}\), you're asked to root \((-2)^6\), i.e. the finished product, or the positive value.

But anyway, what you're doing is absolutely consistent if you consider only the principal root. In this case, it is \(-8\). But if you're to consider all the roots, you'd have to factor in the negative counterpart, which is killed the second you define \(((-2)^6)^{\frac{1}{2}}=\sqrt{(-2)^6}\).

So yeah. By FTA, \(\pm 8\). And this is more apparent if you consider \(x^{2/4}\). This should be the same as \(x^{1/2}\), right? But if you observe carefully, you'll notice that with \(x^{2/4}\) you'll ultimately end up with either four distinct roots (if you choose to root last) or one root, whereas in \(x^{1/2}\) you're guaranteed two (\(x\neq 0\)).

But that "hidden concept" at the end... requires more thought from my end. Thanks for sharing.

@Calvin Lin
–
Maybe you guys should add a little LaTeX tutorial in a similar fashion to Brilliant intro tutorial somewhere around here! That'd be great for the newcomers.
–
John Muradeli
·
2 years, 10 months ago

Log in to reply

Root of a square(or even power) always comes out with modulus

\(\sqrt{x^{2}} = |x| \)
–
Krishna Sharma
·
2 years, 10 months ago

Log in to reply

@Krishna Sharma
–
Why's that? If \(x=-2,\) \(x^2=4\). But according to your expression, \(\sqrt{x^2}=2\). But we know \(x=-2\).

This symbol that you're using, \(\sqrt{\quad}\), is called the principle square root operator. What it does is it takes a a non-negative integer and spits out its principal square root.

\(\sqrt{x^2}\) is not always equal to \(x\).
–
Mursalin Habib
·
2 years, 10 months ago

Log in to reply

@Mursalin Habib
–
Oh really? I never knew that! I knew about principal square roots, but thought they were only subjective to the problem.

Well what a discovery!

But still \(x^{\frac{1}{2}}\) is not taking the principal square root, but it is taking both roots. So I just used the square root symbol as a way to show it.

Thanks for pointing that out, though.
–
John Muradeli
·
2 years, 10 months ago

Log in to reply

@John Muradeli
–
The function \(f(x)=\sqrt{x}\) has its domain in the non-negative reals. \(\sqrt{x^2}=|x|\), so @Krishna Sharma is right.
–
Daniel Liu
·
2 years, 10 months ago

Log in to reply

@Daniel Liu
–
Yes, it does. But taking square roots of specific numbers is not a function, it's an operation. A function is one system of expressing the results of operations. So in its full color, we'd actually be looking at \(y^2=x\), which is a relation, and thus we'd have a tilted parabola, with both positive and negative outputs.

Besides, you should know this from the Fundamental Theorem of Algebra: a polynomial with degree \(n\) has \(n\) roots. Since the polynomial we're interested in is \(y^2\) (whose "roots" we're taking), it will have two roots.

If you'd like a more complete explanation, let me know.

@John Muradeli
–
sir , i wish to hear a complete explanation ....... still confused ; hows that the solution set achieved in first case, got confined to { -8 } and in second case it is {+8 , -8 } ?
–
Sajitha Nair
·
2 years, 9 months ago

If you read what I've written above (the top comment), you can see that both of the transformations are right, but we're just "ordering" different operations. The comment right below mine provides a strong counterargument against this, and it may be right, but the fact that in applied physics we may sometimes want to specifically take \(n\) amount of roots of a quantity raised to some sort of power is unchanged.
–
John Muradeli
·
2 years, 9 months ago

in (-2)^6/2 the square root after multiplying to power 6 may be +ve or -ve
–
Suvir M
·
2 years, 10 months ago

Log in to reply

simply "BODMAS" first division then multiplication . In second case also we have divide first i.e., (1/2) then multiply...........
–
Shivasai Gunda
·
2 years, 10 months ago

Log in to reply

i dun get it , the pink one box :O
–
Maryam Surya
·
2 years, 10 months ago

Log in to reply

Another thing, when you take square root you should have two possible answers....8 and -8. b/c both squared gives you 64. So it works but the 8 is extraneous and you missed the -8
–
Heal Teranaki
·
2 years, 10 months ago

Log in to reply

when you square by odd the negative sign still stays and is possible. However, once you power by an even number, and the number will always be positive no matter what other number power you put it to after. Hence, the parenthesis should be around 6/2 because that should be done first, so the -2 isnt powered by an even exponent.
–
Heal Teranaki
·
2 years, 10 months ago

Log in to reply

You might've not concentrated enough. Look the right side of your question there you've written -
64 to the power 1/2
This is the same as square root of 64.

