Let $p,q,$ and $r$ be constants. One solution to the equation $(x-p)(x-q) = (r-p)(r-q)$ is $x=r$. Find the other soultion in terms of $p,q,$ and $r$.

Question

Let $p,q,$ and $r$ be constants. One solution to the equation $(x-p)(x-q) = (r-p)(r-q)$ is $x=r$. Find the other soultion in terms of $p,q,$ and $r$.

The quadratic in $x$ is $x^2-(p+q)x-pq$. This means the sum of roots $= p+q$. Does that mean either $p =r$ or $q=r$?

Answer 1

You can use Vieta's Theorem. Your equation is equal to $x^2-(p+q)x-r(r-(p+q))=0$. One of the solutions is $r$ hence one root of this polynomial is $r$. Let $y$ be the second root, we get: $r+y=p+q$, $ry =-r(r-(p+q))$

Now set $y=-r+p+q$ and we get $(y-p)(y-q)=(-r+q)(-r+p)=(-1)(r-q)(-1)(r-p)=(r-q)(r-p)$. This shows $y$ is a solution to your equation.

Answer 2

Alternative approach:

Let $f(x) = x^2 + x(-p-q) + (-r^2 + rp + rq).$

You are given that the quadratic equation $f(x) = 0$, has the root $x = r$.

Therefore, you can use polynomial long division, dividing $f(x)$ by $(x - r)$ to obtain the other factor, which is $(x + r - p - q)$.

Therefore, the other root to the quadratic equation $f(x) = 0$ must be $x = (p + q - r).$