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

[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.