If ${a^2}{x^2} + 2bx + c = 0$ has one root greater than unity and the other less than unity, then

Question

The options given are: $$A)\; {a}^{2}+2b+c=0\\ B)\; {a}^{2}+2b+c>0\\ C)\; 2b+c<0\\ D)\; 2b+c > 0$$

How do we proceed with this question?

Answer 1

For the sum $s$ and product $p$ of roots we have $x^2-sx+p=0$.

Thus $s=-\frac{2b}{a^2}$ and $p=\frac c{a^2}$.

Let's write the roots $x_1=(1+u)$ and $x_2=(1-v)$ with $u,v>0$.

We get $s=2+u-v$ and $p=(1+u)(1-v)=1+u-v-uv=s-uv-1$

We arrive to the condition $$s-p=1+uv>1$$

Which is $-(2b+c)>a^2\quad$ and only $(C)$ fulfils this.

Answer 2

Let $p(x)=a^2x^2+2bx+c$. If the roots are $r_1\lt1\lt r_2$, then we must have $p(1)\lt0$, since $p(x)$ is a quadratic with positive lead coefficient $a^2$ -- i.e., it's an upward-pointing parabola. (Note, we can't have $a=0$ since we're told there are two roots.) Thus we have

$$a^2+2b+c\lt0$$

which rules out all options except C).

Alternatively, the quadratic $x^2-2x$, which has roots $0$ and $2$ with $a^2=1$, $b=-1$ and $c=0$, is a counterexample to A), B), and D), leaving only option C).