Advice on solving these simultaneous (quadratic/cubic) equations?












1














I have the following simultaneous equations:



$$a x^2 + (b+2ay)x - c_1 = 0$$



$$ay^2 + (b+2ax)y - c_2 = 0$$



Where I'd like to solve for $x$ and $y$.



Obviously $a,b,c_1,c_2$ are known constants. They have the following restrictions:



$$a in [0,1]$$
$$b > 0$$
$$c_1, c_2 > 0$$



My thinking is that the above equations should give me a bunch of possible solutions, and then all but one can (hopefully) be eliminated according to restrictions that this particular problem imposes (eg. imaginary solutions are not permitted).



The only approach that I can see is to first use the quadratic formula to obtain solutions for $x$ in terms of $y$, and then substitute this into the second equation to then solve for $y$, or vise versa. But this quickly leads to an intractable (at least for me) expression. In particular, I end up with



$$x = frac{-(b+2ay)pm sqrt{(b+2ay)^2 + 4a c_1}}{2a}$$



along with the equation



$$- ay^2 pm sqrt{(b+2ay)^2+4ac_1}y - c_2 = 0$$



The hope is to then solve the above to obtain a solution for $y$, which can then be used to obtain the corresponding solution for $x$. But how on earth can we deal with the square root? Is this problem even solvable without resorting to a numerical solution?






systems-of-equations quadratics symmetric-polynomials







share|cite|improve this question
























  • I have $-ay^2 pm cdots$ in your last line.
    – Doug M
    Jan 4 at 3:24






  • 1




    Because if $$3ay^2 pmsqrt{text{blah}}=0$$ then $$9a^2y^4-text{blah}=0$$. This is just a quartic and can be solved albeit awkwardly.
    – Rhys Hughes
    Jan 4 at 3:26






  • 1




    @Xiaomi Rhys Hughes approach may be best, An alternative approach is to subtract the two eqn's to create a third equation: $a(x^2 - y^2) +b(x-y) - (c_1 - c_2) = 0.$ I would ignore the 2nd equation, apply the quadratic formula to the first equation, expressing $x$ in terms of $y$, and then try to combine that with the third equation. This will result in two equations in two unknowns, one of which (the 3rd) is non-linear. If you still can't solve the problem re Rhys Hughes answer or my comment, respond, and I will take another look at it.
    – user2661923
    Jan 4 at 3:31












  • And it sould be $+ 4ac_1$ under the radical.
    – Doug M
    Jan 4 at 3:32










  • @DougM thanks for catching that mistake. I'll try that first and check back here if I dont maake any progress
    – Xiaomi
    Jan 4 at 4:16


















1














I have the following simultaneous equations:



$$a x^2 + (b+2ay)x - c_1 = 0$$



$$ay^2 + (b+2ax)y - c_2 = 0$$



Where I'd like to solve for $x$ and $y$.



Obviously $a,b,c_1,c_2$ are known constants. They have the following restrictions:



$$a in [0,1]$$
$$b > 0$$
$$c_1, c_2 > 0$$



My thinking is that the above equations should give me a bunch of possible solutions, and then all but one can (hopefully) be eliminated according to restrictions that this particular problem imposes (eg. imaginary solutions are not permitted).



The only approach that I can see is to first use the quadratic formula to obtain solutions for $x$ in terms of $y$, and then substitute this into the second equation to then solve for $y$, or vise versa. But this quickly leads to an intractable (at least for me) expression. In particular, I end up with



$$x = frac{-(b+2ay)pm sqrt{(b+2ay)^2 + 4a c_1}}{2a}$$



along with the equation



$$- ay^2 pm sqrt{(b+2ay)^2+4ac_1}y - c_2 = 0$$



The hope is to then solve the above to obtain a solution for $y$, which can then be used to obtain the corresponding solution for $x$. But how on earth can we deal with the square root? Is this problem even solvable without resorting to a numerical solution?






systems-of-equations quadratics symmetric-polynomials







share|cite|improve this question
























  • I have $-ay^2 pm cdots$ in your last line.
    – Doug M
    Jan 4 at 3:24






  • 1




    Because if $$3ay^2 pmsqrt{text{blah}}=0$$ then $$9a^2y^4-text{blah}=0$$. This is just a quartic and can be solved albeit awkwardly.
    – Rhys Hughes
    Jan 4 at 3:26






