Question

The equations $(q-r)x^2+(r-p)x+p-q=0$ and $(r-p)x^2+(p-q)x+q-r=0$ have a common root $x=a$. Find $a$.

Answer 1

$(q-r)x^2+(r-p)x+p-q=0$

$⟹q{x}^2-r{x}^2+rx-px+p-q=0$

$⟹q(x^2-1)+r(x-x^2)+p(1-x)=0$

$⟹q(x+1)(x-1)+rx(1-x)+p(1-x)=0$

$\therefore q(x+1)-rx-p=0$

$⟹qx+q-rx-p=0$

$⟹x(q-r)=p-q$

$⟹x=\frac{p-q}{q-r}$

$(r-p)x^2+(p-q)x+q-r=0$

$⟹r{x}^2-p{x}^2+px-qx+q-r=0$

$⟹r(x^2-1)+p(x-x^2)+q(1-x)=0$

$⟹r(x+1)(x-1)+px(1-x)+q(1-x)=0$

$\therefore r(x+1)-px-q=0$

$⟹rx+r-px-q=0$

$⟹x(r-p)=q-r$

$⟹x=\frac{q-r}{r-p}$

They have common roots $\alpha +\beta =\frac{p-r}{q-r}$

$\beta =\frac{p-r}{q-r}-\left(\frac{p-q}{q-r}\right)$

$⟹\beta =1$