Simultaneous equations

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  • #1
Mentallic
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I need to solve these 2 equations:

[tex]r^2-a^2=\sqrt{(r^2-a^2)(R^2-(b-d)^2)}[/tex]

[tex]\sqrt{(r^2-a^2)(R^2-(d-b)^2)}=R^2-d^2+bd-ab+ad[/tex]

Now wait, before you run off thinking "screw that - too many letters, too much hassle" I don't actually want you to solve this for me, but I have a question about it.

Firstly, r, R and d are constants and I'm looking to solve for a and b so what I essentially want is:

[tex]a=f(r,R,d)[/tex] and [tex]b=f(r,R,d)[/tex] (I'm not even sure if I have expressed what I want properly, I just took a best guess at this. It translates into "a as a function in terms of r, R and d").

What I thought I'd try is, instead of expanding one equation and solving a quadratic in terms of one of the variables a or b, I'll eliminate the [tex]\sqrt{(r^2-a^2)(R^2-(d-b)^2)}[/tex] so what I end up having is:

[tex]R^2-d^2+bd-ab+ad=r^2-a^2[/tex]

but then this doesn't look like it's helping me solve for either variable. Is the only way I can solve for a and b by going through the long painful process of using the quadratic formula? Any other suggestions on a simpler approach I could take?
 

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  • #2
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Set r2 - a2 = R2 - d2 + bd - ab + ad

The radicals in the two equations are equal; the only difference is that one has a factor of (b - d)2 inside, while the other has a factor of (d - b)2 inside, and these two factors are equal.
 
  • #3
Mentallic
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Yes I realized that and I already had done what you suggested. It's near the end of my OP.

I guess I can solve for either a or b in this 3rd equation and then substitute that back into the 1st or 2nd equation. It doesn't look impossible to solve, but I know it's going to be a pain...

Stay tuned to read about my inevitable failure.
 
  • #4
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Yep, as I thought - my only hope was that many crucial terms would cancel nicely, but there wasn't enough of that happening to help me solve these equations. I have resulted in a 5th order equation in the variable a, so it's quite fruitless to keep going.

It seems like all my most recent math problems have been cut short of being solved simply by becoming too overwhelming to work with. Maybe I should quit with giving myself problems to solve.
 
  • #5
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I need to solve these 2 equations:

[tex]r^2-a^2=\sqrt{(r^2-a^2)(R^2-(b-d)^2)}[/tex]

[tex]\sqrt{(r^2-a^2)(R^2-(d-b)^2)}=R^2-d^2+bd-ab+ad[/tex]

Doesn't look so bad.
The first equation implies r^2-a^2 = R^2-(b-d)^2
[I'm not sure whether the negative of LHS is also implied, let's leave that for another day.]

The second equation when you use the first implies r^2-a^2 = R^2 -d^2+bd-ab+ad
[Once again are there any issues with negatives???]

Equating the RHS of these two equations, and cancelling R^2 you end up with (a-b)(d-b)=0

That is solved if a=b or d=b

Easy enough to check, if you substitute b for a for example.

Is this what you wanted?
 
  • #6
Mentallic
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Doesn't look so bad.
I beg to differ.

The first equation implies r^2-a^2 = R^2-(b-d)^2
How so?

The second equation when you use the first implies r^2-a^2 = R^2 -d^2+bd-ab+ad
You'd be the 3rd person in this thread that has realized this and posted it :tongue:

I'm not sure whether the negative of LHS is also implied, let's leave that for another day.

Once again are there any issues with negatives???
Only a will be negative (although it doesn't have to be the case for the question), the only real restrictions are [itex]r,R>0[/itex]

Equating the RHS of these two equations, and cancelling R^2 you end up with (a-b)(d-b)=0
This would be the case if you hadn't incorrectly made the equality [tex]\sqrt{(r^2-a^2)(R^2-(b-d)^2)}=R^2-(b-d)^2[/tex] I think?

That is solved if a=b or d=b

Easy enough to check, if you substitute b for a for example.

Is this what you wanted?
[itex]a=b[/itex] or [itex]d=b[/itex] iff [itex]r,R,b=0[/itex] and this is a pretty boring case and makes solving the problem quite futile.

EDIT: I now see what you did in the first equation, I'm going to check over everything you've done again to see where else the problem may lie.
 
  • #7
Mentallic
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Haha it seems everything you've done was correct.

But my statement still holds that [itex]a\neq b[/itex] and [itex]b\neq d[/itex] so that can only mean my equations are wrong.

oops :blushing:

Thanks bikengr! :smile:
 
  • #8
D H
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The first equation implies r^2-a^2 = R^2-(b-d)^2
How so?
[tex]r^2-a^2=\sqrt{(r^2-a^2)(R^2-(b-d)^2)}[/tex]

Square both sides:

[tex]\left(r^2-a^2\right)^2=(r^2-a^2)(R^2-(b-d)^2)[/tex]

If [tex]r^2-a^2 \ne 0[/tex] you can clear that common factor, leaving

[tex]r^2-a^2=(R^2-(b-d)^2[/tex]

and, as bikengr showed, this eventually leads to [tex](a-b)(d-b)=0[/tex].


However, that ignores the other solution that results from [tex]r^2-a^2=0[tex].
 
  • #9
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In my frustration at login difficulties, and also general sleep deprivation, I didn't state the solution that I found all that clearly:

The quantities must be chosen to satisfy r^2-a^2 = R^2-(b-d)^2
Furthermore, either a=b or d=b
[I am not precluding other solutions due to positive/negative square roots, and also restrictions to allow the square roots to be real. I simply haven't looked into them.]

These two relations in 5 quantities will generally allow you to choose three variables relatively freely, and solve for two others. For example select d, b and r arbitrarily. The second relation tells you the value of a. The first relation lets you solve for R. Note that there are limits to the choices of d,b,r if you want R to be real. R will not be real if (b-d)^2 and r^2 are small, and a^2 is large.
 
  • #10
D H
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The quantities must be chosen to satisfy r^2-a^2 = R^2-(b-d)^2
Furthermore, either a=b or d=b
Not necessarily. In finding that you implicitly divided by (r2 - a2). This division is not valid if a2=r2. Setting a2=r2 leads to a new family of solutions.
 
  • #11
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Not necessarily. In finding that you implicitly divided by (r2 - a2). This division is not valid if a2=r2. Setting a2=r2 leads to a new family of solutions.
Just so. I tried to be quite explicit that I probably did not have all solutions. My main aim was to break the logjam, in which proposer was tied in knots about high order polynomials. I think I have done so, however cavalierly, and probably won't be adding more.
 
  • #12
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Equating the RHS of these two equations, and cancelling R^2 you end up with (a-b)(d-b)=0
I followed you, up to this point. But how do you cancel R^2? We don't know the relationship between r and R, do we?
 

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