Question

For $a , b \in R -\{0\}$, let $\alpha, \beta$ be roots of the equation $x ^2- ax + b =0$. The quadratic equation whose roots are $(\alpha-a)^{-3}$ and $(\beta-a)^{-3}$ is

• A. $b^3 x^2+\left(a^3-3 a b\right) x+1=0$
• B. $a ^3 x ^2+\left( b ^3-3 ab \right) x +1=0$
• C. $b^3 x^2+\left(a^3-a b\right)+1=0$
• D. $a^3 x^2+\left(b^3-a b\right) x+1=0$

Answer 1

Answer: (A)

Now, $ \alpha^2-a \alpha+b=0 \Rightarrow \alpha(\alpha-a)=-b \Rightarrow(\alpha-a)=\frac{-b}{\alpha}$
Also, $ \beta^2-a \beta+b=0 \Rightarrow \beta(\beta-a)=-b \Rightarrow(\beta-a)=\frac{-b}{\beta}$
Now, sum of roots $=(\alpha-a)^{-3}+(\beta-a)^{-3}$
$=\frac{1}{(\alpha-a)^3}+\frac{1}{(\beta-a)^3}=\frac{-\alpha^3}{b^3}-\frac{\beta^3}{b^3}=\frac{-1}{b^3}\left[(\alpha+\beta)^3-3 \alpha \beta(\alpha+\beta)\right]=\frac{-1}{b^3}\left(a^3-3 b a\right)$
Also, product of roots $=(\alpha-a)^{-3} \cdot(\beta-a)^3$
$=\frac{1}{(\alpha-a)^3} \frac{1}{(\beta-a)^3}=\left(\frac{-\alpha^3}{b^3}\right)\left(\frac{-\beta^3}{b^3}\right)=\frac{\alpha^3 \beta^3}{b^6}=\frac{b^3}{b^6}=\frac{1}{b^3}$
$\therefore $ The required quadratic equation is, $x ^2- Sx + P =0$
$\Rightarrow x ^2+\frac{1}{ b ^3}\left( a ^3-3 ab \right) x +\frac{1}{ b ^3}=0 $
$\Rightarrow b ^3 x ^2+\left( a ^3-3 ab \right) x +1=0 $