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Quadratic Equations

  1. Jan 3, 2005 #1
    I just need to know the basics of 2 ways to excecute a quadratic equation....: Quadratic formula and factoring. Thankyou!!!
     
  2. jcsd
  3. Jan 3, 2005 #2
  4. Jan 3, 2005 #3
    Can you factorise??? (or is this what you mean by factoring???)

    E.g. Could I give you [tex]x^2+5x+6[/tex] and you would end up with [tex](x+2)(x+3)[/tex]???

    The Bob (2004 ©)
     
  5. Jan 3, 2005 #4

    HallsofIvy

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    Yep, us colonials say "factoring" rather than "factorize"!

    (Of course, when I saw "2 ways to execute a quadratic equation" my first thought was "firing squad and hanging"!)
     
  6. Jan 3, 2005 #5
    Sorry. My schools have always said it was factorising.

    LoL. I know. There is an easy way and a way that allows you to do the easy way. :tongue2:

    The Bob (2004 ©)
     
  7. Jan 3, 2005 #6

    qdv

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    u should try factoring first, it's easier if it work, if not use the quadratic equation
     
  8. Jan 3, 2005 #7
    Can anyone explain why when you factorise and use the quadratic equation the answers have different signs?
    Lets say for x^2 + 5x + 4
    If i do it by factoring the answer would be (x+4)(x+1)
    However, if done by the quadratic equation the answer would be (x-4)(x-1) which doesn't work when u multiply them out.
    The signs are reversed.
    Thanks alot! I need to explain this to my sis...Should I tell her to reverse her answers if she does it by the quadratic equation?

    Yawie
     
    Last edited by a moderator: Dec 9, 2008
  9. Jan 3, 2005 #8

    HallsofIvy

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    Do you even KNOW what you mean by "the answer"??
    The correct "answer" to any question does NOT depend on which method you use to answer it!

    What question are you trying to answer?

    IF you are trying to answer the question, "What are the linear factors of
    x2+ 5x+ 4?" then you could note that 4 can be factored as 2*2 or as 4*1. 2+ 2= 4 while 4+1= 5 so you conclude that x2+ 5x+ 4= (x+4)(x+1).
    To use the "quadratic formula", you would convert to the EQUATION x2+ 5x+ 4= 0 to which the quadratic formula gives roots [itex]\frac{-5+/-\sqrt{25-4*4*1}{2*1}= \frac{-5+/- 3}{2}[/itex]= -1 and -4. Of course, you would then recall that roots a and b mean that (x-a)(x-b) are the factors of the polynomial. Since your roots are -1 and -4, your factors are x-(-1)= x+1 and x-(-4)= x+4. x2= (x+1)(x+4) just as you got by factoring.

    IF the question is "What are the roots of x2+ 5x+ 4= 0?" (or, equivalently, "What are the zeroes of x2+ 5x+ 4?", then the quadratic formula gives the roots directly: x= -1 and x= -4 make that polynomial equal to 0.

    By "factoring", you determine that x2+ 5x+ 4= (x+4)(x+1)= 0. Of course, the only way a product of numbers can be equal to 0 is if one or the other is 0: either x+ 4= 0 or x+ 1= 0, from that you conclude that either x= -4 or x= -1, just as with the quadratic formula.

    So: exactly WHAT problem are you trying to solve?
     
    Last edited by a moderator: Dec 9, 2008
  10. Jan 3, 2005 #9
    I didn't really know what the questions were...I was just trying to help my sis...
    but you answered the question....when you said x+4 = 0 is when you factor and when you use the quadratic equation you get x=-4....Now I know how to explain it to her....thanks!
    I didn't realise the questions were differently phrased...
    THanks!
     
  11. Jan 3, 2005 #10

    mathwonk

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    what about completing the square? if you have say x^2 + 5x + 4 = 0, you could write it as x^2 + 5x = -4, then complete the square by adding the square of 5/2 to get

    x^2 + 5x + 25/4 = 25/4 -4 = 9/4. then take square roots of both sides to get

    (x+5/2)^2 = 3/2. so x+5/2 = 3/2 or -3/2. so x = -5/2 + 3/2 = -2/2 = -1, or x = -5/2 - 3/2 = -8/2 = -4.
     
  12. Jan 3, 2005 #11

    DB

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    Parabola
    General form:

    [tex]f(x)=ax^2+bx+c[/tex]

    Quadratic Formula for Zeros:

    [tex]x_1x_2=\frac{-b\pm\sqrt{(b^2)-(4ac)}}{2a}[/tex]

    For Standard form

    [tex]f(x)=a(x-h)^2+k[/tex]

    Formula:
    [tex]x_1x_2=h\pm\sqrt{\frac{-k}{a}}[/tex]

    Just plug in the parameters.
     
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