Question

方程 $x^2+bx+c=0$ 的两根均为大于 $1$ 的实数。令 $s=b+c+1$ 。则 $s$ ：

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

可能小于零

may be less than zero

可能等于零

may be equal to zero

必大于零

must be greater than zero

必小于零

must be less than zero

必介于 $-1$ 与 $1$ 之间

must be between $-1$ and $1$

设两根为 $u$ 和 $v$ ，并用韦达定理表示 $b,c$

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

将 $1-(u+v)+uv$ 因式分解

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

若两根为 $u,v>1$ ，则 $b=-(u+v)$ 且 $c=uv$ 。因此 $\begin{aligned} s&=1-u-v+uv\\ &=(u-1)(v-1)>0 \end{aligned}\text{。}$

因此 $s$ 必大于零，正确答案是 C。

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.