1. **State the problem:** Find the smallest value of $p$ for which the quadratic equation $$x^2 - 2(p+1)x + p^2 = 0$$ has real and equal roots. Then find the roots of the quadratic equation. 2. **Recall the condition for real and equal roots:** A quadratic equation $$ax^2 + bx + c = 0$$ has real and equal roots if its discriminant $$\Delta = b^2 - 4ac = 0$$. 3. **Identify coefficients:** Here, $$a = 1$$, $$b = -2(p+1)$$, and $$c = p^2$$. 4. **Set the discriminant to zero:** $$\Delta = [-2(p+1)]^2 - 4(1)(p^2) = 0$$ 5. **Simplify the discriminant:** $$4(p+1)^2 - 4p^2 = 0$$ Divide both sides by 4: $$(p+1)^2 - p^2 = 0$$ 6. **Expand and simplify:** $$(p^2 + 2p + 1) - p^2 = 0$$ $$2p + 1 = 0$$ 7. **Solve for $p$:** $$2p = -1$$ $$p = -\frac{1}{2}$$ 8. **Find the roots for $p = -\frac{1}{2}$:** Substitute back into the quadratic: $$x^2 - 2\left(-\frac{1}{2} + 1\right)x + \left(-\frac{1}{2}\right)^2 = 0$$ Calculate inside the parentheses: $$-2\left(\frac{1}{2}\right)x + \frac{1}{4} = 0$$ Simplify: $$x^2 - x + \frac{1}{4} = 0$$ 9. **Since roots are equal, root is:** $$x = \frac{-b}{2a} = \frac{-(-1)}{2(1)} = \frac{1}{2}$$ **Final answer:** - Smallest value of $p$ is $-\frac{1}{2}$. - The roots are both $\frac{1}{2}$.