So at last you got your obvious answer as +/- 8. This is according to algebraic rules. Although you've not mentioned -8 there and written only +8.
–
Homo Sapien
·
2 years, 10 months ago

Log in to reply

@Homo Sapien
–
(√-2)^6 =√2i^6 and while doing this directly we are missing iota I so there will be a difference
–
Akshay Raj
·
2 years, 10 months ago

## Comments

Sort by:

TopNewestThe results of raising things into power strictly depends on the writer's interpretation; whereas \(\frac{6}{2}=3\), raising to \(\frac{6}{2}\) is not the same as raising to \(3\). First, let's focus on \(\frac{6}{2}\).

This is because there's a difference between \(\sqrt{(-2)^6}\) and \((\sqrt{-2})^6\).

(i) In one, you're ordering the operation

raise this value into the sixth power and square root it(ii) In another, you're ordering the operation

square root this value and raise it into the sixth power.Here's how they differ: In (i), you will be taking a square root of a single value, and will thus obtain two distinct roots (unless it's 0). In (ii), you will be finding the two distinct square roots, and then raising each of them individually into the sixth power. If your answer was in a complex form, or were being raised into an odd power, you would actually get two distinct answers upon raising into that power; but either way, the results shall differ.

Now, raising to \(3\) is another story. When you choose to simplify \(\frac{6}{2}\) to \(3\), you're killing off roots. Take a look at the following example:

\[x^2=x\].

Now, you may be tempted to cancel the \(x\) on both sides and cheer \(x=1\)!, but then you'd be killing a root:

\[x^2-x=0\]

\[x(x-1)=0\]

\[\boxed{x=0,1}\]

(iii) Here, you're ordering the operation

raise this value into the third powerIn conclusion, it's just a matter of notation. This is one of those confusing wings of mathematics that teachers and professors leave out in high schools and colleges. And this is why generally mathematicians refer to "principal roots." But all of this jibber-jabr isn't for nothing. In real-world applications, depending on what you're trying to calculate, you may want to take the square root of a value to the sixth power to observe symmetry, or just cube it because your model is simplistic. It's all up to you.

Cheers, – John Muradeli · 2 years, 10 months ago

Log in to reply

The right one is :>

\((-2)^{\frac{6}{2}}\)

\(=(-2)^{3}\)

\(=-8\)

But the question is why the other one is wrong.

I will try to explain it ::

\((-2)^{\frac{6}{2}}\)

\(=((-2)^{6})^{\frac{1}{2}}\)

No doubt !!! This step is absolutely right.\(=\sqrt{(-2)^{6}}\)

\(=\sqrt{ (-2).(-2).(-2).(-2).(-2).(-2)}\)

\(=\sqrt{-2}.\sqrt{-2}.\sqrt{-2}.\sqrt{-2}.\sqrt{-2}.\sqrt{-2}.\)

\(=\sqrt{2}.i \sqrt{2}.i \sqrt{2}.i \sqrt{2}.i \sqrt{2}.i \sqrt{2}.i \)

(Here\(i=\sqrt{-1}\) ,i.e. iota )\(=(\sqrt{2})^{6} . i^{6}\)

\(=8 \times -1\)

(Because* \(i^{6}=i^{4}.i^{2}=1 \times -1=-1)\) *Hidden concept here :

\(\sqrt{a} \times \sqrt{b}=\sqrt{ab}\)

if and only if, at least one of a and b is positiveIf both a and b are negative, then:

\(\sqrt{a} \times \sqrt{b}=\sqrt{|a|}.i \times \sqrt{|b|}.i = i^{2}. \sqrt{|a|} \times \sqrt{|b|}=-\sqrt{|a|.|b|}\) – Sandeep Bhardwaj · 2 years, 10 months ago

Log in to reply

But anyway, what you're doing is absolutely consistent if you consider only the

principal root. In this case, it is \(-8\). But if you're to consider all the roots, you'd have to factor in the negative counterpart, which is killed the second you define \(((-2)^6)^{\frac{1}{2}}=\sqrt{(-2)^6}\).So yeah. By FTA, \(\pm 8\). And this is more apparent if you consider \(x^{2/4}\). This should be the same as \(x^{1/2}\), right? But if you observe carefully, you'll notice that with \(x^{2/4}\) you'll ultimately end up with either four distinct roots (if you choose to root last) or one root, whereas in \(x^{1/2}\) you're guaranteed two (\(x\neq 0\)).

But that "hidden concept" at the end... requires more thought from my end. Thanks for sharing.

Cheers, – John Muradeli · 2 years, 10 months ago

Log in to reply

These results are differents because \({ \left( -2 \right) }^{ \frac { 6 }{ 2 } }\neq { \left( { \left( -2 \right) }^{ 6 } \right) }^{ \frac { 1 }{ 2 } }\).

\({ \left( { \left( -2 \right) }^{ 6 } \right) }^{ \frac { 1 }{ 2 } }=\sqrt { { \left( -2 \right) }^{ 6 } } \) ) not \({ \sqrt { \left( -2 \right) } }^{ 6 }\).

