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What Is b^2 - 4ac?

  1. Jan 13, 2016 #1
    As we know that ##x_{1,2}## of a quadratic function can be found with the below formula:

    ##\frac{-b ± \sqrt{b^2 - 4ac}}{2a}##

    What do you call the ##b^2 - 4ac##?
     
  2. jcsd
  3. Jan 13, 2016 #2

    462chevelle

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    It's the discriminant, if I spelled that right.
    Its value lets you know if you have complex or real solutions.
     
  4. Jan 14, 2016 #3

    Mark44

    Staff: Mentor

    Yes, you spelled it correctly.
    It discriminates between two real solutions, one real and repeated solution, and two complex solutions, depending on whether the discriminant is positive, zero, or negative, respectively.
     
  5. Jan 14, 2016 #4
    I wasn't able to find about discriminant in my Calculus textbook.

    What book I can found about this discriminant?
     
  6. Jan 14, 2016 #5
    I would look in an algebra book
     
  7. Jan 14, 2016 #6
    You should be able to find this in any algebra book. Look under quadratic equations.
     
  8. Jan 14, 2016 #7

    symbolipoint

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    Yes, that is right. Any intermediate or college algebra textbook will discuss the discriminant of a quadratic equation or of a quadratic expression.

    The discriminant occurs when you use Completing the Square to generally solve a quadratic equation; as well as if you use Completing the Square to solve a particular quadratic equation.
     
  9. Jan 15, 2016 #8

    Mark44

    Staff: Mentor

    The discriminant shows up in the Quadratic Formula, which is derived by completing the square. If you solve a quadratic equation by completing the square, you won't see the discriminant.

    For example, solve ##x^2 - 4x - 1 = 0##
    1. By Quadratic Formula
    ##\Rightarrow x = \frac{4 \pm \sqrt{4^2 - (4\cdot 1 \cdot (-1)}}{2} = \frac{4 \pm \sqrt{20}}{2} = 2 \pm \sqrt{5}##
    Here the discriminant is ##b^2 - 4ac## = 16 - (-4) = 20

    2. By completing the square
    ##x^2 - 4x - 1 = 0##
    ##\Rightarrow x^2 - 4x + 4 = 1 + 4##
    ##\Rightarrow (x - 2)^2 = 5##
    ##\Rightarrow x - 2 = \pm \sqrt{5}##
    ##\Rightarrow x = 2 \pm \sqrt{5}##

    Notice that the discriminant (20) never explicitly appears in completing the square.
     
  10. Jan 15, 2016 #9

    symbolipoint

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    Mark44 shows the ordinary algebra step behavior, that we usually simplify from one step to the next, and we do not then see the uncomputed expression for the discriminant. If we WANTED to, we could leave that part uncomputed, and finish its computation last.
     
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