Sum of all distinct integral values of alpha | vedclass

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(B) For the quadratic equation $x^2 - \alpha x + \alpha + 1 = 0$ to have integral roots,its discriminant $D$ must be a perfect square.$D = \alpha^2 -

Vedclass · Accepted Answer

$4$

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(B) For the quadratic equation $x^2 - \alpha x + \alpha + 1 = 0$ to have integral roots,its discriminant $D$ must be a perfect square.$D = \alpha^2 - 4(1)(\alpha + 1) = \alpha^2 - 4\alpha - 4$.Let $D = k^2$ for some non-negative integer $k$.$\alpha^2 - 4\alpha - 4 = k^2$$(\alpha^2 - 4\alpha + 4) - 8 = k^2$$(\alpha - 2)^2 - k^2 = 8$$(\alpha - 2 - k)(\alpha - 2 + k) = 8$.Since $8$ is the product of two integers,we consider the factors of $8$: $(1, 8), (2, 4), (-1, -8), (-2, -4)$.Case $1$: $\alpha - 2 - k = 2$ and $\alpha - 2 + k = 4$. Adding these gives $2(\alpha - 2) = 6 \Rightarrow \alpha - 2 = 3 \Rightarrow \alpha = 5$.Case $2$: $\alpha - 2 - k = -4$ and $\alpha - 2 + k = -2$. Adding these gives $2(\alpha - 2) = -6 \Rightarrow \alpha - 2 = -3 \Rightarrow \alpha = -1$.Case $3$: $\alpha - 2 - k = 4$ and $\alpha - 2 + k = 2$. Adding these gives $2(\alpha - 2) = 6 \Rightarrow \alpha = 5$.Case $4$: $\alpha - 2 - k = -2$ and $\alpha - 2 + k = -4$. Adding these gives $2(\alpha - 2) = -6 \Rightarrow \alpha = -1$.Thus,the distinct integral values of $\alpha$ are $5$ and $-1$.The sum of these values is $5 + (-1) = 4$.

Sum of all distinct integral values of $\alpha$ such that the equation $x^2 - \alpha x + \alpha + 1 = 0$ has integral roots,is equal to-

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