#### (New questions of the week)

Two recent questions (five days apart, from high school students in different countries) were about nearly the same thing, and fit nicely together: What do you get when you square a square root, or take the square root of a square, but don’t know the sign of the number ahead of time? And what about other powers? It gets a little subtle!

## Square root of a square: simplifying vs solving

The first question is from Debkanta, who had just learned that \(\sqrt{x}=|x|\), not \(x\):

Dr. today I have encountered a question on finding range of f(x) = x + √(x

^{2}).I have solved this way:

f(x) = x + √(x

^{2})

= x + x

= 2xSo range is -∞ < f(x) < ∞.

When I checked answer I found that

x + √(x

^{2}) = 2x when x≥0

= 0 when x<0Hence range is 0 ≤ f(x) < ∞.

I understood that for all values of x, √(x

^{2}) is +ve.actually √(x^{2}) = Mod x.So, when x≥0, then x + √(x

^{2}) = x + x = 2xand when x<0, then x + √(x

^{2}) = -x + x = 0Now my question is,

while solving equationswe do as followsx

^{2 }= 4taking square root on both sides we get

√(x

^{2}) = √4

or, x = ±2.

Here we don’t use Mod x, we simply write x which can take both +ve and -ve values of x for which given equation is true.Another situation is, I

simplified problems on indicesas follows√(x

^{2}) = x, √(x^{3}) = x√x = x^{3/2}, √(x^{4}) = x^{2}…etc.in these cases

I never used mod x.But now I have understood that

√(x. I am confused. plz help.^{2}) ≠ x when x<0

Although the initial problem was about a range, the *real* question was about the fact that \(\sqrt{x^2}=|x|\), rather than just *x*. (Debkanta, not being from America, calls the absolute value the “modulus”, or “mod”, and calls exponents “indices”.) He recognizes the truth of this fact, but sees contradictions in how we use a square root (a) in solving an equation, and (b) in writing radicals as fractional powers.

### Square roots in solving an equation

Doctor Rick answered:

Hi, Debkanta.

As I understand it, your question is not about the range of a function — you understand that problem now. Rather, it is about square roots of powers, and

what else might not be what you thought, now that you know that √(x^{2}) = |x|, not x.You wrote:

while solving equations we do as follows

x

^{2 }= 4taking square root on both sides we get

√(x

^{2}) = √4

or, x = ±2.When you wrote √4 = ±2, you really meant that there are

two numbers whose square is 4, namely 2 and –2. Using the square-root sign was sort of a shorthand for what you were thinking. But customarily something like √x is taken to mean thenon-negativenumber whose square is x.

The radical represents *the* (principal) square root, but in solving, we want *all* square roots.

Why do we define √x this way? Because if we’re going to use it in an algebraic expression, we’d like it to have a single value! A variable has to have

one value at a time; even when we don’t know what that value is, we assume it’s a number, not two numbers; and when this assumption fails, things can get very confusing. So we can’t say that √4 isboth2 and –2, and there are good reasons to choose thenon-negativevalue as what we call the“principal square root”.

The main idea here is that we want \(\sqrt{x}\) to be a *function*, with *one value*.

Now, this doesn’t really mess up what you are used to doing in solving x

^{2}= 4. You can still take the square root of both sides, but now you do this:√(x

^{2}) = √4

|x| = 2

x = ±2 (or, x ∈ {2, –2} )In other words, the right side has only

one value, but the left side is a function of x, namely the modulus (or absolute value) function, so now we need to solve that equation, using our knowledge of modulus (seeing that the equation hastwo solutions).

I have rarely seen the work of solving written with an absolute value this way; rather, we skip over that line and just write the plus-or-minus version, which is equivalent.

### Absolute values in fractional powers?

You also wrote:

I simplified problems on indices as follows

√(x

^{2}) = x, √(x^{3}) = x√x = x^{3/2}, √(x^{4}) = x^{2}…etc.in these cases I never used mod x.

Often when we have problems about simplifying radicals or expressions with rational exponents, we

assume(maybe without saying it) that the variable ispositive. An expression like x^{3/2}raises some problems when x is negative. Consider these two expressions:√(x

^{3}) = (x^{3})^{1/2}= x^{3 · 1/2}= x^{3/2 }(√x)^{3}= (x^{1/2})^{3}= x^{1/2 · 3}= x^{3/2}So these two expressions are equal, right? But let

x = –3and evaluate each of them directly, and we get (assuming you are familiar withcomplex numbers):√((–3)

^{3}) = √(–27) = 3i√3

(√(–3))^{3}= (i√3)^{3}= –3i√3We get

two different values! This can happen when imaginary numbers appear: in that context,there is no such thing as a “principal square root”— we have to deal with the complexity of “double-valued functions”.

