Question

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

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

Answer 1

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


$f\left(x\right)=x^2+2(a-3)x+9$ is a parabola facing upwards as shown in the figure.
Let the roots be $\alpha$ and $\beta$ with $\alpha <\beta$
If $6$ lies between $\alpha$ and $\beta ,$
$f\left(6\right)<0$
$\Rightarrow (6{)}^2+2(a-3\left)\right(6)+9<0$
$\Rightarrow 12a+9<0$
$\Rightarrow a<-\frac{3}{4}$