If the roots of the equation 6x^2 - 13x + m = | vedclass

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(B) Let the roots of the quadratic equation $ax^2 + bx + c = 0$ be $\alpha$ and $\beta$.Given that the roots are reciprocal of each other,we have $\be

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$ 6 $

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(B) Let the roots of the quadratic equation $ax^2 + bx + c = 0$ be $\alpha$ and $\beta$.Given that the roots are reciprocal of each other,we have $\beta = 1/\alpha$,which implies $\alpha \cdot \beta = 1$.For a quadratic equation $ax^2 + bx + c = 0$,the product of the roots is given by $\alpha \cdot \beta = c/a$.Comparing $6x^2 - 13x + m = 0$ with $ax^2 + bx + c = 0$,we get $a = 6$,$b = -13$,and $c = m$.Substituting these values into the product of roots formula: $1 = m/6$.Therefore,$m = 6$.

If the roots of the equation $6x^2 - 13x + m = 0$ are reciprocal of each other,then $m = $ ...........

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