If the roots of the equation 6x^2 - 7x + k = | vedclass

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(B) For the roots of a quadratic equation $ax^2 + bx + c = 0$ to be rational,the discriminant $D = b^2 - 4ac$ must be a perfect square.Here,$a = 6$,$b

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$1, 2$

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(B) For the roots of a quadratic equation $ax^2 + bx + c = 0$ to be rational,the discriminant $D = b^2 - 4ac$ must be a perfect square.Here,$a = 6$,$b = -7$,and $c = k$.$D = (-7)^2 - 4(6)(k) = 49 - 24k$.For $D$ to be a perfect square,we test values of $k$:If $k = 1$,$D = 49 - 24(1) = 25 = 5^2$ (a perfect square).If $k = 2$,$D = 49 - 24(2) = 49 - 48 = 1 = 1^2$ (a perfect square).Thus,the values of $k$ for which the roots are rational are $1$ and $2$.

If the roots of the equation $6x^2 - 7x + k = 0$ are rational,then $k = .......$

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