Quadratic Formula

Add yours
Algebra - Quadratic - Derive Quadratic Formula

Tip: This isn't the place to ask a question because the teacher can't reply.

1 of 3 videos by AJ Speller

Key Questions

  • Suppose you have #ax^2 + bx + c = 0# where a, b and c is any number. Plug these in the quadratic formula:

    #x= (-b+- sqrt(b^2-4ac))/(2a)# Plug the values and solve for #x#.

  • Since the quadratic formula is derived from the completing the square method, which always works. Note that factoring always works as well, but it is sometimes just very difficult to do it.


    I hope that this was helpful.

  • Your company is going to make frames as part of a new product they are launching.

    The frame will be cut out of a piece of steel, and to keep the weight down, the final area should be # 28 cm^2#

    The inside of the frame has to be 11 cm by 6 cm

    What should the width x of the metal be?

    Area of steel before cutting:

    Area = #(11 + 2x) × (6 + 2x) cm^2#
    Area = #66 + 22x + 12x + 4x^2#
    Area = #4x^2 + 34x + 66#
    Area of steel after cutting out the 11 × 6 middle:

    Area = # + 34x + 66 - 66#
    Area = #4x^2 + 34x#

    Let us solve this one graphically!

    Here is the graph of 4x2 + 34x :
    mATHFUN
    The desired area of 28 is shown as a horizontal line.

    The area equals 28 cm2 when:

    x is about -9.3 or 0.8

    The negative value of x make no sense, so the answer is:

    x = 0.8 cm (approx.)

    You can try it out yourself and find the roots of the quadratic equation using the quadratic formula

    # x = [ -b ± sqrt(b^2 - 4ac) ] / (2a#

    There are many more examples at http://www.mathsisfun.com/algebra/quadratic-equation-real-world.html

  • The quadratic formula is #x= (-b+-sqrt(b^2-4ac))-:2a#.

    1. Identify the values of #a#, #b#, and #c# in #ax^2+bx+c=0#.
    2. Substitute each value into the formula #x= (-b+-sqrt(b^2-4ac))-:2a#.
    3. Simplify for the values of #-b#, #b^2#, #-4a#, and #2a#.
    4. Simplify for the value of #-4ac#.
    5. Add product of #-4ac# to #b^2#.
    6. Simplify for the value of #b^2-4ac#.
    7. Simplify for the value of #-b^2-4ac#. The value remains as it stands.
    8. Simplify for the value of #x= (-b+-sqrt(b^2-4ac))-:2a#. The value for #x# should be determined using value .

    Example:

    #2x^2-16x+32=0#

    1. #a=2, b=-16,c=32#
    2. #x=(-(-16)+-sqrt((-16)^2-4(2)(32)))-:2(2)#
    3. #x=(16+-sqrt(256-8(32)))-:4#
    4. #x=(16+-sqrt(256-256))-:4#
    5. #x=(16+-sqrt(0))-:4#
    6. #x=(16+-0)-:4#
    7. #x=16-:4#
    8. #x=4#
  • This is a little bit tricky but also incredibly elegant!
    You start from your general quadratic:
    #ax^2+bx+c=0#
    Take #c# to the right side:
    #ax^2+bx=-c#
    The idea is now to transform the left side in something like #(a+b)^2#;
    Multiply by #a#;
    #a^2x^2+abx=-ac#
    Multiply by #4#;
    #4a^2x^2+4abx=-4ac#
    Add and subtract #b^2# to the left side:
    #4a^2x^2+4abx+b^2-b^2=-4ac#
    Take the #-b^2# to the right:
    #4a^2x^2+4abx+b^2=b^2-4ac#
    The left side can be written as:
    #(2ax+b)^2=b^2-4ac#
    And:
    #2ax+b=+-sqrt(b^2-4ac)#
    And finally:
    #x=(-b+-sqrt(b^2-4ac))/(2a)#
    Poetry in algebra!

Questions