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 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 -
**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, 4-3i/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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