If the roots of x^2 - bx + c = 0 are two cons | vedclass

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(A) Let the roots be $\alpha$ and $\alpha + 1$.According to the properties of quadratic equations,the sum of the roots is $\alpha + (\alpha + 1) = 2\a

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(A) Let the roots be $\alpha$ and $\alpha + 1$.According to the properties of quadratic equations,the sum of the roots is $\alpha + (\alpha + 1) = 2\alpha + 1 = b$.The product of the roots is $\alpha(\alpha + 1) = c$.We need to find the value of the discriminant $D = b^2 - 4c$.Substituting the expressions for $b$ and $c$:$b^2 - 4c = (2\alpha + 1)^2 - 4(\alpha(\alpha + 1))$Expanding the terms:$b^2 - 4c = (4\alpha^2 + 4\alpha + 1) - (4\alpha^2 + 4\alpha)$Simplifying the expression:$b^2 - 4c = 4\alpha^2 + 4\alpha + 1 - 4\alpha^2 - 4\alpha = 1$.Therefore,$b^2 - 4c = 1$.

If the roots of $x^2 - bx + c = 0$ are two consecutive integers,then $b^2 - 4c$ is

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