Question

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

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

Answer 1

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

Let the roots of the first equation be $\alpha ,\beta$
Hence the roots of the second equation will be $\alpha ,\frac{1}{\beta }$
Therefore the first equation can be re-written as
$(x-\alpha )(x-\beta )=0$
Or
$x^2-(\alpha +\beta )x+\alpha .\beta =0$ ...(i)
And the second equation will be
$x^2-(\alpha +\frac{1}{\beta })x+\frac{\alpha }{\beta }=0$ ...(ii)
Now
$\alpha +\beta =p$ and $\alpha .\beta =q$
And
$\alpha +\frac{1}{\beta }=a$ and $\frac{\alpha }{\beta }=b$
Hence
$\frac{\alpha .\beta }{\frac{\alpha }{\beta }}=\frac{q}{b}$
${\beta }^2=\frac{q}{b}$ ...(i)
And
$\alpha +\frac{1}{\beta }=a$
And
$\alpha +\beta =p$
Hence
$\beta -\frac{1}{\beta }=p-a$
$\frac{{\beta }^2-1}{\beta }=p-a$
Now
${\beta }^2=\frac{q}{b}$
Substituting only ${\beta }^2$ we get
$\frac{q-b}{b(p-a)}=\beta$
Or
$\frac{(q-b{)}^2}{b^2(p-a{)}^2}={\beta }^2$
Or
$\frac{(q-b{)}^2}{b^2(p-a{)}^2}=\frac{q}{b}$
Or
$(q-b{)}^2=bq(a-p{)}^2$.