Question

The range of $a$ for which the equation $x^2+ax-4=0$ has its smaller root in the interval $(-1,2)$ is

• A. $(-\infty ,-3)$
• B. $(0,3)$
• C. $(0,\infty )$
• D. $(-\infty ,-3)\cup (0,\infty )$

Answer 1

The correct option is A $(-\infty ,-3)$

Let $f\left(x\right)=x^2+ax-4$
$f\left(0\right)=-4$


The required conditions are
$f(-1)>0$
$\Rightarrow f(-1)=1-a-4>0\Rightarrow a<-3\cdots \left(1\right)$

$f\left(2\right)<0$
$f\left(2\right)=4+2a-4<0$
$\Rightarrow a<0\cdots \left(2\right)$
From equation $\left(1\right)$ and $\left(2\right)$,
$\Rightarrow a\in (-\infty ,-3)$

Boundary condition : When one root is between $(-1,2)$ and other root is $2$, then
$f\left(2\right)=0\Rightarrow 4+2a-4=0\Rightarrow a=0$
When $a=0$, we get
$f\left(x\right)=x^2-4=0\Rightarrow x=-2,2\notin (-1,2)$

Hence, $a\in (-\infty ,-3)$