Question

If the equation $x^2-px+q=0$ and $x^2-ax+b=0$ have a common root, the other root of the second equation is the reciprocal of the other roots of the first, then

• A. ${(q-b)}^2=bq{(p-a)}^2$
• B. $ap{(q-b)}^2={(p-a)}^2$
• C. ${(q-b)}^2=pq{(p-a)}^2$
• D. None of these

Answer 1

The correct option is A ${(q-b)}^2=bq{(p-a)}^2$

Let $\alpha$ and $\beta$ be the roots of $x^2-px+q=0$

Then, $\alpha +\beta =p$ and $\alpha \beta =q$ ...(1)

And $\alpha$ and $\frac{1}{\beta }$ be the roots of $x^2-ax+b=0$

Then, $\alpha +\frac{1}{\beta }=a$ and $\frac{\alpha }{\beta }=b$ ...(2)

From (1) and (2), we get

Now, ${(q-b)}^2={(\alpha \beta -\frac{\alpha }{\beta })}^2={\alpha }^2{(\beta -\frac{1}{\beta })}^2$

$=\frac{\alpha }{\beta }.\beta \alpha {[(\alpha +\beta )-(\alpha +\frac{1}{\beta })]}^2=bq{(p-a)}^2$

Hence, option (A) is correct.