Coincidence or hidden logic?

Me and my friend are revising for the [UK] STEP exams, and came across this problem:


Given:

   p² + 4q² + 9r² = 729   and   8p + 8q + 3r = 243

Solve p, q, and r.


Now, this question is actually the 3rd part of one question; so I was convinced that we had to make use of one of our previous answers for that question, in order to solve this one (as is usually the case on STEP papers).

However, my friend used the fact that 243 was a factor of 729, and solved it as so:


      3 * (8p + 8q + 3r)  =  3 * 243
      24p + 24q + 9r       =  729

   Seeing as  p² + 4q² + 9r²  also equals 729, then:

      p² + 4q² + 9r²  =  24p + 24q + 9r

   We then move everything from the RHS to the LHS:

      p² - 24p  +  4q² - 24q  +  9r² - 9r  =  0

   Now factor out p, q, and r:

      p(p - 24)  +  q(4q - 24)  +  r(9r - 9)  =  0


Now this is where I had a problem. My mate seemed to think that in this case, each term must each equal 0 -- despite my teacher and I both saying that you can't assume that.

However, we had a bit of a surprise when he done it anyway...

      p(p - 24)   = 0  =>  p=0, p=24
      q(4q - 24) = 0  =>  q=0, q=6
      r(9r - 9)    = 0  =>  r=0, r=1

The p=q=r=0 solution does not work, but p=24, q=6, and r=1 is exactly what me and my teacher achieved when we used a [longer] way which *did* refer back to one of our previous questions.


My teacher (who is actually very good at maths) could not see if there was any sort of logic as to how and why this worked; his first thoughts were that it was coincidence, but perhaps it was intentional?

Can anyone see whether or not there's some logic behind why it worked?


Thanks
(sorry for the long post)
LVL 25
InteractiveMindAsked:
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minjiberCommented:
hi there,

it wasn't a coincedence, because if they were any 2 eqn's with 3 unknows this method will work.

since this is true

p^2 + 4q^2 + 9r^2 = 24p + 24q + 9r

and u have infinite solutions, equating each part of the equation is a solution

that is p^2 = 24p
divide by p
p=24

and so on.

this is similar to what ur friend did, and the reason it worked is coz there are infinite solutions and u only need to find one.
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WaterStreetCommented:
I've been away from this for quite a while, but I remember that I couldn't solve for three variables with only two equations.
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ozoCommented:
There are an infinite number of  solutions where not each term equals 0
for example
p=(272+9i)/10
q=11/4
r=(51-108i)/45
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PaulCaswellCommented:
Hi Rob,

I'd re-read the question. Does it say "find a solution" or "find the solution" or something misleading? There is no unique solution as both WaterStreet and ozo have said.

WaterStreet,

Welcome to the only other page with an & in it (I think). :-)

Paul
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WaterStreetCommented:
Hi Paul,

Then there's the other P&R (Puzzles & Riddles) not to be confused with Philosophy & Religion, but some might think of them as the same.

regards
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InteractiveMindAuthor Commented:
But do all of these other [infinite] solutions fit into the original two equations?
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ozoCommented:
Yes, that's what makes them solutions.
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Infinity08Commented:
p² + 4q² + 9r² = 729                                                                                  (1)
8p + 8q + 3r = 243                                                                                     (2)

p = 243/8 - q - 3r/8                                                                                    (2')

(243/8 - q - 3r/8)^2 + 4q^2 + 9r^2 = 729                                                    (1') : (2') in (1)

243(51) - 243(16)q - 729(2)r + 3(16)qr + 5(64)q^2 + 9(65)(r^2) = 0              (1'')

20q^2 + [3r - 243] q + [9(65)r^2 - 729(2)r + 243(51)]/16 = 0                        (1''')

q = ((243 - 3r) +/-  V[(3r - 243)^2 - 5(9(65)r^2 - 729(2)r + 243(51))] )/40
q = ((243 - 3r) +/-  V[- 4(729)r^2 + 8(729)r - 4(729)] )/40
q = ((243 - 3r) +/-  54 * (r - 1) * i )/40                                                         (3)

so, the (only) real solution :

r = 1             only way (3) can be real with real r
q = 6             (3) with r = 1
p = 24           (2') with r = 1 and q = 6

So, the question was not ambiguous if we stay in the real space.



The way your friend solved this is slightly wrong, as he dropped one equation - he should have kept eg. these two :

    24p + 24q + 9r = 729
    p(p - 24)  +  q(4q - 24)  +  r(9r - 9)  =  0

And then he would have had certainty that his solution is correct (or at least one of the correct solutions). Then he only had to establish that this system has only 1 real solution to be sure that his solution is the one and only correct one.


So, to summarise : it's NOT a coincidence that he arrived at the solution this way, BUT he should have been more complete ...
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BigRatCommented:
>>There are an infinite number of  solutions ...   - ozo

Correct, ozo. The standard way is to introduce a lambda multiplier and multiply and add the two equations together (the example uses lambda=1) and then say that, because lambda is arbitary, the terms must go out one by one. Hence the solution found is when lambda is 1.
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infexCommented:
Just a little remarcable thing :

729 = 3 ^ 6 = 9 ^ 3 = 27 ^ 2
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Harisha M GCommented:
Your question: Coincidence or hidden logic?
My answer: Coincidence.

Consider this set of equations: ( I have initially assumed a, b, c to be 1, 2 and 3 )

a² + b² + c² = 14; a + b + c = 6

3a² - 7a + 3b² - 7b + 3c² - 7c = 0
a(3a - 7) + b(3b - 7) + c(3c - 7) = 0

Clearly, according to your friend, a = b = c = 2.333..., which doesn't fit into the original equation.
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Harisha M GCommented:
> Welcome to the only other page with an & in it (I think). :-)

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