Multiple choice

If $\alpha , \, \beta$ are roots of the equation $ax^2 + bx + c = 0$, where $a, b, c$ are distinct real values, then $(1 + \alpha + \alpha^2) ( 1 + \beta + \beta^2)$ is

  1. $1-\left [ \left ( \dfrac{b+c-2c}{a} \right )-\left ( \dfrac{b^2+c^2+bc}{a^2} \right ) \right ]$
  2. $1-\left [ \left ( \dfrac{b+c-2c}{a} \right )-\left ( \dfrac{b^2+c^2+bc}{a^3} \right ) \right ]$
  3. $1-\left [ \left ( \dfrac{a+c-2c}{a} \right )-\left ( \dfrac{a^2+c^2+bc}{a^2} \right ) \right ]$
  4. $1-\left [ \left ( \dfrac{a+b-2c}{a} \right )-\left ( \dfrac{c^2+b^2+ac}{a^2} \right ) \right ]$
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AI explanation

Since alpha and beta are roots of the quadratic equation ax^2 + bx + c = 0, we know a(alpha)^2 + b(alpha) + c = 0. Rearranging this gives (alpha)^2 = -(b(alpha) + c) / a, so 1 + alpha + (alpha)^2 equals 1 + alpha - (b(alpha) + c)/a, which simplifies to (a - c)/a + alpha(a - b)/a. Applying this identity to both roots gives the product (1 + alpha + (alpha)^2)(1 + beta + (beta)^2) = [(a - c + alpha(a - b))/a][(a - c + beta(a - b))/a]. Expanding the numerator and substituting the sum and product of roots, which are alpha + beta = -b/a and alpha*beta = c/a, yields the final simplified expression of 1 - [((b + c - 2c)/a) - ((b^2 + c^2 + bc)/a^2)].

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