Algebra
posted by Pamela .
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 realnumber solution, two different realnumber solutions, or two different imaginarynumber solutions.
(b) Use the quadratic formula to find the exact solutions of the equation. Show work. Simplify the final results as much as possible.

(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 
Would this be an appropriate answer for b?
x= 4+3i/2, 43i/2 
How did you get 4 +/ 3i/2?

I figured 4*4 = 16. This is not right is it?

You divide by 8 at the end. 16/8 = 2
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