Solving Quadratic Equations with Imaginary Solutions

Key Concept

• Imaginary Unit: ( i )
  • Defined as ( \sqrt{-1} = i )

Examples and Procedures

Example 1: Solving ( x^2 + 16 = 0 )

1. Isolate ( x ):
   • Subtract 16 from both sides to get ( x^2 = -16 )
2. Square Root:
   • Take the square root of both sides: ( x = \pm\sqrt{-16} )
   • Break it down: ( \sqrt{-16} = \sqrt{16} \cdot \sqrt{-1} = 4i )
   • Solutions: ( x = \pm 4i )

Example 2: Solving ( 4x^2 + 49 = 0 )

1. Isolate ( x ):
   • Subtract 49 from both sides: ( 4x^2 = -49 )
   • Divide both sides by 4: ( x^2 = \frac{-49}{4} )
2. Square Root:
   • Take the square root: ( x = \pm\sqrt{\frac{-49}{4}} )
   • Break it down: ( \sqrt{\frac{-49}{4}} = \frac{\sqrt{-49}}{\sqrt{4}} = \frac{7i}{2} )
   • Solutions: ( x = \pm \frac{7i}{2} )

Example 3: Solving ( x^2 + 3x + 7 = 0 )

1. Quadratic Formula:
   • Formula: ( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} )
   • Given: ( a = 1, b = 3, c = 7 )
   • Compute ( b^2 - 4ac ): ( 3^2 - 4(1)(7) = 9 - 28 = -19 )
2. Evaluate:
   • Solution: ( x = \frac{-3 \pm \sqrt{-19}}{2} )
   • Break it down: ( \sqrt{-19} = \sqrt{19} \cdot \sqrt{-1} = \sqrt{19}i )
   • Final form: ( x = \frac{-3}{2} \pm \frac{\sqrt{19}i}{2} )

Example 4: Solving ( 3x^2 + 5x + 10 = 0 )

1. Quadratic Formula:
   • Given: ( a = 3, b = 5, c = 10 )
   • Compute ( b^2 - 4ac ): ( 5^2 - 4(3)(10) = 25 - 120 = -95 )
2. Evaluate:
   • Solution: ( x = \frac{-5 \pm \sqrt{-95}}{6} )
   • Break it down: ( \sqrt{-95} = \sqrt{95} \cdot \sqrt{-1} = \sqrt{95}i )
   • Final form: ( x = \frac{-5}{6} \pm \frac{\sqrt{95}i}{6} )

Important Notes

• Imaginary Numbers appear when taking the square root of negative numbers.
• Standard Form for complex numbers: ( a \pm bi )
• Understand how to apply the quadratic formula in cases involving negative discriminants that lead to imaginary solutions.