, 2019

A quadratic equation has two distinct real solutions exactly when its discriminant is positive.

For the quadratic equation $x^2 - (r+7)x + r+87 = 0$, the discriminant is $$\Delta = (r+7)^2 - 4(1)(r+87) = r^2 + 14r + 49 - 4r - 348 = r^2 + 10r - 299$$ Since $\Delta = r^2 + 10r - 299 = (r+23)(r-13)$ which has roots $r=-23$ and $r=13$, then $\Delta > 0$ exactly when $r>13$ or $r<-23$. (To see this, we could picture the parabola with equation $y = x^2 + 10x - 299 = (x+23)(x-13)$ and see where it lies above the $x$-axis.)

We also want both of the solutions of the original quadratic equation to be negative.

If $r>13$, then the equation $x^2 - (r+7)x + r+87 = 0$ is of the form $x^2 - bx + c = 0$ with each of $b$ and $c$ positive.

In this case, if $x<0$, then $x^2 > 0$ and $-bx > 0$ and $c>0$ and so $x^2 - bx + c > 0$.

This means that, if $r>13$, there cannot be negative solutions.

Thus, it must be the case that $r<-23$. This does not guarantee negative solutions, but is a necessary condition.

So we consider $x^2 - (r+7)x + r+87 = 0$ along with the condition $r<-23$.

This quadratic is of the form $x^2 - bx + c = 0$ with $b<0$. We do not yet know whether $c$ is positive, negative or zero.

We know that this equation has two distinct real solutions.

Suppose that the quadratic equation $x^2 - bx + c = 0$ has real solutions $s$ and $t$.

This means that the factors of $x^2-bx+c$ are $x-s$ and $x-t$.

In other words, $(x-s)(x-t) = x^2 - bx+c$.

Now, $$(x-s)(x-t) = x^2 - tx - sx + st = x^2 - (s+t)x + st$$ Since $(x-s)(x-t) = x^2 - bx + c$, then $x^2 - (s+t)x + st = x^2 - bx + c$ for all values of $x$, which means that $b = (s+t)$ and $c = st$.

Since $b<0$, then it cannot be the case that $s$ and $t$ are both positive, since $b=s+t$.

If $c=0$, then it must be the case that $s=0$ or $t=0$.

If $c<0$, then it must be the case that one of $s$ and $t$ is positive and the other is negative.

If $c=st$ is positive, then $s$ and $t$ are both positive or both negative, but since $b<0$, then $s$ and $t$ cannot both be positive, hence are both negative.

Knowing that the equation $x^2 - bx + c = 0$ has two distinct real roots and that $b<0$, the condition that the two roots are negative is equivalent to the condition that $c>0$.

Here, $c=r+87$ and so $c>0$ exactly when $r>-87$.

Finally, this means that the equation $x^2 - (r+7)x + r+87 = 0$ has two distinct real roots which are both negative exactly when $-87 < r < -23$.

This means that $p=-87$ and $q=-23$ and so $p^2 + q^2 = 8098$.