We can *define* a principal square root in this context, but there is no way to do so that avoids this problem.

For this reason

it’s best just to consider only positive values of the variable. When we do that, of course,we don’t need the modulus.For more on this, you might be interested in this page from the old Ask Dr. Math archive (which requires an account with NCTM, but it’s free):

I’ll look at that question next week. I’m also skipping the rest of the conversation, to avoid repetition.

## What about the square of a square root?

Our second question came five days later, from Myler, again comparing laws of radicals and exponents:

Hello sirs! I have a question regarding this, since it’s really confusing me.

Shouldn’t √(x^{2}) be NOT equal to (√x)^{2}?Here’s why I think it’s not.

√(x, I mean it’s quite obvious. Let’s see for example, x=4, then;^{2}) should be equal to ±x√(4

^{2}) = √16 = ±4Next,

(√x), since;^{2}must be, and only be, +x(√4)

^{2}= (±2)^{2}= 4It’ll be positive no matter if it’s negative 2 or positive 2.

But then, another take on this equation is by proving through the

laws of exponents. Then it would be like this;√(x

^{2}) = (x^{2})^{½}= x^{2•½}= x^{1}= x(√x)

^{ 2}= (x^{½})^{2}= x^{½•2}= x^{1}= xBy trying to evaluate it differently, it shows that √(x

^{2}) = (√x)^{2}, but on the first evaluation, it is √(x^{2}) ≠ (√x)^{2}.

Are there rules on the exponents that I might have missed?Can you please help me understand? Thank you for your time.

We’ll need to deal with the idea of radicals having two values, but that won’t take away the problems.

Doctor Rick answered this, too:

Hi, Myler.

You have done some good thinking, and you are confused about something that really

isconfusing! I’ll try to help.First, on your first argument, that √(x²) is not equal to (√x)²”:

There is one thing to clear up here before we go on: We don’t customarily use √x to mean

bothsquare roots of x. Rather we take it to mean thenon-negativenumber whose square is x.Why do we define √x this way? Because if we’re going to use it in an algebraic expression, we’d like it to have a

single value! A variable has to have one value at a time; even when we don’t know what that value is, we assume it’s a number, nottwonumbers, and when this assumption fails, things can get very confusing. So we can’t say that √4 is both 2 and –2. There are good reasons to choose the non-negative value as what we call the“principal square root”, one reason being just what you are asking about.

So defining a square root *function* is a way to prevent confusion.

With this understanding, I would write

√(4

^{2}) = √16 = 4which is in agreement with

(√4)

^{2}= (2)^{2}= 4But this does not resolve your issue completely, as we’ll see below.

### Clarifying the rules of exponents

Now, on your second argument, based on the laws of exponents:

Yes,

there is something that often gets omittedwhen the laws of exponents are presented.We didn’t see a problem with your example, once we understood √x as the

principalsquare root. But consider what happens withnegative values of the base, x. I don’t know whether you have learned much about imaginary numbers, so I will start with the assumption that you haven’t.

This was also explored in the omitted part of the discussion with Debkanta.

What happens if we let x = –4? Then we have

√((–4)

^{2}) = √16 = 4but √(–4) has

no value. We can’t take the square root of –4; there is no (real) number that, when multiplied by itself, makes –4 (or any negative number, for that matter). Thereforewe can’t evaluate the expression (√(–4)), and the two expressions don’t have the same value — one has a value, the other doesn’t.^{2}

If you’ve learned about imaginary numbers, then what I just said must be modified. In that case, we understand that √(–4) = 2i, but we will run into problems. We have(√(–4))

^{2}= (2i)^{2}= –4So we’ve got a problem again! The order in which we apply the square and the square root (or the exponents 2 and 1/2)

changes the result, in contradiction to the rule(x

^{a})^{b}= x^{ab}which, because multiplication is commutative, says that we can switch a and b without changing the value.

So if we consider only real numbers, the two expressions simply have **different domains**; if we allow complex values, then we actually get **different results**. And this exponent rule fails.

So what do we do?

We add a condition to the rules of exponents. If we’re only thinking about integer exponents, we have no problem; we’re not working with radicals. But if the exponents are allowed to be non-integers like 1/2, then we need to say that the rules apply when thebases are positive.Here is an example of an online resource that includes this restriction while presenting the rules for radicals:

Algebra, Section 1.3: Radicals

You see the condition there:

aandbarepositivereal numbers.

Not all textbooks say this; many initially present the rules as applying to all real numbers, both for radicals and for exponents. Doctor Rick had given an example to Debkanta:

I tried an Intermediate Algebra textbook available on OpenStax.org, and this is what I found:

I can find no comment in that section about negative bases; all examples have positive bases. I assume that a decision was made to ignore the issue for the time being. However, when complex numbers are introduced (Section 8.8 — the above is in Section 8.3), we find a statement that “we cannot use the Product Property for Radicals in order to multiply square roots of negative numbers.”

