2.67 Given that the roots of the equation $a x^{2}+b x+c=0$ are $r$ and $s$, then the equation whose roots are $a r+b$ and $a s+b$ is
(A) $x^{2}-b x-a c=0$.
(B) $x^{2}-b x+a c=0$.
(C) $x^{2}+3 b x+c a+2 b^{2}=0$.
(D) $x^{2}+3 b x-c a+2 b^{2}=0$.
(E) $x^{2}+b(2-a) x+a^{2} c+b^{2}(a+1)=0$.
(12th American High School Mathematics Examination, 1961)
[Solution] Given $r+s=-\frac{b}{a}, r s=\frac{c}{a}$.
Also, $[x-(a r+b)][x-(a s+b)]=0$,
thus $x^{2}-[a(r+s)+2 b] x+a^{2} r s+a b(r+s)+b^{2}=0$,
i.e., $\square$ $x^{2}-\left[a\left(-\frac{b}{a}\right)+2 b\right] x+a^{2}\left(\frac{c}{a}\right)+a b\left(-\frac{b}{a}\right)+b^{2}=0$, or
$$
x^{2}-b x+a c=0 \text {. }
$$

Therefore, the answer is $(B)$.