  • 1




    @Xiaomi Rhys Hughes approach may be best, An alternative approach is to subtract the two eqn's to create a third equation: $a(x^2 - y^2) +b(x-y) - (c_1 - c_2) = 0.$ I would ignore the 2nd equation, apply the quadratic formula to the first equation, expressing $x$ in terms of $y$, and then try to combine that with the third equation. This will result in two equations in two unknowns, one of which (the 3rd) is non-linear. If you still can't solve the problem re Rhys Hughes answer or my comment, respond, and I will take another look at it.
    – user2661923
    Jan 4 at 3:31












  • And it sould be $+ 4ac_1$ under the radical.
    – Doug M
    Jan 4 at 3:32










  • @DougM thanks for catching that mistake. I'll try that first and check back here if I dont maake any progress
    – Xiaomi
    Jan 4 at 4:16
















1












1








1







I have the following simultaneous equations:



$$a x^2 + (b+2ay)x - c_1 = 0$$



$$ay^2 + (b+2ax)y - c_2 = 0$$



Where I'd like to solve for $x$ and $y$.



Obviously $a,b,c_1,c_2$ are known constants. They have the following restrictions:



$$a in [0,1]$$
$$b > 0$$
$$c_1, c_2 > 0$$



My thinking is that the above equations should give me a bunch of possible solutions, and then all but one can (hopefully) be eliminated according to restrictions that this particular problem imposes (eg. imaginary solutions are not permitted).



The only approach that I can see is to first use the quadratic formula to obtain solutions for $x$ in terms of $y$, and then substitute this into the second equation to then solve for $y$, or vise versa. But this quickly leads to an intractable (at least for me) expression. In particular, I end up with



$$x = frac{-(b+2ay)pm sqrt{(b+2ay)^2 + 4a c_1}}{2a}$$



along with the equation



$$- ay^2 pm sqrt{(b+2ay)^2+4ac_1}y - c_2 = 0$$



The hope is to then solve the above to obtain a solution for $y$, which can then be used to obtain the corresponding solution for $x$. But how on earth can we deal with the square root? Is this problem even solvable without resorting to a numerical solution?






systems-of-equations quadratics symmetric-polynomials







share|cite|improve this question















I have the following simultaneous equations:



$$a x^2 + (b+2ay)x - c_1 = 0$$



$$ay^2 + (b+2ax)y - c_2 = 0$$



Where I'd like to solve for $x$ and $y$.



Obviously $a,b,c_1,c_2$ are known constants. They have the following restrictions:



$$a in [0,1]$$
$$b > 0$$
$$c_1, c_2 > 0$$



My thinking is that the above equations should give me a bunch of possible solutions, and then all but one can (hopefully) be eliminated according to restrictions that this particular problem imposes (eg. imaginary solutions are not permitted).



The only approach that I can see is to first use the quadratic formula to obtain solutions for $x$ in terms of $y$, and then substitute this into the second equation to then solve for $y$, or vise versa. But this quickly leads to an intractable (at least for me) expression. In particular, I end up with



$$x = frac{-(b+2ay)pm sqrt{(b+2ay)^2 + 4a c_1}}{2a}$$



along with the equation



$$- ay^2 pm sqrt{(b+2ay)^2+4ac_1}y - c_2 = 0$$



The hope is to then solve the above to obtain a solution for $y$, which can then be used to obtain the corresponding solution for $x$. But how on earth can we deal with the square root? Is this problem even solvable without resorting to a numerical solution?






systems-of-equations quadratics symmetric-polynomials




systems-of-equations quadratics symmetric-polynomials






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 9 hours ago









Harry Peter

5,46111439




5,46111439










asked Jan 4 at 3:15









XiaomiXiaomi

1,021115




1,021115












  • I have $-ay^2 pm cdots$ in your last line.
    – Doug M
    Jan 4 at 3:24






  • 1




    Because if $$3ay^2 pmsqrt{text{blah}}=0$$ then $$9a^2y^4-text{blah}=0$$. This is just a quartic and can be solved albeit awkwardly.
    – Rhys Hughes
    Jan 4 at 3:26






  • 1




    @Xiaomi Rhys Hughes approach may be best, An alternative approach is to subtract the two eqn's to create a third equation: $a(x^2 - y^2) +b(x-y) - (c_1 - c_2) = 0.$ I would ignore the 2nd equation, apply the quadratic formula to the first equation, expressing $x$ in terms of $y$, and then try to combine that with the third equation. This will result in two equations in two unknowns, one of which (the 3rd) is non-linear. If you still can't solve the problem re Rhys Hughes answer or my comment, respond, and I will take another look at it.
    – user2661923
    Jan 4 at 3:31












  • And it sould be $+ 4ac_1$ under the radical.
    – Doug M
    Jan 4 at 3:32










  • @DougM thanks for catching that mistake. I'll try that first and check back here if I dont maake any progress
    – Xiaomi
    Jan 4 at 4:16




















