Solving for b in C = sqrt(a^2 + b^2)

[ProfuselyQuarky]

Hmm, I'm not sure this is totally correct (although I know what you mean).

By convention, $\sqrt a$ represents the positive square root of $a$, where $a$ is a non negative real number.
If $a > 0$, there are two real number satisfying the equation $x² =a$, namely [deleted].

Yes, that's what I meant, although I think you gave away the answer to the OP . . .

 

[Samy_A]

Yes, that's what I meant, although I think you gave away the answer to the OP . . .

You are right, I edited my post.

 

[epenguin]

About the spat, I think students should always post their solutions (see my sig.)
In the matter of courtesy, isn't it a bit much when someone helps to solution and you say I have got it but don't say what it is?
Would you think that is very satisfying to the helper, throwing out help and solutions that fall into an information black hole?
Plus we often 'know' when a student has got it wrong.
So it is just missing out on possible useful help not to give it, I even wonder why this happens,
And even if the student has got the right answer ,it is missing another thing.
Learning and understanding is not about getting right answers!
The helper will often have something to add to a right answer.
There is context. The method or theme is just an aspect of something else, can be carried over to something else. Can be simplified. Can be looked in a different way. Looked at in some way it could have been more obvious at the beginning. Etc. Sometimes we can furnish this, so not giving an answer can be missing something.

Another comment is that solving y = √x , to find x is not really doing an algebraic operation. It is not a calculation, it is merely the recognition of what √ means. It just means y is the number whose square is x. Or to to say again √x is the number, or better, a number, whose square is x. (it is just a particular case of the concept 'inverse function'). Having this clear, instead of thinking of it as an algebraic problem, the student could probably have dealt with the first part of the problem. The trouble is this symbol √, makes it a bit mystical.

It is even a bit of a fraud. I mean we get quadratic equations and others reducible to them, and we write this formula involving √ and say, aha, I have a solution! But we don't really - not unless we know how to calculate square roots. Which a lot of us don't. Were you taught square root extraction at school? We just let someone else do it for us. Nowadays with calculators, but before them I used tables, logarithms or slide rule. I don't say one should do anything else in practice, but one should at least understand how it can be done IMO.

 

[jbriggs444]

Or to to say again √x is the number, or better, a number, whose square is x. (it is just a particular case of the concept 'inverse function').

This particular statement misses the crucial point being earlier in this thread. See, for instance, #35 by Samy_A. $\sqrt{x}$ does not denote a number whose square is x. It denotes the number which is not negative and whose square is x.

 

[epenguin]

OK I was doing some overcorrection trying to get that down in a hurry. My essential point is that at least for the first part of the calculation is not doing anything except the recall definition.

 

[Einstein's Cat]

OK I was doing some overcorrection trying to get that down in a hurry. My essential point is that at least for the first part of the calculation is not doing anything except the recall definition.

Am I right in thinking that graphically representing the equation would mean that it would intercept an axis twice and thus two solutions? This is based on my understanding of quadratic equations however and so probably doesn't apply...

 

[Samy_A]

Am I right in thinking that graphically representing the equation would mean that it would intercept an axis twice and thus two solutions? This is based on my understanding of quadratic equations however and so probably doesn't apply...

So, you already gave one solution: $b = \sqrt{c^2 - a^2}$.
What could be the second solution?

 

[Einstein's Cat]

So, you already gave one solution: $b = \sqrt{c^2 - a^2}$.
What could be the second solution?

Perhaps b= - square root(c^2 - a^2) ?

 

[Samy_A]

Perhaps b= - square root(c^2 - a^2) ?

Indeed.

 

[ProfuselyQuarky]

Great job, Einstein's Cat! Square roots can have two solutions: $\pm \sqrt {x}$

Remember that and you'll be just fine :)

 

[micromass]

Great job, Einstein's Cat! Square roots can have two solutions: $\pm \sqrt {x}$

Remember that and you'll be just fine :)

Also remember that $\sqrt{b^2}= |b|$.

 

[ProfuselyQuarky]

Ah, yes, thanks @micromass. You're always so thorough :P

 

[Mark44]

Square roots can have two solutions: $\pm \sqrt {x}$

Quadratic equations can have two solutions, but a square root represents a single number (considering the square root of a nonnegative real number).

