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
 
Thread 'Erroneously finding discrepancy in transpose rule'

Thread 'Erroneously finding discrepancy in transpose rule'

Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...
View full post »

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