Question

If both the roots of the equation $a{x}^2+2xa+1+a^2-16=0$ are opposite in sign, then the range of $a$ is

• A. $(-\infty ,-4)\cup (4,\infty )$
• B. $(-4,4)$
• C. $(-\infty ,-4)\cup (0,4)$
• D. $(0,4)$

Answer 1

The correct option is C $(-\infty ,-4)\cup (0,4)$

$a{x}^2+2ax+1+a^2-16=0$ [Ref. image 1]

$\left(a\right)f\left(0\right)<0\Rightarrow \left(a\right)(1+a^2-16)<0$

$\left(a\right)(a+4)(a-4)<0$

$\therefore aϵ(-\infty ,-4)\cup (0,4)$ [Ref. image 2]

$\therefore$ option C