1. The problem is to solve or analyze a pre-calculus question, but since no specific problem was given, let's consider a common pre-calculus topic: solving a quadratic equation. 2. The general form of a quadratic equation is $$ax^2 + bx + c = 0$$ where $a \neq 0$. 3. The quadratic formula to find the roots is: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ 4. Important rules: - The discriminant $\Delta = b^2 - 4ac$ determines the nature of roots. - If $\Delta > 0$, two distinct real roots. - If $\Delta = 0$, one real root (repeated). - If $\Delta < 0$, two complex roots. 5. Example: Solve $$2x^2 - 4x - 6 = 0$$ 6. Identify coefficients: $a=2$, $b=-4$, $c=-6$. 7. Calculate discriminant: $$\Delta = (-4)^2 - 4 \times 2 \times (-6) = 16 + 48 = 64$$ 8. Since $\Delta = 64 > 0$, two distinct real roots. 9. Apply quadratic formula: $$x = \frac{-(-4) \pm \sqrt{64}}{2 \times 2} = \frac{4 \pm 8}{4}$$ 10. Calculate roots: - $$x_1 = \frac{4 + 8}{4} = \frac{12}{4} = 3$$ - $$x_2 = \frac{4 - 8}{4} = \frac{-4}{4} = -1$$ 11. Final answer: The roots are $x=3$ and $x=-1$. This example illustrates how to solve quadratic equations using the quadratic formula in pre-calculus.