Question

All possible values of $a$, so that $6$ lies between the roots of the equation $x^2+2(a-3)x+9=0$ is

• A. $(-\infty ,-2)\cup (2,\infty )$
• B. $(-\infty ,-\frac{3}{4})$
• C. $(2,\infty )$
• D. none of these

Answer 1

The correct option is B $(-\infty ,-\frac{3}{4})$

Given equation,
$x^2+2(a-3)x+9=0$
Condition for a number 6 to lie between the roots is
$f\left(6\right)<0\dots \dots (\because a.f(k)<0)$
$\Rightarrow 36+12(a-3)+9<0$
$\Rightarrow 12a+9<0$
$\Rightarrow a<-\frac{3}{4}$
$a\in (-\infty ,-\frac{3}{4})$