The correct way to solve this in the 2nd step in the pink box is \({ \left( { \left( -2 \right) }^{ \frac { 1 }{ 2 } } \right) }^{ 6 } \)because \({ \left( { \left( -2 \right) }^{ \frac { 1 }{ 2 } } \right) }^{ 6 }={ \sqrt { \left( -2 \right) } }^{ 6 }={ \left( -2 \right) }^{ \frac { 6 }{ 2 } }\) – Victor Paes Plinio · 2 years, 10 months ago

Log in to reply

Log in to reply

– Calvin Lin Staff · 2 years, 10 months ago

Note: You can edit your comment directly by selecting the pencil in the top right.Log in to reply

– Victor Paes Plinio · 2 years, 10 months ago

Thanks! I go edit now.Log in to reply

– John Muradeli · 2 years, 10 months ago

Maybe you guys should add a little LaTeX tutorial in a similar fashion to Brilliant intro tutorial somewhere around here! That'd be great for the newcomers.Log in to reply

Root of a square(or even power) always comes out with modulus

\(\sqrt{x^{2}} = |x| \) – Krishna Sharma · 2 years, 10 months ago

Log in to reply

Thus, \(\sqrt{x^2}=\pm x\).

Let's have another look:

\((-2)^2=4\).

\(\sqrt{(-2)^2}=\sqrt{4}=\pm 2\).

Since \(x\neq +2\), \(\boxed{x=-2}\).

(\(\pm \rightarrow\) plus OR minus).

Cheers, – John Muradeli · 2 years, 10 months ago

Log in to reply

notequal to \(-2\).\(\sqrt{(-2)^2}\) is equal to \(2\).

This symbol that you're using, \(\sqrt{\quad}\), is called the principle square root operator. What it does is it takes a a non-negative integer and spits out its

principalsquare root.\(\sqrt{x^2}\) is not always equal to \(x\). – Mursalin Habib · 2 years, 10 months ago

Log in to reply

Well what a discovery!

But still \(x^{\frac{1}{2}}\) is not taking the principal square root, but it is taking both roots. So I just used the square root symbol as a way to show it.

Thanks for pointing that out, though. – John Muradeli · 2 years, 10 months ago

Log in to reply

@Krishna Sharma is right. – Daniel Liu · 2 years, 10 months ago

The function \(f(x)=\sqrt{x}\) has its domain in the non-negative reals. \(\sqrt{x^2}=|x|\), soLog in to reply

Besides, you should know this from the Fundamental Theorem of Algebra: a polynomial with degree \(n\) has \(n\) roots. Since the polynomial we're interested in is \(y^2\) (whose "roots" we're taking), it will have two roots.

If you'd like a more complete explanation, let me know.

Cheers, – John Muradeli · 2 years, 10 months ago

Log in to reply

– Sajitha Nair · 2 years, 9 months ago

sir , i wish to hear a complete explanation ....... still confused ; hows that the solution set achieved in first case, got confined to { -8 } and in second case it is {+8 , -8 } ?Log in to reply

If you read what I've written above (the top comment), you can see that both of the transformations are right, but we're just "ordering" different operations. The comment right below mine provides a strong counterargument against this, and it may be right, but the fact that in applied physics we may sometimes want to specifically take \(n\) amount of roots of a quantity raised to some sort of power is unchanged. – John Muradeli · 2 years, 9 months ago

Log in to reply

– Sajitha Nair · 2 years, 9 months ago

ThanksLog in to reply

in (-2)^6/2 the square root after multiplying to power 6 may be +ve or -ve – Suvir M · 2 years, 10 months ago

Log in to reply

simply "BODMAS" first division then multiplication . In second case also we have divide first i.e., (1/2) then multiply........... – Shivasai Gunda · 2 years, 10 months ago

Log in to reply

i dun get it , the pink one box :O – Maryam Surya · 2 years, 10 months ago

Log in to reply

Another thing, when you take square root you should have two possible answers....8 and -8. b/c both squared gives you 64. So it works but the 8 is extraneous and you missed the -8 – Heal Teranaki · 2 years, 10 months ago

Log in to reply

when you square by odd the negative sign still stays and is possible. However, once you power by an even number, and the number will always be positive no matter what other number power you put it to after. Hence, the parenthesis should be around 6/2 because that should be done first, so the -2 isnt powered by an even exponent. – Heal Teranaki · 2 years, 10 months ago

Log in to reply

You might've not concentrated enough. Look the right side of your question there you've written - 64 to the power 1/2 This is the same as square root of 64.

So at last you got your obvious answer as +/- 8. This is according to algebraic rules. Although you've not mentioned -8 there and written only +8. – Homo Sapien · 2 years, 10 months ago

Log in to reply

– Akshay Raj · 2 years, 10 months ago

(√-2)^6 =√2i^6 and while doing this directly we are missing iota I so there will be a differenceLog in to reply