Find the value of p so that the equation x^2+10x+21=0 and x^2+9x+p=0 may have a common root. Find also the equation formed by the other root.
Observe that $\displaystyle f(x) = x^2+10x+21 = (x+3)(x+7)$. Hence roots of $\displaystyle f(x) = 0$ are $\displaystyle x=-3$ and $\displaystyle x=-7$.
Observe that roots of $\displaystyle g(x) = x^2+9x+p =0$ are given by $\displaystyle x_0 = \frac{-9\pm \sqrt{81-4p}}{2}$.
Thus if you find the values of $\displaystyle p$ where $\displaystyle x_0 = -3$ or $\displaystyle x_0 = -7$ you're done.
Sorry. This can be done easier ;p.
Since you must find p such that $\displaystyle (x-a)(x+3) =x^2+(3-a)x-3a = x^2+9x+p$. Observe the only possibility is $\displaystyle a = -\frac{p}{3}$.
Thus now we must find p such that: $\displaystyle x^2+(3+\frac{p}{3})x+ p= x^2+9x+p$. This gives $\displaystyle p = 18$
Now you can do the same trick for $\displaystyle f(x) = (x-a)(x+7)$ to find another value of p