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 $