Negative roots when simplifying surds

NeedingWD40

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Something that puzzles me:

The square roots of 100 are + or -10, yes? So why when one simplifies surds are we able to ignore negative roots?

e.g. ⎷300 -⎷48 =⎷100x3 -⎷16x3 = 10⎷3 - 4⎷3 = 6⎷3

what about the negative root of 100, and the negative root of 16?

Thanks in advance for your help.
 

Dr.Peterson

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Something that puzzles me:

The square roots of 100 are + or -10, yes? So why when one simplifies surds are we able to ignore negative roots?

e.g. ⎷300 -⎷48 =⎷100x3 -⎷16x3 = 10⎷3 - 4⎷3 = 6⎷3

what about the negative root of 100, and the negative root of 16?

Thanks in advance for your help.
When we write the radical sign, it stands for the square root function, which means it has only one value. We call this the principal value, namely the non-negative root. The symbol does not represent both roots, but just one.

So although 100 has two square roots, ⎷100 is just 10. And that is what you are given when you simplify a surd, so you ignore the negative roots.

Have you noticed that if you chose always to keep both roots, an expression like yours would have more and more values? To take a simpler example, ⎷9 -⎷4 would have to be (±3) - (±2), with no requirement that the two signs are correlated, so its values would be 3 - 2 = 1, 3 - -2 = 5, -3 - 2 = -5, and -3 - -2 = -1. More radicals would make it even more confusing! That's why we insist on a square root function.
 

Jomo

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Something that puzzles me:

The square roots of 100 are + or -10, yes? So why when one simplifies surds are we able to ignore negative roots?

e.g. ⎷300 -⎷48 =⎷100x3 -⎷16x3 = 10⎷3 - 4⎷3 = 6⎷3

what about the negative root of 100, and the negative root of 16?

Thanks in advance for your help.
By definition, \(\displaystyle \sqrt{x}\) is > 0 providing x > 0. If x<0, then \(\displaystyle \sqrt{x}\) has no real solution.
 

NeedingWD40

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When we write the radical sign, it stands for the square root function, which means it has only one value. We call this the principal value, namely the non-negative root. The symbol does not represent both roots, but just one.

So although 100 has two square roots, ⎷100 is just 10. And that is what you are given when you simplify a surd, so you ignore the negative roots.

Have you noticed that if you chose always to keep both roots, an expression like yours would have more and more values? To take a simpler example, ⎷9 -⎷4 would have to be (±3) - (±2), with no requirement that the two signs are correlated, so its values would be 3 - 2 = 1, 3 - -2 = 5, -3 - 2 = -5, and -3 - -2 = -1. More radicals would make it even more confusing! That's why we insist on a square root function.
Ah! That's why the formula for solving a quadratic equation explicitly says + or -⎷(b2-4ac), otherwise only the positive root would be found. Thank you very much, I'm clearer now :)
 

JeffM

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Something that puzzles me:

The square roots of 100 are + or -10, yes? So why when one simplifies surds are we able to ignore negative roots?

e.g. ⎷300 -⎷48 =⎷100x3 -⎷16x3 = 10⎷3 - 4⎷3 = 6⎷3

what about the negative root of 100, and the negative root of 16?

Thanks in advance for your help.
You are absolutely correct that

\(\displaystyle (-\ 10)^2 = 100 \text { and } (+\ 10)^2 = 100.\)

But the technical definition of the square root function of a non-negative real number is:

\(\displaystyle \text {If } a \ge 0, \text { then } \sqrt{a} \ge 0 \text { and } \sqrt{a} * \sqrt{a} = a.\)

That is why if we get an equation that looks like

\(\displaystyle x^2 = 100 \implies x = \pm \sqrt{100} \implies x = 10 \text { or } x = -\ 10.\)

That plus or minus sign is there for a reason.
 

NeedingWD40

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By definition, \(\displaystyle \sqrt{x}\) is > 0 providing x > 0. If x<0, then \(\displaystyle \sqrt{x}\) has no real solution.
Thank you, I'm clearer now.
 

NeedingWD40

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You are absolutely correct that

\(\displaystyle (-\ 10)^2 = 100 \text { and } (+\ 10)^2 = 100.\)

But the technical definition of the square root function of a non-negative real number is:

\(\displaystyle \text {If } a \ge 0, \text { then } \sqrt{a} \ge 0 \text { and } \sqrt{a} * \sqrt{a} = a.\)

That is why if we get an equation that looks like

\(\displaystyle x^2 = 100 \implies x = \pm \sqrt{100} \implies x = 10 \text { or } x = -\ 10.\)

That plus or minus sign is there for a reason.
So, I'm thinking, if a problem includes the square root function, but no plus or minus, you assume the primary root, however, if you create a square root when solving a problem you need to be aware of the negative root in the context of the question....
 

tkhunny

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So, I'm thinking, if a problem includes the square root function, but no plus or minus, you assume the primary root, however, if you create a square root when solving a problem you need to be aware of the negative root in the context of the question....
"problem includes ... no plus or minus" - Not a fan.

