1. The problem is to rewrite the quadratic equation $x^2 - x - r^2 = 0$ in the form $(x - a)(x + b) = 0$. 2. The general form of a quadratic equation is $x^2 + bx + c = 0$. When factored as $(x - a)(x + b) = 0$, it expands to $x^2 + (b - a)x - ab = 0$. 3. Comparing coefficients from $x^2 - x - r^2 = 0$ and $x^2 + (b - a)x - ab = 0$, we get: - Coefficient of $x$: $b - a = -1$ - Constant term: $-ab = -r^2$ which implies $ab = r^2$ 4. We need to find $a$ and $b$ such that: $$b - a = -1$$ $$ab = r^2$$ 5. From $b = a - 1$, substitute into $ab = r^2$: $$a(a - 1) = r^2$$ $$a^2 - a - r^2 = 0$$ 6. Solve this quadratic for $a$ using the quadratic formula: $$a = \frac{1 \pm \sqrt{1 + 4r^2}}{2}$$ 7. Then $b = a - 1$. 8. Therefore, the factorization is: $$(x - a)(x + b) = 0$$ where $$a = \frac{1 + \sqrt{1 + 4r^2}}{2} \text{ or } \frac{1 - \sqrt{1 + 4r^2}}{2}$$ and $$b = a - 1$$ This completes the factorization of the quadratic equation.