Question

For real numbers \( \alpha \) and \( \beta \), let \( p(x)=x^{2}-(\alpha+\beta)x+\alpha\beta \) and \( q(x)=x^{2}-(\alpha+\beta+2)x+(\alpha+1)(\beta+1) \). Which of the following statements is true?}

• A. If \( p(28)=0, \) then \( q(-91)=0 \)
• B. If \( q(34)=0 \) then \( p(35)=0 \)
• C. If \( q(-97)=0 \), then \( p(-98)=0 \)
• D. If \( p(74)=0 \) then \( q(73)=0 \)

Hint:

Identify the shift in roots. If \( q(x) \) has roots \( r+k \), then \( q(x) = p(x-k) \). Here \( k=1 \), so if \( q(y)=0 \), then \( p(y-1)=0 \).

Answer 1

The Correct Option is C

Strategy:

The polynomial \( p(x) \) factors as \( (x-\alpha)(x-\beta) \). The roots are \( \alpha \) and \( \beta \). The polynomial \( q(x) \) has coefficients: Sum of roots = \( \alpha+\beta+2 = (\alpha+1) + (\beta+1) \). Product of roots = \( (\alpha+1)(\beta+1) \). Therefore, the roots of \( q(x) \) are \( \alpha+1 \) and \( \beta+1 \).
If \( x \) is a root of \( p(x) \), then \( x+1 \) is a root of \( q(x) \). Conversely, if \( y \) is a root of \( q(x) \), then \( y-1 \) is a root of \( p(x) \).
Statement: "If \( q(-97)=0 \), then \( p(-98)=0 \)". If \( q(-97)=0 \), then \(-97\) is a root of \( q(x) \). Therefore, \( -97 - 1 = -98 \) must be a root of \( p(x) \). So \( p(-98)=0 \). This is logically sound.
Option C is the correct implication.