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Math Help - 4 non-linear equations and 4 unknowns

  1. #1
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    4 non-linear equations and 4 unknowns

    I've already found the solution using an online calculator, but I'm wondering if one of you is able to work it out by hand and show me the steps: A: p + r = 6; B: pr + q + s = 16; C: ps + qr = 22; D: qs = 15. The solution is: p=2, q=3, r=4, s=5. I've tried the following steps: First, solve A for p in terms of r: p = -r + 6. Then, solve D for s in terms of q: s = 15/q. Next, I plugged those into B like so: (-r + 6)r + q + 15/q = 16. Then, I solved that new equation for r in terms of q and got the following: r = [3q +- sqrt[q(-q^2+25q-15)]] / q. Finally, I tried subbing everything into equation C (including that long, hairy expression for r) and solving for q. After several pages of meticulous algebraic manipulation, it seems like a wild goose chase. Perhaps there's a better approach? Thanks for any help.
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    Re: 4 non-linear equations and 4 unknowns

    $p + r = 6 \implies p = 6 - r.$ I'm with you so far.

    $qs = 15 \implies s = \dfrac{15}{q}.$ Still humming along.

    $pr + q + s = 16 \implies r(6 - r) + q + \dfrac{15}{q} = 16 \implies 6r - r^2 + \dfrac{q^2 + 15}{q} = 16 \implies$

    $r^2 - 6r + 9 + 7 = \dfrac{q^2 + 15}{q} \implies (r - 3)^2 = \dfrac{q^2 - 7q + 15}{q} \implies r = 3 \pm \sqrt{\dfrac{q^2 - 7q + 15}{q}}.$

    Best check my algebra, but I think you slipped up on step 3. I agree that step 4 will be tedious and prone to error in the best case, but if step 3 is messed up, you are doomed. Nothing wrong with your thought process.
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    Re: 4 non-linear equations and 4 unknowns

    That guidance is helpful; thanks, Jeff. Because I'm new here, I have a quick off-topic question about how you formatted your post: did you use a separate equation editor and then paste it in, or did you use the editing functionality built into this forum's reply page? A brief FAQ search on this didn't help me...
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    Re: 4 non-linear equations and 4 unknowns

    Quote Originally Posted by NicholasDeMaio View Post
    That guidance is helpful; thanks, Jeff. Because I'm new here, I have a quick off-topic question about how you formatted your post: did you use a separate equation editor and then paste it in, or did you use the editing functionality built into this forum's reply page? A brief FAQ search on this didn't help me...
    There are a couple LaTex interpreters built into the site. Read some of the posts in the LaTex Help forum for instructions on how to use this.
    Thanks from topsquark
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    Re: 4 non-linear equations and 4 unknowns

    Quote Originally Posted by NicholasDeMaio View Post
    That guidance is helpful; thanks, Jeff. Because I'm new here, I have a quick off-topic question about how you formatted your post: did you use a separate equation editor and then paste it in, or did you use the editing functionality built into this forum's reply page? A brief FAQ search on this didn't help me...
    I used the LaTeX editor. There are sources on the internet to explain LaTeX, but the easiest way to start learning it is to use reply with quote for posts that use LaTeX. You will then see exactly how it is invoked on this site and learn many of the conventions involving its use. Then the internet sources can fill in the gaps.

    $For\ example,\ click\ on\ reply\ with\ quote\ to\ see\ how\ to\ write\ \displaystyle \int_a^bf(x)\ dx.$
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    Re: 4 non-linear equations and 4 unknowns

    By the way, if you are interested in whether there is a solution in integers to your problem, you can solve your problem analytically very quickly once you have step 3.

    $q,\ r \in \mathbb Z\ and r = 3 \pm \sqrt{\dfrac{q^2 - 7q + 15}{q}} = \implies r =3 \pm \sqrt{q - 7 + \dfrac{15}{q}} \implies q - 7 + \dfrac{15}{q}\ is\ a\ perfect\ square \implies q = 3\ or\ q = 5.$

    $q = 5 \implies \sqrt{q - 7 + \dfrac{15}{q}} = \sqrt{5 - 7 + 3} = \sqrt{1} = 1 \implies r = 3 + 1 = 4\ or\ r = 3 - 1 = 2.$

    $q = 3 \implies \sqrt{q - 7 + \dfrac{15}{q}} = \sqrt{3 - 7 + 5} = \sqrt{1} = 1 \implies r = 3 + 1 = 4\ or\ r = 3 - 1 = 2.$

    So there are four possibilities to explore further.

    q = 5 and r = 2. So p = 6 - r = 6 - 2 = 4. And s = 15 / q = 15 / 5 = 3. Let's try ps + qr = 4 * 3 + 5 * 2 = 12 + 10 = 22. A solution in positive integers.

    q = 5 and r = 4. So p = 2 and s = 3. ps + qr = 2 * 3 + 5 * 4 = 6 + 20 = 26. Not a solution.

    q = 3 and r = 2. So p = 4 and s = 5. ps + qr = 4 * 5 + 3 * 2 = 26. Not a solution.

    q = 3 and r = 4. So p = 2 and s = 5. ps + qr = 2 * 5 + 3 * 4 = 22. Another solution in integers.

