Two Equations, Two UnknownsDate: 7/27/96 at 2:17:49 From: Anonymous Subject: Two Equations, Two Unknowns For the equations 4 4 4 x + y = b x + y = a solve for x and y in terms of a and b... I've tried it this way: 1. from the second equation : a-x = y 2. subs into the first and I get the polynomial equation: 4 3 2 3 4 4 2x - 4x + 6ax + - 4a x + a - b = 0 but I could not solve it... I know that there have to be 4 answers and the sum of them must be 2 and the product is 0.5 (a^4 - b^4) but then I get stuck. Please help me. Date: 7/27/96 at 14:38:15 From: Doctor Pete Subject: Re: Two Equations, Two Unknowns Let R = a^4 + b^4 , S = 3a^2 +- 2*Sqrt[2*R] . (the "+-" symbol is the usual 'plus or minus' notation.) Then {(x,y)} = {((a+Sqrt[-S])/2, (a-Sqrt[-S])/2)} . Since the original system is symmetric in x and y, I have not written out the full solution set (which should be obvious). Thus, we have a total of four possible solutions, two of which arise due to symmetry. This result was obtained through Mathematica, though I am fairly certain it can be solved by hand, looking at the simplicity of the solution. To be honest, I haven't really thought about this problem; if you're more interested in how to obtain the solution rather than the result, then I might suggest trying to solve the above quartic in x by assuming that it has a factorization of the form (x^2 + px + q)(x^2 + rx + s) = 0, and then finding p, q, r, and s by expanding and equating coefficients. If successful (I haven't worked it out yet), then you get two quadratics (which, looking from the solution, will most likely be related) which are easily solved using the quadratic formula. Then again, it might not work.... ;) Will post again if I figure it out. -Doctor Pete, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 7/28/96 at 0:55:37 From: Anonymous Subject: Re: Two Equations, Two Unknowns why do you let > R = a^4 + b^4 , > S = 3a^2 +- 2*Sqrt[2*R] ? I don't quite understand. And how do you solve for x and y from these two equations? Thanks again! Date: 7/28/96 at 17:7:15 From: Doctor Pete Subject: Re: Two Equations, Two Unknowns Well, last time I mailed you, I was only able to find the solution, not the *method*, using a computer program called Mathematica. So R and S are only labels to make writing the solution easier. I'm very sorry for the confusion. In any case, I have found a means of solving these equations. From equation 2, x+y = a ==> y = a-x substituting into equation 1 gives x^4 + (a-x)^4 = b^4 ==> 2x^4 - 4ax^3 + 6a^2 x^2 - 4a^3 x + a^4 - b^4 = 0 ==> x^4 - 2ax^3 + 3a^2 x^2 - 2a^3 x + (a^4 - b^4)/2 = 0 , as you've noted already. Now, here's the nasty part. We will factor the above 4th degree polynomial by assuming that it factors into two quadratics; that is, (x^2 + px + q)(x^2 + rx + s) = x^4 - 2ax^3 + 3a^2 x^2 - 2a^3 x + (a^4 - b^4)/2 = 0 , and our goal then is to multiply out the two quadratics on the left and solve for the unknowns p, q, r, and s by equating the coefficients on the left with those on the right. So (x^2 + px + q)(x^2 + rx +s) = x^4 + (p+r)x^3 + (pr+q+s)x^2 + (ps+ qr)x + qs , and if we match up the coefficients, we see that p + r = -2a [equation 1] pr + q + s = 3a^2 [equation 2] ps + qr = -2a^3 [equation 3] qs = (a^4 - b^4)/2 . [equation 4] Now, at first this seems even more difficult than solving the original problem, since there are more equations, but it's not. Take the first two equations in this system; p + r = -2a ==> r = -(2a+p) , pr + q + s = 3a^2 ==> pr = 3a^2 - (q+s) . Substituting the first into the second, we eliminate r and solve for p: -(2a+p)p = 3a^2 - (q+s) ==> p^2 + 2ap + 3a^2 - (q+s) = 0 ==> p = (-2a +- Sqrt[4a^2 - 4(3a^2 - (q+s))])/2 = -a +- Sqrt[a^2 - 3a^2 + q+s] = -a +- Sqrt[-2a^2 + q+s] . This gives us two solutions. Note that p and r must have the same solution set (similarly, q and s also have the same solution set), because the quadratics they represent are interchangeable. This is important, because from this we know that either choice of p will work. So we may let p = -a + Sqrt[-2a^2 + q+s] , and thus r = -a - Sqrt[-2a^2 + q+s] . (Note this works, since p+r = -2a, which is equation 1.) Now, take equation 3, and substitute: ps + qr = -2a^3 ==> (-a + Sqrt[-2a^2 + q+s])s + (-a - Sqrt[-2a^2 + q+s])q = -2a^3 ==> (-a)(q+s) + (s-q)Sqrt[-2a^2 + q+s] = -2a^3 ==> (s-q)Sqrt[-2a^2 + q+s] = -2a^3 + a(q+s) (s-q)Sqrt[-2a^2 + q+s] = a(-2a^2 + q+s) . Now, note that if q+s = 2a^2, both sides of this equation will be zero. Thus q+s = 2a^2 is a condition that solves equation 3. Thus p = -a + Sqrt[-2a^2 + q+s] = -a + Sqrt[-2a^2 + 2a^2] = -a r = -a - Sqrt[-2a^2 + q+s] = -a - Sqrt[-2a^2 + 2a^2] = -a, so p and r both equal -a. Finally, equation 4 allows us to solve for q and s: qs = (a^4 - b^4)/2 q+s = 2a^2 ==> s = 2a^2 - q ==> q(2a^2 - q) = (a^4 - b^4)/2 ==> q^2 - 2a^2 q + (a^4 - b^4)/2 = 0 ==> q = (2a^2 +- Sqrt[4a^4 - 4(a^4-b^4)/2])/2 = a^2 +- Sqrt[a^4 - (a^4-b^4)/2] = a^2 +- Sqrt[(a^4+b^4)/2] . Remembering that q and s must have the same solution sets (again due to symmetry), it follows that we may have q = a^2 + Sqrt[(a^4+b^4)/2] , s = a^2 - Sqrt[(a^4+b^4)/2] , and thus our original polynomial has the factorization (x^2 - ax + q)(x^2 - ax + s) = 0 . Thus we may solve for x by solving each quadratic individually; we have x = {(a +- Sqrt[a^2 - 4q])/2 , (a +- Sqrt[a^2 - 4s])/2} a +- Sqrt[-3a^2 +- 2*Sqrt[2(a^4+b^4)]] = -------------------------------------- , 2 which is precisely what Mathematica found. And because of symmetry, this solves y automatically as well. So for one particular solution, choose x by picking the first plus/minus to be plus, and the second plus/minus to be minus; then y will be the same as above except you pick the first plus/minus to be *minus*, and the second to be *minus*. That is, you choose opposite signs on the outer plus/minus, and the * same* sign on the inner one. We denote this by saying a +- Sqrt[-3a^2 +- 2*Sqrt[2(a^4+b^4)]] x = -------------------------------------- , 2 a -+ Sqrt[-3a^2 +- 2*Sqrt[2(a^4+b^4)]] y = -------------------------------------- . 2 Note I have switched the order of the outer plus/minus symbol on y, so it reads "minus/plus." This completes the solution of your system. (Whew!) Now, I'd consider this quite difficult for a high-school level problem, primarily because you have to make several observations, one of which is assuming that the quartic you had found has a nice factorization. The second difficult step is being able to solve for p, q, r, and s - this is the most computational part of the problem. For a somewhat easier problem, can you solve x^3 + y^3 = b^3 x + y = a ? (Hint: factor x^3 + y^3.) For a *very* nasty problem, do you think that there is a general method to solve x^n + y^n = b^n x + y = a for positive integer n? Hope I made sense - if I didn't, please don't hesitate to point out the unclear parts of my solution. -Doctor Pete, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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