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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-> 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
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.
