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(B) Given the quadratic equation $7x^2 - 13x + a = 0$. For the roots to be rational,the discriminant $D = b^2 - 4ac$ must be a perfect square. Here,$D

Vedclass · Accepted Answer

$6$

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(B) Given the quadratic equation $7x^2 - 13x + a = 0$. For the roots to be rational,the discriminant $D = b^2 - 4ac$ must be a perfect square. Here,$D = (-13)^2 - 4(7)(a) = 169 - 28a$. For $D$ to be a perfect square,$169 - 28a = k^2$ for some non-negative integer $k$. Testing the given options: If $a = 5$,$D = 169 - 28(5) = 169 - 140 = 29$ (not a perfect square). If $a = 6$,$D = 169 - 28(6) = 169 - 168 = 1 = 1^2$ (a perfect square). Since $a=6$ satisfies the condition and is the smallest value among the options,the smallest possible value of $a$ is $6$.

If $a$ is a positive integer such that the roots of the equation $7x^2 - 13x + a = 0$ are rational numbers,then the smallest possible value of $a$ is

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