# Algebra

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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