  • I have $-ay^2 pm cdots$ in your last line.
    – Doug M
    Jan 4 at 3:24






  • 1




    Because if $$3ay^2 pmsqrt{text{blah}}=0$$ then $$9a^2y^4-text{blah}=0$$. This is just a quartic and can be solved albeit awkwardly.
    – Rhys Hughes
    Jan 4 at 3:26






  • 1




    @Xiaomi Rhys Hughes approach may be best, An alternative approach is to subtract the two eqn's to create a third equation: $a(x^2 - y^2) +b(x-y) - (c_1 - c_2) = 0.$ I would ignore the 2nd equation, apply the quadratic formula to the first equation, expressing $x$ in terms of $y$, and then try to combine that with the third equation. This will result in two equations in two unknowns, one of which (the 3rd) is non-linear. If you still can't solve the problem re Rhys Hughes answer or my comment, respond, and I will take another look at it.
    – user2661923
    Jan 4 at 3:31












  • And it sould be $+ 4ac_1$ under the radical.
    – Doug M
    Jan 4 at 3:32










  • @DougM thanks for catching that mistake. I'll try that first and check back here if I dont maake any progress
    – Xiaomi
    Jan 4 at 4:16


















I have $-ay^2 pm cdots$ in your last line.
– Doug M
Jan 4 at 3:24




I have $-ay^2 pm cdots$ in your last line.
– Doug M
Jan 4 at 3:24




1




1




Because if $$3ay^2 pmsqrt{text{blah}}=0$$ then $$9a^2y^4-text{blah}=0$$. This is just a quartic and can be solved albeit awkwardly.
– Rhys Hughes
Jan 4 at 3:26




Because if $$3ay^2 pmsqrt{text{blah}}=0$$ then $$9a^2y^4-text{blah}=0$$. This is just a quartic and can be solved albeit awkwardly.
– Rhys Hughes
Jan 4 at 3:26




1




1




@Xiaomi Rhys Hughes approach may be best, An alternative approach is to subtract the two eqn's to create a third equation: $a(x^2 - y^2) +b(x-y) - (c_1 - c_2) = 0.$ I would ignore the 2nd equation, apply the quadratic formula to the first equation, expressing $x$ in terms of $y$, and then try to combine that with the third equation. This will result in two equations in two unknowns, one of which (the 3rd) is non-linear. If you still can't solve the problem re Rhys Hughes answer or my comment, respond, and I will take another look at it.
– user2661923
Jan 4 at 3:31






@Xiaomi Rhys Hughes approach may be best, An alternative approach is to subtract the two eqn's to create a third equation: $a(x^2 - y^2) +b(x-y) - (c_1 - c_2) = 0.$ I would ignore the 2nd equation, apply the quadratic formula to the first equation, expressing $x$ in terms of $y$, and then try to combine that with the third equation. This will result in two equations in two unknowns, one of which (the 3rd) is non-linear. If you still can't solve the problem re Rhys Hughes answer or my comment, respond, and I will take another look at it.
– user2661923
Jan 4 at 3:31














And it sould be $+ 4ac_1$ under the radical.
– Doug M
Jan 4 at 3:32




And it sould be $+ 4ac_1$ under the radical.
– Doug M
Jan 4 at 3:32












@DougM thanks for catching that mistake. I'll try that first and check back here if I dont maake any progress
– Xiaomi
Jan 4 at 4:16






@DougM thanks for catching that mistake. I'll try that first and check back here if I dont maake any progress
– Xiaomi
Jan 4 at 4:16












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You can make life simpler avoiding radicals but you will still arrive to a nasty quartic equation.



Considering $$a x^2 + (b+2ay)x - c_1 = 0 tag 1$$
$$ay^2 + (b+2ax)y - c_2 = 0 tag 2$$
eliminate $y$ from $(1)$to get (assuming $aneq 0$ and $xneq 0$)
$$y=frac{-a x^2-b x+c_1}{2 a x}tag 3$$
Plug this result in $(2)$, simplify as much as you can to get by the end
$$3 a^2 x^4+4a bx^3+(b^2+4 a c_2-2 a c_1 ) x^2 -c_1^2=0 tag 4$$



Now, have fun !



I think that considering some numerical method (such as Newton) could be a good idea looking for the zero's of function
$$f(x)=3 a^2 x^4+4a bx^3+(b^2+4 a c_2-2 a c_1 ) x^2 -c_1^2$$
Since $f(0)=-c_1^2 <0$, you could catch a first root (so you know that there is at least a second real root); deflate to a cubic equation (using long division) and this would be easier to handle.






