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

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(B) Let the roots of the quadratic equation be $\alpha$ and $\frac{1}{\alpha}$.For a quadratic equation $ax^2 + bx + c = 0$,the product of the roots i

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$1$

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(B) Let the roots of the quadratic equation be $\alpha$ and $\frac{1}{\alpha}$.For a quadratic equation $ax^2 + bx + c = 0$,the product of the roots is given by $\frac{c}{a}$.Here,$a = 2k + 1$,$b = -(7k + 3)$,and $c = k + 2$.Since the roots are reciprocal,their product is $\alpha 	imes \frac{1}{\alpha} = 1$.Therefore,$\frac{c}{a} = 1 \Rightarrow \frac{k + 2}{2k + 1} = 1$.Solving for $k$: $k + 2 = 2k + 1$.$2 - 1 = 2k - k$.$k = 1$.

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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