Posted by Pamela on Tuesday, July 7, 2009 at 7:02pm.
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.

Algebra  Marth, Tuesday, July 7, 2009 at 7:07pm
(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  Pamela, Tuesday, July 7, 2009 at 7:22pm
Would this be an appropriate answer for b?
x= 4+3i/2, 43i/2

Algebra  Marth, Tuesday, July 7, 2009 at 7:29pm
How did you get 4 +/ 3i/2?

Algebra  Pamela, Tuesday, July 7, 2009 at 7:32pm
I figured 4*4 = 16. This is not right is it?

Algebra  Marth, Wednesday, July 8, 2009 at 3:24pm
You divide by 8 at the end. 16/8 = 2
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