Question

If the equations $x^2-3px+2q=0$ and $x^2-3ax+2b=0$ have a common root and the other root of the second equation is the reciprocal of the other root of the first equation, then $(2q-2b{)}^2$ is equal to

• A. $36pa(q-b{)}^2$
• B. $18pa(q-b{)}^2$
• C. $36bq(p-a{)}^2$
• D. $18bq(p-a{)}^2$

Answer 1

The correct option is C $36bq(p-a{)}^2$

Given,
$x^2-3px+2q=0$ and $x^2-3ax+2b=0$
Let $\alpha$ be the common roots and $\beta ,\frac{1}{\beta }$ are the other roots of respective equations.
$\alpha \beta =2q$ and $\frac{\alpha }{\beta }=2b$
$\alpha +\beta =3p$ and $\alpha +\frac{1}{\beta }=3a$
$\alpha \beta \times \frac{\alpha }{\beta }=4qb={\alpha }^2$
Now,$(2q-2b{)}^2$
$=(\alpha \beta -\frac{\alpha }{\beta }{)}^2$
$={\alpha }^2(\beta -\frac{1}{\beta }{)}^2$
$=4qb(\alpha +\beta -(\alpha +\frac{1}{\beta }){)}^2$
$=4qb(3p-3a{)}^2=36qb(p-a{)}^2$
Hence option 'C' is correct.