Solving Quadratic Equations using the Quadratic Formula

Introduction

• The lecture discusses how to solve quadratic equations using the quadratic formula.

The Quadratic Formula

• Formula: ( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} )
• Components:
  • (a): Coefficient in front of (x^2)
  • (b): Coefficient in front of (x)
  • (c): Constant term

Example 1: Solving (2x^2 + 3x - 2 = 0)

1. Identify a, b, c
   • (a = 2), (b = 3), (c = -2)
2. Substitute into the formula
   • (x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-2)}}{2(2)})
   • (x = \frac{-3 \pm \sqrt{9 + 16}}{4})
   • (x = \frac{-3 \pm 5}{4})
3. Calculate solutions
   • (x = \frac{-3 + 5}{4} = \frac{2}{4} = \frac{1}{2})
   • (x = \frac{-3 - 5}{4} = \frac{-8}{4} = -2)
4. Verification
   • Substitute (x = -2) into the original equation to check.
   • Equation holds true, confirming the solution.

Example 2: Solving (6x^2 - 17x + 12 = 0)

1. Identify a, b, c
   • (a = 6), (b = -17), (c = 12)
2. Substitute into the formula
   • (x = \frac{17 \pm \sqrt{(-17)^2 - 4(6)(12)}}{2(6)})
   • (x = \frac{17 \pm \sqrt{289 - 288}}{12})
   • (x = \frac{17 \pm 1}{12})
3. Calculate solutions
   • (x = \frac{17 + 1}{12} = \frac{18}{12} = \frac{3}{2})
   • (x = \frac{17 - 1}{12} = \frac{16}{12} = \frac{4}{3})

Conclusion

• The quadratic formula effectively solves quadratic equations by finding the values of (x) that satisfy the equation.
• Practice solving quadratic equations by identifying (a), (b), (c) and applying the quadratic formula.