The sum of all distinct integral values of al | vedclass

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

Vedclass · Accepted Answer

$4$

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(B) Given the quadratic equation $x^2 - \alpha x + \alpha + 1 = 0$.For the roots to be integral,the discriminant $D$ must be a perfect square,say $\beta^2$,where $\beta \ge 0$.$D = (-\alpha)^2 - 4(1)(\alpha + 1) = \alpha^2 - 4\alpha - 4 = \beta^2$.Completing the square: $(\alpha - 2)^2 - 8 = \beta^2$.Rearranging gives: $(\alpha - 2)^2 - \beta^2 = 8$,which factors as $(\alpha - 2 - \beta)(\alpha - 2 + \beta) = 8$.Since $\alpha$ and $\beta$ are integers,we look for factor pairs of $8$ (where the product is $8$ and both factors have the same parity):Case $1$: $(\alpha - 2 - \beta) = 2$ and $(\alpha - 2 + \beta) = 4$. Adding these gives $2(\alpha - 2) = 6$ $\Rightarrow \alpha - 2 = 3$ $\Rightarrow \alpha = 5$.Case $2$: $(\alpha - 2 - \beta) = -4$ and $(\alpha - 2 + \beta) = -2$. Adding these gives $2(\alpha - 2) = -6$ $\Rightarrow \alpha - 2 = -3$ $\Rightarrow \alpha = -1$.Case $3$: $(\alpha - 2 - \beta) = -2$ and $(\alpha - 2 + \beta) = -4$. Adding these gives $2(\alpha - 2) = -6 \Rightarrow \alpha = -1$.Case $4$: $(\alpha - 2 - \beta) = 4$ and $(\alpha - 2 + \beta) = 2$. Adding these gives $2(\alpha - 2) = 6 \Rightarrow \alpha = 5$.The distinct integral values for $\alpha$ are $-1$ and $5$.The sum of these values is $-1 + 5 = 4$.

The 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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