$2 \cdot 23$ When solving a certain quadratic equation, student A made a mistake in the constant term of the equation, and thus obtained the two roots of the equation as 8 and 2. Student B made a mistake in the coefficient of the linear term, and thus obtained the two roots of the equation as -9 and -1. Then the correct equation is
(A) $x^{2}-10 x+9=0$.
(B) $x^{2}+10 x+9=0$.
(C) $x^{2}-10 x+16=0$.
(D) $x^{2}-8 x-9=0$.
(E) None of the above.
(4th American High School Mathematics Examination, 1953)
[Solution] The equation solved by A is $(x-8)(x-2)=x^{2}-10 x+16=0$. The equation solved by B is $(x+9)(x+1)=x^{2}+10 x+9=0$.
The constant term of the former equation is incorrect; the coefficient of the linear term of the latter equation is incorrect, so the correct equation is
$$
x^{2}-10 x+9=0 \text {. }
$$

Therefore, the answer is $(A)$.