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Algebra Difficulty 4.8 AIME Find the answer

2⋅622 \cdot 62 The roots of the equation x2+bx+c=0x^{2}+b x+c=0 are both real numbers greater than 1. Let s=b+c+1s=b+c+1, then ss
（A）may be less than zero.
（B）may be equal to zero.
(C) should be greater than zero.
(D) should be less than zero.
(E) is between -1 and 1.
（10th American High School Mathematics Examination, 1959）

This was a multiple-choice question, but the options didn't survive into the source we have. The answer given is C, and the solution below works it through.

Solution

[Solution] Let the two roots of the equation be 1+α1+\alpha and 1+β1+\beta, where α>0\alpha>0 and β>0\beta>0. By Vieta's formulas,
{(1+α)+(1+β)=−b,(1+α)(1+β)=c. \left\{\begin{array}{l} (1+\alpha)+(1+\beta)=-b, \\ (1+\alpha)(1+\beta)=c . \end{array}\right.

Then,
s=b+c+1=−(α+β+2)+(1+α)(1+β)+1=αβ>0. \begin{aligned} s & =b+c+1 \\ & =-(\alpha+\beta+2)+(1+\alpha)(1+\beta)+1 \\ & =\alpha \beta>0 . \end{aligned}

Therefore, the answer is (C)(C).

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.