text: # Algebra
  
  posted by on .
  
  Consider the equation
  
  4x^2 – 16x + 25 = 0.
  
  (a) Show how to compute the discriminant, b^2 – 4ac, and then state whether there is one real-number solution, two different real-number solutions, or two different imaginary-number solutions.
  
  (b) Use the quadratic formula to find the exact solutions of the equation. Show work. Simplify the final results as much as possible.
  
  • Algebra - ,
  
  (a) The equation is in the form ax^2 + bx + c
  
  Use those values for b^2 - 4ac.
  
  (-16)^2 - 4*4*25
  = 256 - 400
  = -144
  
  The determinant is negative, so there will be two imaginary solutions.
  
  (b) The quadratic equation is x = (-b +/- sqrt(determinant))/(2a)
  
  Using the values of a, b, and c from the equation, we get
  
  x = (16 +/- sqrt(-144))/(2*4)
  x = (16 +/- 12i)/8
  x = 2 +/- 3i/2
  
  • Algebra - ,
  
  Would this be an appropriate answer for b?
  
  x= 4+3i/2, 4-3i/2
  
  • Algebra - ,
  
  How did you get 4 +/- 3i/2?
  
  • Algebra - ,
  
  I figured 4*4 = 16. This is not right is it?
  
  • Algebra - ,
  
  You divide by 8 at the end. 16/8 = 2