1961 AHSME Problems/Problem 29

Problem

Let the roots of $ax^2+bx+c=0$ be $r$ and $s$. The equation with roots $ar+b$ and $as+b$ is:

$\textbf{(A)}\ x^2-bx-ac=0\qquad \textbf{(B)}\ x^2-bx+ac=0 \qquad\\ \textbf{(C)}\ x^2+3bx+ca+2b^2=0 \qquad \textbf{(D)}\ x^2+3bx-ca+2b^2=0 \qquad\\ \textbf{(E)}\ x^2+bx(2-a)+a^2c+b^2(a+1)=0$

Solution

From Vieta's Formulas, $r+s=\frac{-b}{a}$ and $rs=\frac{c}{a}$ in the original quadratic.

The sum of the roots in the new quadratic is \[ar + b + as + b\] \[a(r+s) + 2b\] \[b\] The product of the roots in the new quadratic is \[(ar + b)(as + b)\] \[a^2rs + arb + asb + b^2\] \[a^2rs + ab(r+s) + b^2\] \[ac - b^2 + b^2\] \[ac\] Thus, the new quadratic is $x^2-bx+ac$. The answer is $\boxed{\textbf{(B)}}$.

Solution 2

Let $f(x)=ax^2+bx+c$ and $g(x)$ be the desired function with roots $ar+b$ and $as+b$. Applying graphical transformations, we have \[f(r)=f(s)=0\] \[f(\frac{(ar+b)-b}{a})=f(\frac{(as+b)-b}{a})=0\] Thus $ar+b$ and $as+b$ are the roots to \[g(x)=f(\frac{x-b}{a})\] Plugging $x=\frac{x-b}{a}$ into $f(x)$ gets $g(x)=x^2-bx+ac \boxed{B}$.

~ Nafer

See Also

1961 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 28
Followed by
Problem 30
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