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

[Einstein's Cat]

Say if
C = square root (a^2 + b^2),
How could one make the term "b" the subject of the equation? Thank you for your help.

 

[ProfuselyQuarky]

What's the opposite of a square root?

 

[Ssnow]

you can start doing the square of your equation (both side), so $C^{2}=a^{2}+b^{2}$.

 

[ProfuselyQuarky]

you can start doing the square of your equation (both side), so $C^{2}=a^{2}+b^{2}$.

The OP was supposed to answer

 

[Ssnow]

Only an input ... , now it is your turn ...

 

[ProfuselyQuarky]

$C^2=a^2+b^2$
$b^2=C^2-a^2$

Now, @Einstein's Cat what's the last step?

 

[symbolipoint]

Let me play dim-witted for a few seconds:
There is a variable and an expression and these are related by equality. How would any of that be or have a "subject"?

Now let me play as if I know what is really wanted:
Just use INVERSE operations to undo what has been done to b, so you solve the equation for b, to have a formula to show what is an expression for b.

 

[ProfuselyQuarky]

Let me play dim-witted for a few seconds:
There is a variable and an expression and these are related by equality. How would any of that be or have a "subject"?

Aw, you knew what he/she meant

 

[Einstein's Cat]

Jesus! I just realized how stupid my question was; numerous apologies for diminishing your lives

 

[ProfuselyQuarky]

Jesus! I just realized how stupid my question was; numerous apologies for diminishing your lives

No prob! It was amusing.

 

[symbolipoint]

Let me play dim-witted for a few seconds:
There is a variable and an expression and these are related by equality. How would any of that be or have a "subject"?

Now let me play as if I know what is really wanted:
Just use INVERSE operations to undo what has been done to b, so you solve the equation for b, to have a formula to show what is an expression for b.

Aw, you knew what he/she meant

Jesus! I just realized how stupid my question was; numerous apologies for diminishing your lives

No prob! It was amusing.

Maybe I should not have said the way I did say.
The way "subject" is used seems to expect that the equation is a sentence and works according to the grammar of the language, but such is for HUMAN languages, and not for an extremely precise written language such as "Algebra". According to that, treating a number within an equation as any element of a sentence or as any part of speech part of sentence structure seems out-of-place; but still the question YOU ask suggests that you want to solve the equation for b, and to have a formula for b. Maybe this is a cultural thing; or maybe I have not seen a few new trends in mathematics education - I thought I would have seen this already if it were a normal way of discussing (subject, as part of numbers in an equation).

 

[SteamKing]

Jesus! I just realized how stupid my question was; numerous apologies for diminishing your lives

It's OK. You're only a Cat.

 

[Mark44]

you can start doing the square of your equation (both side), so $C^{2}=a^{2}+b^{2}$.

$C^2=a^2+b^2$
$b^2=C^2-a^2$

Now, @Einstein's Cat what's the last step?

If the OP hadn't already seen these posts I would have deleted them. Giving too much help is a violation of the rules of this forum.

 

[micromass]

Jesus! I just realized how stupid my question was; numerous apologies for diminishing your lives

Can you post what you think is the solution. It is very likely that what you think is incorrect.

 

[ProfuselyQuarky]

If the OP hadn't already seen these posts I would have deleted them. Giving too much help is a violation of the rules of this forum.

Sorry! :(

 

[Einstein's Cat]

Can you post what you think is the solution. It is very likely that what you think is incorrect.

Denouncing the mathematical abilities of others is obscenely unnecessary and by doing so, your personal interpretation of the extent of your ability is not enhanced. Thus, I advise you cease denouncing other's abilities and instead designate time to advancing your own.

 

[micromass]

Denouncing the mathematical abilities of others is obscenely unnecessary and by doing so, your personal interpretation of the extent of your ability is not enhanced. Thus, I advise you cease denouncing other's abilities and instead designate time to advancing your own.

What is your solution?

 

[Einstein's Cat]

What is your solution?