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    You can make life simpler avoiding radicals but you will still arrive to a nasty quartic equation.



    Considering $$a x^2 + (b+2ay)x - c_1 = 0 tag 1$$
    $$ay^2 + (b+2ax)y - c_2 = 0 tag 2$$
    eliminate $y$ from $(1)$to get (assuming $aneq 0$ and $xneq 0$)
    $$y=frac{-a x^2-b x+c_1}{2 a x}tag 3$$
    Plug this result in $(2)$, simplify as much as you can to get by the end
    $$3 a^2 x^4+4a bx^3+(b^2+4 a c_2-2 a c_1 ) x^2 -c_1^2=0 tag 4$$



    Now, have fun !



    I think that considering some numerical method (such as Newton) could be a good idea looking for the zero's of function
    $$f(x)=3 a^2 x^4+4a bx^3+(b^2+4 a c_2-2 a c_1 ) x^2 -c_1^2$$
    Since $f(0)=-c_1^2 <0$, you could catch a first root (so you know that there is at least a second real root); deflate to a cubic equation (using long division) and this would be easier to handle.






    share|cite|improve this answer


























      1














      You can make life simpler avoiding radicals but you will still arrive to a nasty quartic equation.



      Considering $$a x^2 + (b+2ay)x - c_1 = 0 tag 1$$
      $$ay^2 + (b+2ax)y - c_2 = 0 tag 2$$
      eliminate $y$ from $(1)$to get (assuming $aneq 0$ and $xneq 0$)
      $$y=frac{-a x^2-b x+c_1}{2 a x}tag 3$$
      Plug this result in $(2)$, simplify as much as you can to get by the end
      $$3 a^2 x^4+4a bx^3+(b^2+4 a c_2-2 a c_1 ) x^2 -c_1^2=0 tag 4$$



      Now, have fun !



      I think that considering some numerical method (such as Newton) could be a good idea looking for the zero's of function
      $$f(x)=3 a^2 x^4+4a bx^3+(b^2+4 a c_2-2 a c_1 ) x^2 -c_1^2$$
      Since $f(0)=-c_1^2 <0$, you could catch a first root (so you know that there is at least a second real root); deflate to a cubic equation (using long division) and this would be easier to handle.






      share|cite|improve this answer
























        1












        1








        1






        You can make life simpler avoiding radicals but you will still arrive to a nasty quartic equation.



        Considering $$a x^2 + (b+2ay)x - c_1 = 0 tag 1$$
        $$ay^2 + (b+2ax)y - c_2 = 0 tag 2$$
        eliminate $y$ from $(1)$to get (assuming $aneq 0$ and $xneq 0$)
        $$y=frac{-a x^2-b x+c_1}{2 a x}tag 3$$
        Plug this result in $(2)$, simplify as much as you can to get by the end
        $$3 a^2 x^4+4a bx^3+(b^2+4 a c_2-2 a c_1 ) x^2 -c_1^2=0 tag 4$$



        Now, have fun !



        I think that considering some numerical method (such as Newton) could be a good idea looking for the zero's of function
        $$f(x)=3 a^2 x^4+4a bx^3+(b^2+4 a c_2-2 a c_1 ) x^2 -c_1^2$$
        Since $f(0)=-c_1^2 <0$, you could catch a first root (so you know that there is at least a second real root); deflate to a cubic equation (using long division) and this would be easier to handle.






        share|cite|improve this answer












        You can make life simpler avoiding radicals but you will still arrive to a nasty quartic equation.



        Considering $$a x^2 + (b+2ay)x - c_1 = 0 tag 1$$
        $$ay^2 + (b+2ax)y - c_2 = 0 tag 2$$
        eliminate $y$ from $(1)$to get (assuming $aneq 0$ and $xneq 0$)
        $$y=frac{-a x^2-b x+c_1}{2 a x}tag 3$$
        Plug this result in $(2)$, simplify as much as you can to get by the end
        $$3 a^2 x^4+4a bx^3+(b^2+4 a c_2-2 a c_1 ) x^2 -c_1^2=0 tag 4$$



        Now, have fun !



        I think that considering some numerical method (such as Newton) could be a good idea looking for the zero's of function
        $$f(x)=3 a^2 x^4+4a bx^3+(b^2+4 a c_2-2 a c_1 ) x^2 -c_1^2$$
        Since $f(0)=-c_1^2 <0$, you could catch a first root (so you know that there is at least a second real root); deflate to a cubic equation (using long division) and this would be easier to handle.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 4 at 6:15









        Claude LeiboviciClaude Leibovici

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