Question

The roots of $x^2+bx+c=0$ are both real and greater than $1.$ Let $s=b+c+1.$ Then $s:$

may be less than zero

may be equal to zero

must be greater than zero

must be less than zero

must be between $-1$ and $1$

Call the two roots $u$ and $v$ and express $b,c$ using Vieta’s formulas

Factor $1-(u+v)+uv$

If the roots are $u,v>1,$ then $b=-(u+v)$ and $c=uv.$ Therefore $\begin{aligned} s&=1-u-v+uv\\ &=(u-1)(v-1)>0. \end{aligned}$

Thus $s$ must be greater than zero, and the correct answer is C.