$x^2+\left (a+p+\frac{2(b-q)}{p-a}\right )x+\left (a+\frac{b-q}{p-a}\right )\left (p+\frac{b-q}{p-a}\right )$

claserken wrote: $x^2+\left (a+p+\frac{2(b-q)}{p-a}\right )x+\left (a+\frac{b-q}{p-a}\right )\left (p+\frac{b-q}{p-a}\right )$ Please show your solution also.

Well, if $\alpha $ be the common root, then, $(\alpha^2+a\alpha+b)-(\alpha^2+p\alpha+q)=0-0=0$ $\Rightarrow \alpha=\frac{b-q}{p-a}$. Also, if $\alpha,\beta$ be roots of 1st eqn and $\alpha,\gamma$ be that of 2nd eqn,then we have $\alpha+\beta=-a$, $\alpha+\gamma=-p$. So the required eqn is $=x^2-(\beta+\gamma)x+\beta \gamma$ =the @bove

TST for Kosovo is very easy

Why did we do all that work @2 above? Let the other roots be $\beta$ and $\gamma$. We want a quadratic with a sum of roots $\beta + \gamma$ and a product of roots $\beta \gamma$. So we will eventually get down to this monic: $$x^2 - (\beta + \gamma)x + \beta\gamma$$