Question

Find the value of a,so that 6 lies between the roots of the equation $x^2+2(a-3)x+9=0$.

• A. undefined
• B. undefined
• C. undefined
• D. undefined

Answer 1

$x^2+2(a-3)x+9=0$, Consider graph,
The graph is concave upward since $a>0$ (+ve).

So, conditions for 6 to be in between the roots are,
(i) $f\left(6\right)<0$ and (ii) $D>0$ (for the roots to be real)
(i) $f\left(6\right)<0\Rightarrow (6{)}^2+2(a-3)6+9<0\Rightarrow 36-36+12a+9<0\Rightarrow a<-\frac{3}{4}$
(ii) $D>0\Rightarrow \left[2\right(a-3{)}^2-]4\times 9\times 1>0\Rightarrow (a-3{)}^2-9>0\Rightarrow (a-3{)}^2-(3{)}^2>0$
$\Rightarrow (a-6)a>0$

$\Rightarrow a<0$ and $a>6$
Containing (i) and (ii),

common part is $a<-\frac{3}{4}$.
Thus, value of a, so that 6 lies between the roots is $a<-\frac{3}{4}$.