A very common mistake that we often see here is the erroneous statement that, say, $\sqrt{4} = \pm 2$.

 

[micromass]

Quadratic equations can have two solutions, but a square root represents a single number (considering the square root of a nonnegative real number).

A very common mistake that we often see here is the erroneous statement that, say, $\sqrt{4} = \pm 2$.

I don't know. It really depend. In multiple sources, they say that $4$ has two square roots, namely $2$ and $-2$. There is only one principal square root though, and that is the one with which we use the symbol $\sqrt{4}=2$.

 

[ProfuselyQuarky]

Apologies if my comment over simplified things. For the OP, confusion can come because many always assume that if $b=a^2$, then $a=\sqrt {b}$. As @micromass stated, it is, more conventionally, if $b=a^2$, then $|a|=\sqrt {b}$. Thus, $b=a^2=(-a)^2$. Nevertheless, $\sqrt {b}=|a|\neq {-|a|}$. That would be the same as saying that $2=-2$ because $2=\sqrt {4}=-2$. Obviously, this is incorrect. For this reason, $b=a^2$ gives you two solutions whereas $a=\sqrt {b}$ has one, generally.

A very common mistake that we often see here is the erroneous statement that, say, $\sqrt{4} = \pm 2$.

I don't think that that is entirely erroneous, though. $2^2=4$ but $(-2)^2=4$, also. The problem can also be a matter of notation, perhaps, because $\sqrt {x}$ usually refers to the positive solution whereas $-\sqrt {x}$ would more specifically refer to the negative solution.

All this talk about such a simple operation is starting to make me confused, too :P

 

[Einstein's Cat]

Cheers for all your help! Apologies for the length it took...

 

[jbriggs444]

I don't think that that is entirely erroneous, though. $2^2=4$ but $(-2)^2=4$, also. The problem can also be a matter of notation, perhaps, because $\sqrt {x}$ usually refers to the positive solution whereas $-\sqrt {x}$ would more specifically refer to the negative solution.

Indeed, it is a matter of notation entirely. In the context of the real numbers, the notation $\sqrt{x}$ always denotes the principle square root, i.e. the non-negative one.

 

[Mark44]

I don't know. It really depend. In multiple sources, they say that $4$ has two square roots, namely $2$ and $-2$. There is only one principal square root though, and that is the one with which we use the symbol $\sqrt{4}=2$.

My concern was with this statement by ProfuselyQuarky:

Square roots can have two solutions: $\pm \sqrt {x}$

I interpret this to mean that, for example, $\sqrt{4} = \pm 2$.
A positive number has two square roots, one positive and one negative, but as you say, $\sqrt{4} = 2$, the principal square root of 4.

 

[ProfuselyQuarky]

My concern was with this statement by ProfuselyQuarky:
I interpret this to mean that, for example, $\sqrt{4} = \pm 2$.
A positive number has two square roots, one positive and one negative, but as you say, $\sqrt{4} = 2$, the principal square root of 4.

I understand that. Sometimes the difference between the principle square root and other negative "solution" is not identified in basic algebra courses, though. Is there any correction that you could make in my last post, however?

 

[micromass]

I understand that. Is there any correction that you could make in my last post, however?

I don't think you said anything strictly wrong, but it could be interpreted very easily as being wrong. I would say that the equation $x^2 = 4$ has two solutions in $\mathbb{R}$, namely $\pm \sqrt{4}$, but that $\sqrt{4}$ indicates the unique positive solution in all cases.

 

[ProfuselyQuarky]

Apologies if my comment over simplified things. For the OP, confusion can come because many always assume that if $b=a^2$, then $a=\sqrt {b}$. As @micromass stated, it is, more conventionally, if $b=a^2$, then $|a|=\sqrt {b}$. Thus, $b=a^2=(-a)^2$. Nevertheless, $\sqrt {b}=|a|\neq {-|a|}$. That would be the same as saying that $2=-2$ because $2=\sqrt {4}=-2$. Obviously, this is incorrect. For this reason, $b=a^2$ gives you two solutions whereas $a=\sqrt {b}$ has one, generally.

Thank you, but I was referring to this paragraph. I want to know if there are any faults with the statements made.

 

[micromass]

That seems to be correct.