1961 AMC 12 Problem 29

Attempt Problem 29 of the 1961 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1961 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

29.

Let the roots of ax2+bx+c=0ax^2+bx+c=0 be rr and s.s. The equation with roots ar+bar+b and as+bas+b is:
x2−bx−ac=0x^2-bx-ac=0
x2−bx+ac=0x^2-bx+ac=0
x2+3bx+ca+2b2=0x^2+3bx+ca+2b^2=0
x2+3bx−ca+2b2=0x^2+3bx-ca+2b^2=0
x2+bx(2−a)+a2c+b2(a+1)=0\begin{gathered}x^2+bx(2-a)\\{}+a^2c+b^2(a+1)=0\end{gathered}
Answer: B
Concepts:Vieta’s Formulasquadraticalgebraic manipulation
Small hint:
Use r+s=−bar+s=-\frac{b}{a} and rs=cars=\frac{c}{a}
Big hint:
Compute the sum and product of ar+bar+b and as+bas+b
Solution:
The transformed roots have sum a(r+s)+2b=−b+2b=b a(r+s)+2b=-b+2b=b and product (ar+b)(as+b)=a2rs+ab(r+s)+b2=ac. \begin{gathered} (ar+b)(as+b)\\ =a^2rs+ab(r+s)+b^2\\ =ac. \end{gathered} Therefore their monic equation is x2−bx+ac=0.x^2-bx+ac=0.
Thus, the correct answer is B.
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