What is \(\displaystyle \sqrt{4}\)? It is 2

I do not ever recall seeing such a simplification problem as: What is \(\displaystyle \pm\sqrt{4}\)? It is +/- 2.

Let the context be your guide.

Don't make up things that don't exist. \(\displaystyle \sqrt{4} = 2\) and that is NOT the same as "+/-2".
Don't miss things that do exist. \(\displaystyle Solve\;x^{2} = 4\). \(\displaystyle x = +2\;or\;x = -2\) and that is the same as "x = +/-2".

Also, keep your eyes on that "+/-" symbol. It means several different things in different contexts.
 

Otis

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I'm not sure whether you've yet learned anything about solving algebraic equations or the concept of a function (you posted on the Pre-Algebra board), but here's my comments, anyways.

When you are given a surd, it always represents the positive root (only is exception is √0).

When you are solving an equation, sometimes you need to introduce surds (like the algebraic step of taking the square root of each side of an equation). In this case, you need to consider both the positive and negative root. Otherwise, you might miss solution(s).

As Jeff wrote, we express a positive root as √n, and when we need to express the negative root we write -√n. Cheers :cool:
 

JeffM

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So, I'm thinking, if a problem includes the square root function, but no plus or minus, you assume the primary root, however, if you create a square root when solving a problem you need to be aware of the negative root in the context of the question....
Please read the other responses to this post because they are good and will give you different perspectives.

\(\displaystyle \sqrt{a} \ge 0\) by definition if we are dealing with real numbers.

So, we do not assume that the square root is the primary root (meaning a non-negative number), we know that it is a non-negative number by definition.

However, if the square root of a is positive, we know that the square of the square root's additive inverse is also a. In notation,

\(\displaystyle \sqrt{a} * \sqrt{a} = a = (-\ \sqrt{a}) * (-\ \sqrt{a}).\)

So if we have an equation like

\(\displaystyle x^2 = c \ge 0\),

we have no way to know from the equation itself whether

\(\displaystyle x = \sqrt{c} \text { or } x = - \ \sqrt{c}.\)

It may be that either solution works or that only one of the two works, but that must be decided based on information not contained in the equation itself.

Uncertainty about which answer applies does not mean that there is any uncertainty about what the surd means. To clarify where the uncertainty lies we say

\(\displaystyle x = \pm \sqrt{c}.\)

The sign of the surd is certain. It is the sign of x that is uncertain.
 
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pka

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Something that puzzles me:
The square roots of 100 are + or -10, yes? So why when one simplifies surds are we able to ignore negative roots?
e.g. ⎷300 -⎷48 =⎷100x3 -⎷16x3 = 10⎷3 - 4⎷3 = 6⎷3
what about the negative root of 100, and the negative root of 16?
I have had issues with this my whole active research & teaching life.
I had problems with students in my complex variables classes with the use of the radical: \(\displaystyle \sqrt{z}\).
Now I belong to a school of analyst who think that the symbol \(\displaystyle \sqrt~\) must be applied only to a non-negative real number.
So that means \(\displaystyle \sqrt{-1}\) has no meaning whatsoever. So what is \(\displaystyle \bf{i}~?\)
It is so simple, \(\displaystyle \bf{i}\) is a solution of the equation of the equation \(\displaystyle z^2+1=0\)
Now clearly the number \(\displaystyle 100\) has two square roots, \(\displaystyle \pm 10\).
BUT \(\displaystyle \sqrt{100}=10\) that is one number. Moreover, \(\displaystyle -\sqrt{100}=-10\) that is one number.
Thus it is absolutely incorrect to write that \(\displaystyle \sqrt{100}=\pm 10\)

 

JeffM

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I have had issues with this my whole active research & teaching life.
I had problems with students in my complex variables classes with the use of the radical: \(\displaystyle \sqrt{z}\).
Now I belong to a school of analyst who think that the symbol \(\displaystyle \sqrt~\) must be applied only to a non-negative real number.
So that means \(\displaystyle \sqrt{-1}\) has no meaning whatsoever. So what is \(\displaystyle \bf{i}~?\)
It is so simple, \(\displaystyle \bf{i}\) is a solution of the equation of the equation \(\displaystyle z^2+1=0\)
Now clearly the number \(\displaystyle 100\) has two square roots, \(\displaystyle \pm 10\).
BUT \(\displaystyle \sqrt{100}=10\) that is one number. Moreover, \(\displaystyle -\sqrt{100}=-10\) that is one number.
Thus it is absolutely incorrect to write that \(\displaystyle \sqrt{100}=\pm 10\)

I like this convention that the surd can only be applied to non-negative real numbers. It would avoid a lot of confusion.

Unfortunately, conventions are necessarily extra-individual. And the surd is frequently used in a broader sense.

