If the sum of the roots of the equation x2 − x = λ(2x − 1) is zero, then λ =

The given quadric equation is $x^{2}-x=\lambda(2 x-1)$, and roots are zero.

Then find the value of $\lambda$.

Here,

$x^{2}-x=\lambda(2 x-1)$

$x^{2}-x-2 \lambda x+\lambda=0$

$x^{2}-(1+2 \lambda) x+\lambda=0$

$a=1, b=-(1+2 \lambda)$ and, $c=\lambda$

As we know that $D=b^{2}-4 a c$

Putting the value of $a=1, b=-(1+2 \lambda)$ and, $c=\lambda$

$=\{-(1+2 \lambda)\}^{2}-4 \times 1 \times \lambda$

$=1+4 \lambda+4 \lambda^{2}-4 \lambda$

$=1+4 \lambda^{2}$

The given equation will have zero roots, if $D=0$

$1+4 \lambda^{2}=0$

$4 \lambda^{2}=-1$

$\lambda^{2}=\frac{-1}{4}$

$\lambda=\sqrt{\frac{-1}{4}}$

$=\frac{-1}{2}$

Therefore, the value of $\lambda=-\frac{1}{2}$

Thus, the correct answer is $(c)$