The discussion focuses on solving a system of equations involving four variables: a, b, c, and d. The equations provided are: a + b = 8, ab + c + d = 23, ad + bc = 28, and cd = 12. Despite multiple hints and attempts, no definitive solution has been reached, leading to a consensus that the problem remains unsolved. Participants express frustration over the lack of progress in finding a satisfactory answer.
PREREQUISITESMathematicians, students studying algebra, and anyone interested in solving complex systems of equations.
kaliprasad said:
kaliprasad said:
Thank you so much, kaliprasad for the hint and also for sharing with us of this great challenge! I could tell I've fallen in love with this problem at first sight!Hahaha...anemone said:My solution:
If we form a quartic equation as the product of two quadratic equation as follow, we have:
$\begin{align*}(x^2+ax+c)(x^2+bx+d)&=x^4+(a+b)x^3+(ab+c+d)x^2+(ad+bc)x+cd\\&=x^4+8x^3+23x^2+28x+12\\&\overset{I}{=}((x+1)(x+2))((x+2)(x+3))=(x^2+3x+2)(x^2+5x+6)\\&\overset{II}{=}((x+1)(x+3))((x+2)(x+2))=(x^2+4x+3)(x^2+4x+4)\\&\overset{III}{=}((x+2)(x+3))((x+1)(x+2))=(x^2+5x+6)(x^2+3x+2)\\&\overset{IV}{=}((x+2)(x+2))((x+1)(x+3))=(x^2+4x+4)(x^2+4x+3)\end{align*}$
Hence, $(a,\,b,\,c,\,d)=(3,\,5,\,2,\,6),\,(4,\,4,\,3,\,4),\,(5,\,3,\,6,\,2),\,(4,\,4,\,4,\,3)$.
But...I'm not proud of myself for this solution because...
Without the hint, I could have never solved this great challenge!