Anyway, I do like your school's approach. But is there not an inconsistency when it comes to

\(\displaystyle \sqrt[3]{-\ 8}\)

or is that prohibited as well. If so, is there a convenient notation to deal with it?
 

pka

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Anyway, I do like your school's approach. But is there not an inconsistency when it comes to
\(\displaystyle \sqrt[3]{-\ 8}\)
or is that prohibited as well. If so, is there a convenient notation to deal with it?
No, because the any real number has a real root if the index is odd. Example \(\displaystyle \large\sqrt[5]{-32}=-2\)
 

JeffM

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No, because the any real number has a real root if the index is odd. Example \(\displaystyle \large\sqrt[5]{-32}=-2\)
Ahh so your school's rule is

\(\displaystyle \sqrt[a]{b} \text { is defined only if (1) } a \in \mathbb Z \text {, (2) } a \ge 1 \text {, (3) } \ b \in \mathbb R,\)

\(\displaystyle \text {and (4) } a \text { even } \implies b \ge 0.\)
 

pka

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Ahh so your school's rule is
\(\displaystyle \sqrt[a]{b} \text { is defined only if (1) } a \in \mathbb Z \text {, (2) } a \ge 1 \text {, (3) } \ b \in \mathbb R,\)
\(\displaystyle \text {and (4) } a \text { even } \implies b \ge 0.\)
.
At meetings we would have new jokes about exponents. Prof Paterson has address this above.
The objection is this \(\displaystyle \sqrt{-16}=\pm 4\bf{i}\). I dare say you can some such in many basic mathematics textbooks.
It is true that many in my tradition are hyper sensitive to vocabulary abuse.

There are four fourth roots of \(\displaystyle -16\bf{i}\) but there is just one answer to \(\displaystyle \large\sqrt[4]{-16\bf{i}}=~?\)
 

lookagain

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NeedingWD40 said:
e.g. ⎷300 -⎷48 =⎷100x3 -⎷16x3 = 10⎷3 - 4⎷3 = 6⎷3
The above is incorrect. You need grouping symbols, such as below. I used asterisks for the products.


e.g.

⎷300 - ⎷48 =

⎷(100*3) - ⎷(16*3) =

10⎷3 - 4⎷3 =

6⎷3
 
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topsquark

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.
At meetings we would have new jokes about exponents. Prof Paterson has address this above.
The objection is this \(\displaystyle \sqrt{-16}=\pm 4\bf{i}\). I dare say you can some such in many basic mathematics textbooks.
It is true that many in my tradition are hyper sensitive to vocabulary abuse.

There are four fourth roots of \(\displaystyle -16\bf{i}\) but there is just one answer to \(\displaystyle \large\sqrt[4]{-16\bf{i}}=~?\)
When I see stuff like "i"s under a radical I go straight over to the exponential format.
\(\displaystyle \sqrt[4]{-16 i } = \sqrt[4]{16} \cdot \sqrt[4]{-i}\)

\(\displaystyle = 2 \cdot \left ( e^{ 3i \pi/2 + 2 i \pi n } \right ) ^{1/4}\)

\(\displaystyle = 2 \cdot e^{3i \pi /8 + i \pi n / 2}\)

\(\displaystyle = 2 \cdot e^{3i \pi / 8} \cdot i^n\)

\(\displaystyle = 2 (i ^n) \left ( cos ( 3 \pi /8 ) + i~sin (3 \pi / 8) \right )\)
where n is an integer.

Perhaps it's a bit messy but I don't have to think so hard about any negative signs.

-Dan
 

pka

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When I see stuff like "i"s under a radical I go straight over to the exponential format.
\(\displaystyle \sqrt[4]{-16 i } = \sqrt[4]{16} \cdot \sqrt[4]{-i}\)
-Dan
I did a very poor job explaining our objections. No one would ever need to use \(\displaystyle \sqrt[4]{-16 i }. \)
On the other hand there could be a need to solve \(\displaystyle z^4+16\bf{i}=0 \)
There are four roots of that equation. If \(\displaystyle \rho=2\exp\left(\frac{-\pi\bf{i}}{8}\right) \) then \(\displaystyle \rho^4+16\bf{i}=0 \), in other words one solution.
Let \(\displaystyle \zeta=\exp\left(\frac{2\pi\bf{i}}{4}\right) \ \) now the collection \(\displaystyle \rho\cdot\zeta^k,~k=1,2,3 \) are the other three roots(solutions).

 

NeedingWD40

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Thanks all, one clear message I've got is I need to clarify my sometimes foggy thought processes!

So, to attempt to summarise:

⎷a is always positive.
The solution to an equation involving a square root may have two solutions:⎷a or -⎷a
The context of the question should clarify if the negative of the root gives a valid solution.
 

JeffM

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Thanks all, one clear message I've got is I need to clarify my sometimes foggy thought processes!

So, to attempt to summarise:

⎷a is always positive.
The solution to an equation involving a square root may have two solutions:⎷a or -⎷a
The context of the question should clarify if the negative of the root gives a valid solution.
That is very good. I'd make one slight emendation to your third point. I'd phrase it as

The context of the equation should clarify if both solutions are valid or else which one of the two is valid.

I would not assume that the negative solution is the only one that may not apply. In some cases, what is classified as positive and negative is arbitrary. It is of course true that we usually classify in a way that ensures a valid answer is non-negative, but that is not guaranteed.

Very minor point. You seem to have it. Good job.
 
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