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(A) For the roots of the quadratic equation $\lambda x^2 + 13x + 7 = 0$ to be rational,the discriminant $D$ must be a perfect square of a rational num

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$4$

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(A) For the roots of the quadratic equation $\lambda x^2 + 13x + 7 = 0$ to be rational,the discriminant $D$ must be a perfect square of a rational number. Since $\lambda$ is an integer,$D$ must be a perfect square of an integer.$D = b^2 - 4ac = (13)^2 - 4(\lambda)(7) = 169 - 28\lambda$.Given $\lambda \in (-3, 7)$,the possible integer values for $\lambda$ are $\{-2, -1, 0, 1, 2, 3, 4, 5, 6\}$.We test these values:If $\lambda = -2$,$D = 169 - 28(-2) = 169 + 56 = 225 = (15)^2$ (Perfect square).If $\lambda = -1$,$D = 169 - 28(-1) = 197$ (Not a perfect square).If $\lambda = 0$,the equation becomes $13x + 7 = 0$,which gives $x = -7/13$ (Rational).If $\lambda = 1$,$D = 169 - 28 = 141$ (Not a perfect square).If $\lambda = 2$,$D = 169 - 56 = 113$ (Not a perfect square).If $\lambda = 3$,$D = 169 - 84 = 85$ (Not a perfect square).If $\lambda = 4$,$D = 169 - 112 = 57$ (Not a perfect square).If $\lambda = 5$,$D = 169 - 140 = 29$ (Not a perfect square).If $\lambda = 6$,$D = 169 - 168 = 1 = (1)^2$ (Perfect square).Thus,the set $S = \{-2, 0, 6\}$.The sum of the elements in $S$ is $-2 + 0 + 6 = 4$.

Let $S$ be the set of all possible integral values of $\lambda$ in the interval $(-3, 7)$ for which the roots of the quadratic equation $\lambda x^2 + 13x + 7 = 0$ are all rational numbers. Then the sum of the elements in $S$ is

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