MHB Solve System of Equations: a+b, ab+c+d, ad+bc, cd

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The discussion revolves around 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 attempts to find a solution, no definitive answer has been reached, and participants express a lack of confidence in their solutions. The conversation indicates a struggle with the complexity of the equations, leading to a sense of closure without resolution. The thread highlights the challenges of solving nonlinear systems of equations.
kaliprasad
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solve the following system of equations in real a,b,c,d

$a+ b =8 $

$ab + c +d = 23 $

$ad + bc = 28$

$cd = 12$
 
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No answer yet

hint

form a quartic equation
 
kaliprasad said:
solve the following system of equations in real a,b,c,d

$a+ b =8 $

$ab + c +d = 23 $

$ad + bc = 28$

$cd = 12$

kaliprasad said:
No answer yet

hint

form a quartic equation

My solution:

If we form a quartic equation as the product of two quadratic equations 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!:oThank 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...
 
Last edited:
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!:oThank 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...

Nothing more to write to make it right(pun intended)

Hence it is closed
 

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