Question

If $x^2-ax+b=0$ and $x^2-px+q=0$ have a root in common and the second equation has equal roots, show that $b+q=\frac{ap}{2}.$

Answer 1

Given equations are: $x^2-ax+b=0$ (i)
and $x^2-px+q=0$ (ii)

Let $\alpha$ be the common root. Then roots of equation (ii) will be $\alpha$ and $\alpha .$ Let $\beta$ be the other root of equation (i).

Thus roots of equation (i) are $\alpha ,\beta$ and those of equation (ii) are $\alpha ,\alpha$

Now $\alpha +\beta =a$ (iii)

$\alpha \beta =b$ (iv)

$2\alpha =p$ (v)

${\alpha }^2=q$ (vi)

L.H.S.$=b+q=\alpha \beta +{\alpha }^2=\alpha (\alpha +\beta )$ (vii)

and R.H.S.$=\frac{ap}{2}=\frac{(\alpha +\beta )2\alpha }{2}=\alpha (\alpha +\beta )$ (viii)

from (vii) and (viii), L.H.S.=R.H.S.