Question

The complex solution to a quadratic equation is x equals start fraction negative 24 plus or minus square root of negative 288 end square root over four end fraction full stop Write this solution in standard form, x = a ± bi, where a and b are real numbers. Justify your answer by showing your work.

Answer 1

The complex solution to the quadratic equation written in standard form, x = a ± bi, is:

• 
  `x = -6\ \pm\ 3i√(2)`
• a = -6
• 
  `b = 3√(2)`

How to write the complex solution in standard form?

From the question given, the complex solution is written as:

• Complex solution of quadratic equation:
  `x = (-24\ \pm\ √(-288))/(4)`
• Solution of quadratic equation in standard form =?

`x = (-24\ \pm\ √(-288))/(4)\\\\Recall:\\\\√(-288) = √(-1\ *288)\\\\√(-288) = √(-1)\ *\ √(288)\\\\But,\\\\i = √(-1)\\\\Thus,\ we\ have\\\\√(-288) = i\ *\ √(288)\\\\Now,\\\\√(288) = 12√(2)\\\\Therefore,\\\\√(-288) = 12i√(2)\\\\This\ means\ that:\\\\x = (-24\ \pm\ √(-288))/(4) = (-24\ \pm\ 12i√(2))/(4)\\\\x = -6\ \pm\ 3i√(2)`

Therefore, we have a and b to be:

• a = -6
• 
  `b = 3√(2)`