SOLUTION: Please help me solve this problem : Let M and N be the root of an equation x^2+ax+b=0 and let O and P be the roots of x^2+cx+d=0. Express (M-O)(N-O)(M-P)(N-P) in terms of the

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Question 1075157: Please help me solve this problem :
Let M and N be the root of an equation x^2+ax+b=0 and
let O and P be the roots of x^2+cx+d=0. Express
(M-O)(N-O)(M-P)(N-P) in terms of the coefficients
a,b,c,d.
Found 2 solutions by Edwin McCravy, ikleyn:
Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
Let M and N be the root of an equation x^2+ax+b=0 and
let O and P be the roots of x^2+cx+d=0. Express
(M-O)(N-O)(M-P)(N-P) in terms of the coefficients
a,b,c,d. 








Therefore:





I agree it isn't simplified, but it does express 
(M-O)(N-O)(M-P)(N-P) in terms of the coefficients a,b,c,d.
Nothing was mentioned about "in simplest form".

Edwin
Answer by ikleyn(52794)   (Show Source): You can put this solution on YOUR website!
.
1.  (M-O)*(N-O) =  

                =           (I replaced MN by b based on Vieta's formulas)

                =      (1)      (I replaced M+N by -c  based on Vieta's formulas)



2.  (M-P)*(N-P) =  

                =           (I replaced MN by b based on Vieta's formulas)

                =      (2)      (I replaced M+N by -c  based on Vieta's formulas)



3.  (M-O)*(N-O)*(M-P)*(N-P) =  =  (I used here (1) and (2) )

    =  = 

    =  = 

    =  = 

    = 


         ( I replaced here P+O by -c,  by , and PO by d. )

    =  =  = .     ( !!! )

Done.


Check my Math. I could make a mistake.

But the major idea is VERY CLEAR: 

     make multiplications and use everywhere the Vieta's formulas for quadratic polynomials: 


         the sum of the roots is the coef. at x with the opposite sign; the product of the roots is the constant term.


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