Curves and Solutions: Quadratic Equations
Quadratic equations have the form ax² + bx + c = 0, where the highest power of the variable is 2. Their solutions — called roots — represent where a parabola crosses the x-axis. Quadratics model projectile motion, area optimization, profit calculations, and many other situations where relationships are not simply linear.
Components of Quadratic Equations
This section covers the main methods for solving quadratics:
- Factoring Method: Write the quadratic as a product of two binomials and set each factor equal to zero.
- Quadratic Formula: Use x = (-b ± √(b² - 4ac)) / 2a to find solutions for any quadratic equation.
- Completing the Square: Rewrite the equation so the left side forms a perfect-square trinomial, then take the square root of both sides.
- The Discriminant: The value b² - 4ac determines the number and type of solutions — positive means two real roots, zero means one, negative means none (real).
Examples of Quadratic Equations
Factoring Method Examples
- Solve x² + 7x + 12 = 0: Factor as (x + 3)(x + 4) = 0, so x = -3 or x = -4.
- Solve x² - 5x = 0: Factor as x(x - 5) = 0, so x = 0 or x = 5.
- Solve 2x² - 8x + 6 = 0: Divide by 2 to get x² - 4x + 3 = 0, factor as (x - 1)(x - 3) = 0, so x = 1 or x = 3.
Quadratic Formula Examples
- Solve x² + 2x - 8 = 0: a = 1, b = 2, c = -8. Compute b² - 4ac = 4 + 32 = 36. Then x = (-2 ± 6) / 2, giving x = 2 or x = -4.
- Solve 3x² - x - 2 = 0: a = 3, b = -1, c = -2. Compute b² - 4ac = 1 + 24 = 25. Then x = (1 ± 5) / 6, giving x = 1 or x = -2/3.
- A ball is thrown upward with height h = -16t² + 48t. It hits the ground when h = 0: -16t² + 48t = 0, so t(−16t + 48) = 0, giving t = 0 or t = 3 seconds.
Completing the Square Examples
- Solve x² + 6x = 7: Half of 6 is 3, and 3² = 9. Add 9 to both sides: x² + 6x + 9 = 16, so (x + 3)² = 16, x + 3 = ±4, giving x = 1 or x = -7.
- Solve x² - 4x = 5: Half of -4 is -2, and (-2)² = 4. Add 4: x² - 4x + 4 = 9, so (x - 2)² = 9, x - 2 = ±3, giving x = 5 or x = -1.
- Solve x² + 8x + 10 = 0: Move 10: x² + 8x = -10. Add 16: (x + 4)² = 6, so x = -4 ± √6.
Discriminant Examples
- For x² + 3x + 5 = 0: b² - 4ac = 9 - 20 = -11 (negative), so there are no real solutions.
- For x² - 6x + 9 = 0: b² - 4ac = 36 - 36 = 0, so there is exactly one repeated root: x = 3.
- For x² + x - 6 = 0: b² - 4ac = 1 + 24 = 25 (positive), so there are two distinct real roots.