Solving Quadratic Equations Using the Quadratic Formula

Overview

• Quadratic equations can be solved using the quadratic formula.
• The formula: ( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} )
  • ( a ), ( b ), and ( c ) are coefficients from the equation in the form ( ax^2 + bx + c = 0 ).

Example 1

• Equation: ( 2x^2 + 3x - 2 = 0 )
  • ( a = 2 ), ( b = 3 ), ( c = -2 )
• Steps:
  1. Identify ( a ), ( b ), and ( c ) from the equation.
  2. Plug values into the quadratic formula:
     • ( x = \frac{-3 \pm \sqrt{3^2 - 4 \times 2 \times (-2)}}{2 \times 2} )
     • Simplify under the square root: ( 9 + 16 = 25 )
     • ( \sqrt{25} = 5 )
  3. Calculate the two possible values for ( x ):
     • ( x = \frac{-3 + 5}{4} = \frac{2}{4} = \frac{1}{2} )
     • ( x = \frac{-3 - 5}{4} = \frac{-8}{4} = -2 )
• Solutions: ( x = \frac{1}{2} ) or ( x = -2 )
• Verification:
  • Substitute ( x = -2 ) back into the original equation to confirm it equals 0.

Example 2

• Equation: Coefficients are ( a = 6 ), ( b = -17 ), ( c = 12 )
• Steps:
  1. Calculate using the quadratic formula:
     • ( x = \frac{-(-17) \pm \sqrt{(-17)^2 - 4 \times 6 \times 12}}{2 \times 6} )
     • Simplify under the square root: ( 289 - 288 = 1 )
     • ( \sqrt{1} = 1 )
  2. Calculate the two possible values for ( x ):
     • ( x = \frac{17 + 1}{12} = \frac{18}{12} = \frac{3}{2} )
     • ( x = \frac{17 - 1}{12} = \frac{16}{12} = \frac{4}{3} )
• Solutions: ( x = \frac{3}{2} ) or ( x = \frac{4}{3} )

Conclusion

• Quadratic formula provides a method for solving any quadratic equation.
• Always verify solutions by substituting back into the original equation.

Practice

• Repeat similar steps for different quadratic equations to master the use of the quadratic formula.