For the equations $x^2 + bx + c = 0$ and $2x^2 + (b + 1)x + c + 1 = 0$ select the correct alternative
-
both the equations can have integral roots
-
both the equations cant have integral roots simultaneously
-
none of the equations can have integral roots
-
nothing can be said
If the first equation has integral roots, then b and c are integers. If the second also has integral roots, its root sum and product require b and c to be odd. But integral roots of the first equation with odd sum must have opposite parity, giving an even product, which contradicts c being odd. Therefore, both equations cannot have integral roots simultaneously.
Assume both equations have integer roots, which means their discriminants must be perfect squares. For the first equation, let b^2 - 4c = p^2, and for the second, let (b+1)^2 - 8(c+1) = q^2, where p and q are non-negative integers. Multiplying the first condition by 2 and subtracting it from the second eliminates the variable c, resulting in the equation 2q^2 - p^2 = (b-3)^2 - 6. An analysis of this Diophantine equation modulo 8 reveals a contradiction because a square modulo 8 can only be 0, 1, or 4, proving the initial assumption is impossible. Therefore, both equations cannot have integral roots simultaneously.