Question

The roots of the equation $x^2-3x+a=0$ are $\alpha ,\beta$ and roots of the equation $x^2-12x+b=0$ are $\gamma ,\delta$ and $\alpha ,\beta ,\gamma ,\delta$ are in GP, then

• A. $a=2,b=32$
• B. $a=3,b=12$
• C. $a=4,b=16$
• D. $a=12,b=3$

Answer 1

Answer: (A)

Let $\beta =\alpha r,=\alpha r^2,\delta =\alpha r^3,r>1$
For equation $x^2-3x+a=0$ $\alpha +\beta =3$
$\Rightarrow$ $\alpha +\alpha r=3$
$\Rightarrow$ $\alpha (1+r)=3$ ...(i)
and $\alpha .\beta =a$
$\Rightarrow$ $a=\alpha .\alpha r={\alpha }^{2}r$ ..(ii)
For equation $x^2-12x+b=0$
$\gamma +\delta =12$
$\Rightarrow$ $\alpha r^2+\alpha r^3=12$
$\Rightarrow$ $\alpha r^2(1+r)=12$ ..(iii) and
$\gamma \delta =b$
$\Rightarrow$ $b=\alpha r^2.\alpha r^3={\alpha }^2{r}^5$ ...(iv)
From Eqs. (i) and (iii), $r^2.3=12$
$\Rightarrow$ $r^2=4$
$\Rightarrow$ $r=2$
Then, $\alpha (1+2)=3$
$\Rightarrow$ $\alpha =1$ On putting the values of $\alpha$ and r in Eqs. (ii) and (iv),
we get $a={\alpha }^2.r={(1)}^2\times 2=2$ and $b={(1)}^2{(2)}^5=32$