    Two solutions in integers.
    p = 4, q = 5, r = 2, s = 3.

    p = 2, q = 3, r = 4, s = 5.
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    Re: 4 non-linear equations and 4 unknowns

    Quote Originally Posted by JeffM View Post
     r = 3 \pm \sqrt{\dfrac{q^2 - 7q + 15}{q}}

    Best check my algebra, but I think you slipped up on step 3. I agree that step 4 will be tedious and prone to error in the best case, but if step 3 is messed up, you are doomed. Nothing wrong with your thought process.
    Thanks so much for the help. I've checked your algebra for r in terms of q, and I agree with it. I've made some more progress with pencil and paper but haven't yet reached the solution for q. I'll post it once I get the chance to enter it into LaTex. Thanks again.
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    Re: 4 non-linear equations and 4 unknowns

    Now that I know p, r, and s all in terms of q, I can sub all of those into C, leaving only q to solve for:

     \left( 6 - \left( 3 + \sqrt { \frac {q^2-7q+15}{q} } \right) \right) \left( \frac {15}{q} \right) + q \left( 3 + \sqrt { \frac {q^2-7q+15}{q} } \right) = 22

    More to come.
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    Re: 4 non-linear equations and 4 unknowns

    Remember that there could be multiple distinct solutions. (As I showed above, there are two distinct solutions in integers alone.) Because more than one equation is of degree 2, there may be even more than two possible solutions. Solving simultaneous non-linear equations is not something I ever studied formally. For example, I can imagine two ellipses intersecting in four points.
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    Re: 4 non-linear equations and 4 unknowns

    Right, thanks for the reminder.
    Last edited by NicholasDeMaio; April 22nd 2014 at 11:16 AM.
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    Re: 4 non-linear equations and 4 unknowns

    JeffM, your analytical approach was quite helpful - I wouldn't have thought of it.
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    Re: 4 non-linear equations and 4 unknowns

    Here's how the problem came about. I'm exploring a way to split up a quartic polynomial into two quadratic factors. The Rational Root Theorem would normally be the first place to start, but what about quartics whose roots are all irrational and/or imaginary?

    I'm assuming this has been done before, and it probably has a name. I'm kinda re-inventing the wheel just for the sake of it.

    So, let's say I have a 4th degree function like so:  f(x) = x^4 + ax^3 + bx^2 +cx + d . I thought I'd take two quadratics and multiply them together and then reverse the process.

     (x^2 + px + q)(x^2 + rx + s) = x^4 + rx^3 + sx^2 + px^3 + prx^2 + psx + qx^2 + qrx + qs = x^4 + (p+r)x^3 + (pr+q+s)x^2 + (ps+qr)x + qs

    So it stands to reason that the following system of equations could be used to find the coefficients of the quartic's two quadratic factors:

     x^4+ax^3+bx^2+cx+d=(x^2+px+q)(x^2+rx+s) where  p+r = a, pr+q+s=b, pr+qs=c , and qs=d

    If there aren't any solutions to this system, then the quartic isn't factorable into two quadratics.

    The initial problem above is an example I made up:

     x^4+6x^3+16x^2+22x+15 = (x^2+2x+3)(x^2+4x+5)

    because

     2+4=6, (2)(4)+3+5=16, (2)(5)+(3)(4)=22 , and  (3)(5)=15

    When the two quadratic factors have all integer coefficients, the problem is apparently much easier. But what if they have rational coefficients? I'd need to solve the system for p, q, r, and s in terms of a, b, c, and d. That'll be the cumbersome and error-prone problem that I'm not sure I currently have the time and energy for.

    If you have any thoughts on these ideas, let me know.
    Last edited by NicholasDeMaio; May 5th 2014 at 12:11 PM.
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    Re: 4 non-linear equations and 4 unknowns

    Mods: It may be necessary to move this thread over the Peer Review forum, but I'm not sure how original this work is. I've googled the topic of factoring quartics into two quadratics and have found lots of info on the Quartic Formula and some other things. But I haven't seen anything written on this specific approach. Having said that, I haven't searched extensively.
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    Re: 4 non-linear equations and 4 unknowns

    Every polynomial with real coefficients of degree n can be factored into (n / 2) quadratics with real coefficients if n is even and into ((n - 1) / 2) quadratics with real coefficients and one linear factor with real coefficients if n is odd. This is a result of the fundamental theorem of algebra, which took about two hundred years to prove. The modern version is that any polynomial of degree n with complex coefficients can be factored into n linear terms with complex coefficients. Fundamental theorem of algebra - Wikipedia, the free encyclopedia

    It does not mean such factorings are easy to find.

    However, for cubics and quartics, there are formulas that let you find the factorings. The formulas were discovered in the 16th century. The formulas are ugly and complex, and I am certainly not going to memorize them.

    Cubic function - Wikipedia, the free encyclopedia

    Quartic function - Wikipedia, the free encyclopedia

    This process breaks down with quintics as was proved early in the 19th century.

    Abel?Ruffini theorem - Wikipedia, the free encyclopedia
    Last edited by JeffM; May 5th 2014 at 01:17 PM. Reason: added missing phrase
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    Re: 4 non-linear equations and 4 unknowns

    Quote Originally Posted by NicholasDeMaio View Post
    Mods: It may be necessary to move this thread over the Peer Review forum, but I'm not sure how original this work is. I've googled the topic of factoring quartics into two quadratics and have found lots of info on the Quartic Formula and some other things. But I haven't seen anything written on this specific approach. Having said that, I haven't searched extensively.
    Actually there were some problems in the Math Challenge Forum recently that show some elegant solutions to problems involving cubics, one by prasum and one by soroban.

    Problem #3 - Cubic Expression

    Problem #4 - Solving a cubic equation
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