Question

The value of ${}^{\prime }{a}^{\prime }$ so that $6$ lies between the roots of the equation $x^2+2(a-3)x+9=0$ are

• A. $a>-\frac{3}{4}$
• B. $a<-\frac{3}{4}$
• C. $a>\frac{3}{4}$
• D. $a<\frac{3}{4}$

Answer 1

The correct option is B $a<-\frac{3}{4}$

Let $f\left(x\right)=x^2+2(a-3)x+9$. If $6$ lies between the roots of $f\left(x\right)=$, then we must have the following:
(1) $D>0$
(2) $f\left(6\right)<0$ $(\because$ coefficient of $x^2$ is positive $)$.
Now $D>0\Rightarrow 4{(a-3)}^2-36\Rightarrow {(a-3)}^2-9>0$
$\Rightarrow a^2-6a>0\Rightarrow a(a-6)>0\Rightarrow a<0$ or $a>6$ ...(i)
and $f\left(6\right)<0\Rightarrow 36+12(a-3)+9<0\Rightarrow 12a+9<0\Rightarrow a<-\frac{3}{4}$ ...(ii)
From (i) and (ii) we get
$a<-\frac{3}{4}$