SOLUTION: suppose the roots of ax^2 + bx + c =0 are r and s. which one of the following has roots ar + b and as + b ?
(A) x^2 - bx - ac =0
(B) x^2 - bx + ac =0
(C) x^2 + 3bx + ca
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Question 523322: suppose the roots of ax^2 + bx + c =0 are r and s. which one of the following has roots ar + b and as + b ?
(A) x^2 - bx - ac =0
(B) x^2 - bx + ac =0
(C) x^2 + 3bx + ca + 2b^2 =0
(D) x^2 + 3bx - ca + 2b^2 =0
(E) not given
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
Recall that, if z is a root of a polynomial, then x-z is a factor of the polynomial.
Hence, we can write our new polynomial equation as (x - (ar + b))(x - (as + b)) = 0. Expanding the LHS, this yields
 - x(ar + b) + (ar + b)(as + b) = 0)
 + (a^2rs + ab(r+s) + b^2) = 0)
Recall that by Vieta's formulas, r+s = -b/a and rs = c/a. We can substitute these expressions in:
) + (a^2(\frac{c}{a}) + ab(-\frac{b}{a}) + b^2) = 0)
 = 0)
b^2 cancels, so you will see that this equation is the same as in answer choice B.
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