If the roots of the given equation (2k + 1)x^ | vedclass

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(B) For a quadratic equation of the form $ax^2 + bx + c = 0$,if the roots are reciprocal to each other,then their product is equal to $1$.Given the eq

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(B) For a quadratic equation of the form $ax^2 + bx + c = 0$,if the roots are reciprocal to each other,then their product is equal to $1$.Given the equation $(2k + 1)x^2 - (7k + 3)x + k + 2 = 0$,we have $a = 2k + 1$ and $c = k + 2$.The product of the roots is given by $\frac{c}{a}$.Since the roots are reciprocals,$\frac{c}{a} = 1$.Therefore,$\frac{k + 2}{2k + 1} = 1$.Multiplying both sides by $(2k + 1)$,we get $k + 2 = 2k + 1$.Rearranging the terms,$2 - 1 = 2k - k$,which gives $k = 1$.

If the roots of the given equation $(2k + 1)x^2 - (7k + 3)x + k + 2 = 0$ are reciprocal to each other,then the value of $k$ will be

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