Subjects algebra

Quadratic Roots 137D08

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1. You asked for more questions on further math. Let's consider a common algebra problem: Solve the quadratic equation $ax^2 + bx + c = 0$. 2. The formula to find the roots of a quadratic equation is the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ 3. Important rules: - The discriminant $\Delta = b^2 - 4ac$ determines the nature of the roots. - If $\Delta > 0$, there are two distinct real roots. - If $\Delta = 0$, there is one real root (a repeated root). - If $\Delta < 0$, there are two complex roots. 4. Example: Solve $2x^2 - 4x - 6 = 0$. 5. Calculate the discriminant: $$\Delta = (-4)^2 - 4 \times 2 \times (-6) = 16 + 48 = 64$$ 6. Since $\Delta = 64 > 0$, there are two distinct real roots. 7. Apply the quadratic formula: $$x = \frac{-(-4) \pm \sqrt{64}}{2 \times 2} = \frac{4 \pm 8}{4}$$ 8. Calculate each root: - $x_1 = \frac{4 + 8}{4} = \frac{12}{4} = 3$ - $x_2 = \frac{4 - 8}{4} = \frac{-4}{4} = -1$ 9. Final answer: The solutions are $x = 3$ and $x = -1$.