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Quadratic equation

  1. Jun 15, 2014 #1
    Question 1

    (a) Solve the quadratic equation

    x^2 + 4x -5 = 0

    (b) Factorise its left hand side.

    (c) Find interval(s) of x where the left hand side is positive


    Q1
    (a)
    (x+-1) (x+5)
    x= 1 x=-5
    is that solving the equation?

    (b)
    didnt i already factorise the left hand side of the equation when i solved for in (a)?
    (x+-1)?

    (c)
    not sure on this one
     
  2. jcsd
  3. Jun 15, 2014 #2
    Hi there, you have solved the equation in a). Yes you did already factorise the equation, but perhaps the question assumed you might have solved it a different way (by the quadratic equation). For part c you need to use the fact the two negative numbers multiplied together will give you a positive number etc.
     
  4. Jun 15, 2014 #3

    Ray Vickson

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    Your notation x+-1 is meaningless, and should never be used. If x=1 is a root then one of the factors is x-1; if x = -1 is a root, one of the factors is x+1.
     
  5. Jun 15, 2014 #4

    Mentallic

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    You should try graphing the quadratic

    [tex]y=x^2+4x-5[/tex]

    You should know that the quadratic is convex (in the shape of a u) instead of concave (in the shape of an n) because the coefficient of x2 (multiple of x2) is 1 and hence positive (if it were negative the quadratic would be concave).

    You've also found that it crosses the x-axis at x=1 and x=-5 because this is where the quadratic =0 hence y=0 for those x values.

    Putting these two ideas together, it's quite obvious what the domain is (values of x) for where the quadratic is positive.
     
  6. Jun 15, 2014 #5

    NascentOxygen

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    Yes, you have found the values of the unknowns which will make the right side equal to the left side.

    You are trying to determine your answer without sketching the graph? ✻shame✻

    Always sketch the graph, so you can see what you are dealing with.
     
  7. Jun 15, 2014 #6
    For c), besides sketching the graph you can analyse the sign of the terms individually.

    (x-1)(x+5)

    When is (x-1) positive? When is it negative?
    Same to (x+5).

    What must be the signs of (x-1) and (x+5) so that (x-1)(x+5) > 0?
     
  8. Jun 16, 2014 #7
    quadratic

    ok so i have graphed the quadratic and circled in red all the positive values of x, how would i write this?

    [PLAIN]http://i.imgur.com/WwvsteW.jpg?1[/PLAIN]
     
  9. Jun 16, 2014 #8

    SammyS

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    What you're asked for is:

    All of the x values for which the y values are positive .
     
  10. Jun 16, 2014 #9
    so that would be everything before -5 and everything after 1 right?
     
  11. Jun 16, 2014 #10

    Matterwave

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    Yes. So just turn that into a mathematical expression. What's "everything before -5" and "everything after 1" in mathematical terms?
     
  12. Jun 16, 2014 #11
    x < -5 & x > 1
     
  13. Jun 21, 2014 #12
    hmm. In part (a), "solve" does seem confusing in this problem. There are many other ways of solving. Sometimes textbooks want you to graph first. You could also use a rational root test, or something like that. Some books give problems with ##ax^2 + bx + c = N##. The concept of "solving" is to graph this as two functions and find their intersection points. Of course, it seems like a lot more work than necessary.

    There is a concern I have with what you wrote in part (a). The expression ##(x + -1)(x+5)## does not mean ##x = 1## or ##x = -5##. You have to state ##(x + -1)(x+5) = 0## to find those solutions for ##x##. I think you KNOW that you are doing this, you just didn't write it. If you did that on a test, a teacher might take points off.

    This is also a big idea that connects with part (c). In part (a) you are using the zero property: Zero times a number is zero. In part (c) you are using the following concepts about multiplication of real numbers: "a positive times a positive is a positive"; "a negative times a negative is a positive"; etc. Again these positive & negative multiplication rules are for real numbers, not the imaginary numbers.
     
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