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(B) For the quadratic equation $x^2+(2m+1)x+m=0$ to have equal roots,the discriminant $D$ must be equal to $0$.The discriminant is given by $D = b^2 -

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$0$

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(B) For the quadratic equation $x^2+(2m+1)x+m=0$ to have equal roots,the discriminant $D$ must be equal to $0$.The discriminant is given by $D = b^2 - 4ac$.Here,$a=1$,$b=(2m+1)$,and $c=m$.Substituting these values into $D=0$:$(2m+1)^2 - 4(1)(m) = 0$$4m^2 + 4m + 1 - 4m = 0$$4m^2 + 1 = 0$$4m^2 = -1$$m^2 = -\frac{1}{4}$Since the square of any real number $m$ must be non-negative $(m^2 \ge 0)$,there are no real values of $m$ that satisfy this equation.Therefore,the number of real values of $m$ is $0$.

The number of real values of $m$ such that the equation $x^2+(2m+1)x+m=0$ has equal roots is

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