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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

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