Question

The set of values of $a$ for which $6$ lies between the roots of the equation $x^2+2(a-3)x+9=0$, is

• A. $(-\infty ,0)\cup (6,\infty )$
• B. $(-\infty ,-\frac{3}{4})$
• C. $(-\frac{3}{4},\frac{3}{4})$
• D. $(\frac{3}{4},\infty )$

Answer 1

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

Let $\alpha$ and $\beta$ be the roots of equation $f\left(x\right)=x^2+2(a-3)x+9=0.$

Figure:



As $6$ lies between the roots,
$\therefore f\left(6\right)<0$
$\Rightarrow 36+12(a-3)+9<0$
$\Rightarrow 12a+9<0$
$\Rightarrow a\in (-\infty ,-\frac{3}{4})$