1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Factoring a quadratic: strange issues

  1. Jul 6, 2015 #1
    1. The problem statement, all variables and given/known data
    Hello!

    One of the easiest rules (when possible to apply) to factor a quadratic is to find both x-s by

    x1 + x2 = b
    x1 * x2 = c

    2. Relevant equations
    Please, take a look at what is written in the book. I can't grasp why x1 = -2 and x2 = 3, and not, as I thought, x1 = -3 and x2 = 2.

    3. The attempt at a solution
    What is wrong in my understanding?

    x1 + x2 = -1
    x1 * x2 = -6
    Therefore, x1 = -3 and x2 = 2; and the equation (x + 3 )(x - 2) =0

    I would be grateful for the help.
    Thank you!
     

    Attached Files:

  2. jcsd
  3. Jul 6, 2015 #2

    epenguin

    User Avatar
    Homework Helper
    Gold Member

    Can see only two lines of your quote so not sure what your problem is, but could it be this
    x1 + x2 = b

    Which should be -b. ?
     
  4. Jul 6, 2015 #3
    Thank you for your reply. There is a screenshot attached, and my equations.
    Do you say that x1 + x2 equal -b and not b? Mathisfun lied to me )))
     
  5. Jul 6, 2015 #4

    epenguin

    User Avatar
    Homework Helper
    Gold Member

    Check that again, but better and faster would be to look see if roots are a and b so that they satisfy

    (x - a)(x - b) = 0

    what is the coefficient of x in the quadratic you get expanding that?
     
  6. Jul 6, 2015 #5

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    This page from Math is Fun gives the correct relationship for the sum and product of the roots of a quadratic:

    http://www.mathsisfun.com/algebra/polynomials-sums-products-roots.html

    You can always check with Wikipedia for quadratic equations.

    BTW, these formulas are called Vieta's formulas, for the famous French mathematician.
     
  7. Jul 7, 2015 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You can multiply the left side to see that [itex](x+ a)(x+ b)= x^2+ (a+ b)x+ ab[/itex]. In your given example, [itex]x^2- x- 6[/itex], so we must have a+ b= -1. And ab= -6. Obviously, 3 and -2 multiply to give -6. But -2+ 3= 1, not -1. We must have a=- 2 and b= 3. Once we have that we have the two factors (x+ (-2)(x+ 3)= (x- 2)(x+3)= 0, we must have either x- 2= 0 or x+ 3= 0. From those, x= 2 and x= -3 are the roots of the equation [itex]x^2- x- 6= 0[/itex].
     
  8. Jul 7, 2015 #7
    There are several ways to find the roots of a quadratic:

    Factoring
    Quadratic formula
    Completing the square
    Graphing

    If a problem requires factoring and you can't figure it out that way, you can always use another method to find the roots and work backwards.

    Suppose you use the quadratic formula to find roots f and g. The factors are then:

    (x - f) and (x - g).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Factoring a quadratic: strange issues
  1. Factoring a quadratic (Replies: 7)

  2. Factoring quadratics (Replies: 9)

  3. Factoring a Quadratic (Replies: 15)

Loading...