So you will not always find the rules stated with their complete conditions, but they are there. Resuming his answer to Myler:

So I have given you two answers for the dilemma you encountered.

One:understand √x as the non-negative number whose square is x; this solves the problem for positive numbers under the radical, or as bases of powers.If the exponent isn’t restricted to the integers, then restrict the base to positive numbers.Two:I hope this helps!

In subsequent discussion, Doctor Rick made a couple additional comments:

If an operation could yield an imaginary number, then we have to be careful— we can’t assume that the laws of exponents will apply; they might, but we’d need to check the specific case. (If you read on in the page from Paul’s Online Notes that I showed you, it says: “Note that on occasion we can allow a or b to be negative and still have these properties work. When we run across those situations we will acknowledge them.”)

For example, we can multiply radicals if only one radicand is negative; but if you don’t know that to be true, you can’t do it.

We should make a distinction: sometimes we want to talk about “

asquare root of a”, meaning any solution of the equation x^{2}= a. In this sense, –2 is a square root of 4. But we reserve the notation √a, “thesquare root of a”, to mean the non-negative root only. When we want to talk aboutbothsquare roots of x, we can write: ±√x.

### What to do about negative bases

After settling these ideas, Myler had a follow-up question:

So let me clarify first, the rule of exponent ( ²√x = x½ ) only applies to positive bases, and

if it is negative, what do we do?Since you mentioned about a problem of the order of operations [√(x²) and (√x)²] yielding different results (with a negative x).(It’s actually just a follow up question, my first question is now answered)

Doctor Rick replied, using Debkanta’s examples:

I’m glad I have answered your question.

I would just work out a particular problem step by step rather than trying to rewrite it using rules that we can’t be sure of.

I’m not sure what types of problem you might be thinking about, since we have finished with this one. Another student who asked about this recently, said that he had learned to simplify the following expressions this way:

√(x

^{2}) = x√(x

^{3}) = x√x√(x

^{4}) = x^{2}After our discussion he understood, as you do now, that

the first of these should be |x|when x can be negative. What about the other two? Do they need to be changed? I’d like you to think about this, perhaps just by trying some positive and negative numbers.You might also consider whether it’s true that

√(x

^{3}) = (√x)^{3}And what about cube roots? Is this true?

^{3}√(x^{2}) = (^{3}√x)^{2}

Exploration can be the best way to learn.

Myler answered:

I’ll try finding it out

x = -4

√(-4)³ = (√-4)³

√-64 = (2i)³

8i = (2i)(2i)(2i)

8i = -4(2i)

8i ≠ -8i

Does this mean that

√(x³) and (√x)³ doesn’t necessarily need to be equal? But is equal just for a positive x? So to summarize, we’ll just need to becarefulof the parentheses when it comes to a negative x and just do as it was shown. I hope I got it right now:]

Doctor Rick answered:

What you have found is that

√((-4)³) = √(–64) = 8i

(√-4)³ = (2i)

^{3}= –8iwhere we are following the direction in many textbooks to write √(–1) = i, rather than ±i (the “double-valued function” approach I mentioned). You’re correct; these are different, which shows that we did the right thing in not trusting the multiplication property of exponents to work in this case. Good work!

As for the three simplifications I showed, wondering if they had to be modified,

√(x

^{2}) = x√(x

^{3}) = x√x√(x

^{4}) = x^{2}we already know that

the first should be modifiedto√(x

^{2}) = |x|If I try a negative value for x in the second one, I find:

√((–4)

^{3}) = √(–64) = 8i–4√(–4) = (–4)(2i) = –8i

We see that

it isn’t true that √(x— but we could say^{3}) = x√x for all x√(x

^{3}) = |x| √xas you will see by trying x = –4 in this identity. (Trying a specific example does not

provethat it’s an identity, but you can see that it would work foranypositive numberandforanynegative number.)If you’re still interested in doing a bit more work (I realize that we are well past what you originally asked about), try out the last simplification problem I showed: is this true for all real x (that is,

is it an identity over the reals?)√(x

^{4}) = x^{2}

Myler answered:

As for the √(x⁴)=x² , to me yes it is true. Since it was mentioned in the first example √x²=|x|, I’d just distribute the radical to parts of the radicand and substitute.

√(x²)(x²) = √(x²)√(x²) = |x||x| = x²

Some examples to further verify this is if we substitute x to -2

√(-2⁴) = (-2)²

√16 = 4

4 = 4

So, yep it’s true:]

Thank you for your help, it’s all clear now!

Doctor Rick closed:

Very good, you’re welcome and I hope we’ll hear from you again. It was a good question.

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