The square root of c^2 - a^2

 

[micromass]

The square root of c^2 - a^2

Exactly like I thought. That is the wrong solution. This is why I asked. Instead of berating me for being arrogant, you should trust my skills in knowing when a student like has a flaw in his/her thinking! I rarely go wrong there.

To give a hint where you go wrong: $\sqrt{b^2} \neq b$.

 

[ProfuselyQuarky]

Mmm… @Einstein's Cat even if you’re personally annoyed with @micromass for showing how you are wrong, it would only benefit yourself to try at it again

 

[Einstein's Cat]

I

Mmm… @Einstein's Cat even if you’re personally annoyed with @micromass for showing how you are wrong, it would only benefit yourself to try at it again

I would like to clarify that I very much appreciate the provided help, it is however the dismissive attitude that was adopted that irritated me. Also, I am struggling with the concept of how root b^2 is not equal to b. Once again any help would be appreciated.

 

[micromass]

I

I would like to clarify that I very much appreciate the provided help, it is however the dismissive attitude that was adopted that irritated me. Also, I am struggling with the concept of how root b^2 is not equal to b. Once again any help would be appreciated.

You should be happy I posted, otherwise you would have a wrong solution and you would not have learned something new! Dislike me all you want, but I did provide you valuable help!

 

[berkeman]

Also, I am struggling with the concept of how root b^2 is not equal to b

It's equal to more than just "b". Think for a bit and work some examples... I bet you can figure it out.

 

[symbolipoint]

Can you post what you think is the solution. It is very likely that what you think is incorrect.

Denouncing the mathematical abilities of others is obscenely unnecessary and by doing so, your personal interpretation of the extent of your ability is not enhanced. Thus, I advise you cease denouncing other's abilities and instead designate time to advancing your own.

Hey! I agree with micromass about this.

 

[Mark44]

Can you post what you think is the solution. It is very likely that what you think is incorrect.

Denouncing the mathematical abilities of others is obscenely unnecessary and by doing so, your personal interpretation of the extent of your ability is not enhanced. Thus, I advise you cease denouncing other's abilities and instead designate time to advancing your own.

We get a lot of people who misunderstand square roots. micromass was extrapolating from a very large sample to make a prediction about your answer.

The square root of c^2 - a^2

As an equation, that would be $b = \sqrt{c^2 - a^2}$. It's not wrong, per se, but it is incomplete.

 

[Einstein's Cat]

We get a lot of people who misunderstand square roots. micromass was extrapolating from a very large sample to make a prediction about your answer.

As an equation, that would be $b = \sqrt{c^2 - a^2}$. It's not wrong, per se, but it is incomplete.

May I inquire why it is complete

 

[Samy_A]

$5=\sqrt{3²+( -4 )^2}$

 

[Mark44]

May I inquire why it is complete

What I said was that your solution was incomplete, the opposite of complete. There are two solutions to the equation, not the one that you showed.

 

[Einstein's Cat]

What I said was that your solution was incomplete, the opposite of complete. There are two solutions to the equation, not the one that you showed.

Sorry about that! I meant "incomplete" but typed otherwise and I now understand.

 

[ProfuselyQuarky]

Sorry about that! I meant "incomplete" but typed otherwise and I now understand.

So you know what the answer is now?

 

[Einstein's Cat]

So you know what the answer is now?

B= square root( +/- c^2 - +/- a^2) I believe

 

[Samy_A]

B= square root( +/- c^2 - +/- a^2) I believe

Still an incomplete answer ...

As $(-c)²=(+c)²$, this answer is not different from your previous one.

 

[ProfuselyQuarky]

Try again. You can do it!

 

[ProfuselyQuarky]

$\sqrt {c^2-a^2}$ has two answers. Think how the same equation (with a square root) can have two solutions.

 

[Samy_A]

$\sqrt {c^2-a^2}$ has two answers. Think how the same equation (with a square root) can have two solution.

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.

 

[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