1. **State the problem:** Solve the quadratic equation using the quadratic formula. 2. **Quadratic formula:** For any quadratic equation $ax^2 + bx + c = 0$, the solutions are given by $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $b^2 - 4ac$ is called the discriminant. 3. **Important rules:** - If the discriminant $b^2 - 4ac > 0$, there are two distinct real solutions. - If the discriminant $b^2 - 4ac = 0$, there is exactly one real solution (a repeated root). - If the discriminant $b^2 - 4ac < 0$, there are no real solutions (the solutions are complex). 4. **Apply the quadratic formula to the first exercise:** $x^2 - 5x + 4 = 0$ - Identify coefficients: $a=1$, $b=-5$, $c=4$ - Calculate discriminant: $$b^2 - 4ac = (-5)^2 - 4 \times 1 \times 4 = 25 - 16 = 9$$ - Since $9 > 0$, there are two real solutions. 5. **Calculate the solutions:** $$x = \frac{-(-5) \pm \sqrt{9}}{2 \times 1} = \frac{5 \pm 3}{2}$$ 6. **Find each root:** - First root: $$x = \frac{5 + 3}{2} = \frac{8}{2} = 4$$ - Second root: $$x = \frac{5 - 3}{2} = \frac{2}{2} = 1$$ 7. **Show cancellation step for clarity:** $$x = \frac{\cancel{-(-5)} \pm \sqrt{9}}{\cancel{2 \times 1}} = \frac{5 \pm 3}{2}$$ **Final answer:** The solutions to $x^2 - 5x + 4 = 0$ are $x=4$ and $x=1$.