1. **Problem statement:** We are given the quadratic equation $$r^2x^2 + 2r(2q - p)x + (p - 2q)^2 = 0$$ where $p$, $q$, and $r$ are nonzero coefficients. We need to determine the nature of its roots. 2. **Recall the quadratic formula and discriminant:** For a quadratic equation $$ax^2 + bx + c = 0$$ the roots are given by $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ and the discriminant $$\Delta = b^2 - 4ac$$ determines the nature of roots: - If $\Delta > 0$, two distinct real roots. - If $\Delta = 0$, two equal real roots. - If $\Delta < 0$, complex roots. 3. **Identify coefficients:** - $a = r^2$ - $b = 2r(2q - p)$ - $c = (p - 2q)^2$ 4. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = \left(2r(2q - p)\right)^2 - 4(r^2)(p - 2q)^2$$ 5. **Simplify the discriminant:** $$\Delta = 4r^2(2q - p)^2 - 4r^2(p - 2q)^2$$ 6. **Notice that $(2q - p)^2 = (p - 2q)^2$ because squaring removes sign:** $$\Delta = 4r^2\cancel{(2q - p)^2} - 4r^2\cancel{(p - 2q)^2} = 0$$ 7. **Interpretation:** Since the discriminant is zero for any nonzero $p$, $q$, and $r$, the quadratic equation has two equal real roots always. **Final answer:** There are two equal real roots for any value of $p$, $q$, and $r$.