The Quadratic Formula

Key Questions

  • What is the quadratic formula used for?

    The quadratic formula is used to get the roots of a quadratic equation, if the roots exists at all.

    We usually just perform factorization to get the roots of a quadratic equation. However, this is not always possible (especially when the roots are irrational)

    The quadratic formula is

    #x = (-b +- root 2 (b^2 - 4ac))/(2a)#

    Example 1:

    #y = x^2 -3x - 4#
    #0 = x^2 -3x - 4#

    #=> 0 = (x - 4)(x + 1)#
    #=> x = 4, x = -1#

    Using the quadratic formula, let's try to solve the same equation

    #x = (-(-3) +- root 2 ((-3)^2 - 4*1*(-4)))/(2 * 1)#
    #=> x = (3 +- root 2 (9 + 16))/2#
    #=> x = (3 +- root 2 (25))/2#
    #=> x = (3 + 5)/2, x = (3 - 5)/2#
    #=> x = 4, x = -1#

    Example 2:

    #y = 2x^2 -3x - 5#
    #0 = 2x^2 - 3x - 5#

    Performing factorization is a little hard for this equation, so let's jump straight to using the quadratic formula

    #x = (-(-3) +- root 2 ((-3)^2 - 4 * 2 * (-5)))/(2 * 2)#

    #x = (3 +- root 2 (9 + 40))/4#

    #x = (3 +- root 2 49)/4#

    #x = (3 + 7)/4, x = (3 - 7)/4#

    #x = 5/2, x = -1#

    seph · · Oct 19 2014
  • What is an example of using the quadratic formula?

    Suppose that you have a function represented by #f(x) = Ax^2 + Bx + C#.

    We can use the quadratic formula to find the zeroes of this function, by setting #f(x) = Ax^2 + Bx + C = 0#.

    Technically we can also find complex roots for it, but typically one will be asked to work only with real roots. The quadratic formula is represented as:

    #(-B +- sqrt(B^2-4AC))/(2A) = x#

    ... where x represents the x-coordinate of the zero.

    If #B^2 -4AC <0#, we will be dealing with complex roots, and if #B^2 - 4AC >=0#, we will have real roots.

    As an example, consider the function #x^2 -13x + 12#. Here,

    #A = 1, B = -13, C = 12.#

    Then for the quadratic formula we would have:

    # x = (13 +- sqrt ((-13)^2 - 4(1)(12)))/(2(1))# =

    #(13 +- sqrt (169 - 48))/2 = (13+-11)/2#

    Thus, our roots are #x=1# and #x=12#.

    For an example with complex roots, we have the function #f(x) =x^2 +1#. Here #A = 1, B = 0, C = 1.#

    Then by the quadratic equation,

    #x = (0 +- sqrt (0^2 - 4(1)(1)))/(2(1)) = +-sqrt(-4)/2 = +-i#

    ... where #i# is the imaginary unit, defined by its property of #i^2 = -1#.

    In the graph for this function on the real coordinate plane, we will see no zeroes, but the function will have these two imaginary roots.

    Psykolord1989 . · 1 · Nov 2 2014
  • What is the discriminant and how do you find it?

    The discriminant is part of the quadratic formula.

    Quadratic Formula

    #x=(-b+-sqrt(b^2-4ac))/(2a)#

    Discriminant

    #b^2-4ac#

    The discriminant tells you the number and types of solutions to a quadratic equation.

    #b^2-4ac = 0#, one real solution

    #b^2-4ac > 0#, two real solutions

    #b^2-4ac < 0#, two imaginary solutions

    AJ Speller · · Oct 22 2014
  • What is the quadratic formula?

    Answer:

    #x=(-b+-sqrt(b^2-4ac))/(2a)#

    Explanation:

    Negative b plus minus the square root of b squared minus 4*a*c over 2*a. To plug something into the quadratic formula the equation needs to be in standard form (#ax^2 + bx^2 +c #).

    hope this helps!

    Lauren · 1 · Mar 26 2018

Questions

Linear and Quadratic Functions