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Prove that the equation

(x - p) (x - q) = r^{2}

has at least one real solution in $x$ for any choice of $p, q, r \in R .$

Hint

Collect it into a quadratic in terms of $x$ and consider the discriminant.

Solution

We begin by expanding and collecting

\begin{aligned} x^{2} - p x - q x + p q & = r^{2} \\ x^{2} - (p + q) x + p q - r^{2} & = 0 \end{aligned}

The discriminant is

\begin{aligned} (p + q)^{2} - 4 (p q - r^{2}) \\ = & p^{2} + 2 p q + q^{2} - 4 p q + 4 r^{2} \\ = & p^{2} - 2 p q + q^{2} + 4 r^{2} \\ = & (p - q)^{2} + 4 r^{2} \end{aligned}

This expression is the sum of two squares which must be positive for all choices of $p, q, r \in R$ .