1. Limited time only! Sign up for a free 30min personal 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!

Algebra II Correct Form to Answer Discrimants Problems

  1. Aug 7, 2012 #1
    Hello everyone! I'm back again. I hope you enjoyed your vacation from my questions! *teasing*

    So, I'm working on discriminants right now. Although, I've been "working" on them for a while via the quadratic formula I just didn't know it.

    I came across a point in my notes that I didn't fully understand. And this might not be the correct place to post the thread. If it isn't, I'm sorry mods! I just want to check to see if I fully understand this.

    1. The problem statement, all variables and given/known data
    The equation [itex] kx^2 + 12x + 9k = 0[/itex] has different roots for different values of k. Find the values of k for which the equation has the following:
    a. a real double root
    b. two different real roots
    c. imaginary roots

    My problem is with b & c...

    b.Equation has real roots if D > 0:
    144 - 36k^2 > 0
    -36k^2 > -144
    k^2 < 4
    |k| < 2, or

    -2<k<2


    c. Equation has imgainary roots if D < 0:
    144 - 36k^2 < 0
    -36k^2 < -144
    k^2 > 4
    |k| > 2 or

    k>2 or k<-2





    2. Relevant equations
    D = b^2 - 4ac where it is given that the quadratic equation has real coefficients and of the form ax^2 + bx + c = 0.


    3. The attempt at a solution

    Okay, I don't quite completely understand the last parts of b & c from the absolute value on. My book does caution me that it has to be that way because you need it to ultimately satisfy k^2 < 4 or k^2 > 4. I get that... I was just wondering if I could make it a general rule that if |x| < 2 it will always be with the x in between such as in answer b; and if |x| > 2 it will always be like answer c.

    I didn't 100% understand this when I went over that lesson eons ago, but that was pre-PF, so that is why I'm asking now.

    Thanks in advance for your help! :shy:
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 7, 2012 #2

    Mentallic

    User Avatar
    Homework Helper

    But that is the rule! If [itex]|x|<a[/itex] for some value [itex]a\geq 0[/itex] then it is always true that [itex]-a<x<a[/itex], and if [itex]|x|>a[/itex] for some [itex]a\geq 0[/itex] then [itex]x>a[/itex] or [itex]x<-a[/itex].
    The reason I mention that [itex]a\geq 0[/itex] in both cases is because if [itex]a<0[/itex] then |x| cannot possibly be less than a, it just doesn't make sense, because |x| is defined as the length of x from the origin if you will, and negative length is meaningless.
    Also if we have [itex]|x|>a[/itex] and [itex]a<0[/itex] then that is true for all x.
     
  4. Aug 7, 2012 #3
    Remember with absolute value that it leads to two separate equations; |f(x)| = g(x) leads to f(x) = g(x) or f(x) = -g(x),
    This also applies to inequalities.
    |x| < a is the same as x < a or x > -a, which is condensed into -a < x < a.
    Similarly, |x| > a is the same as x > a or x < -a.
    With inequalities, make sure to note the the direction is flipped for the second statement.
     
    Last edited: Aug 8, 2012
  5. Aug 8, 2012 #4

    Mark44

    Staff: Mentor

    One minor correction: |x| > a is the same as x > a OR x < -a. x cannot simultaneously be less than -a and larger than a, which is the implication when the connector is "and".
     
  6. Aug 8, 2012 #5
    Very true.
    I edited my post to avoid future confusion.
     
  7. Aug 11, 2012 #6
    Woe. I actually got something right. What a pleasant surprise!

    Thanks to Mentallic, Villyer, and Mark44! All of your explanations really helped and make a lot of sense. For some reason, this is something that I understand while reading, but it seeps out of my mind when I go to do the work. I think I'll keep this thread up when solving those problems! :D

    Thanks once again everyone! I have a test due Monday, so you may or may not be hearing a lot from me! Hopefully, I don't hit any snags!
     
  8. Aug 11, 2012 #7

    Mentallic

    User Avatar
    Homework Helper

    Good luck with your exam!
     
  9. Aug 13, 2012 #8

    Mark44

    Staff: Mentor

    A pleasant surprise should not fill you with woe...:tongue:
     
  10. Aug 20, 2012 #9
    @Mentallic: Thank you very much. I spent a long time on it, but I think I did well. We'll see. *fingers crossed*

    @Mark44: Ha ha! Good one!

    Thanks everyone once again. I've moved on to logarithms which I have never encountered before, so I'll keep at it. I just wanted to say thanks again for the help.

    All of my errors that have been pointed out and all of the different ways that everyone phrases things really help me to solidify the math principles and to grow better as a mathematician. I'm actually seeing improvement and catching a lot of my own mistakes! :D I know I owe a lot of that to PF, so I just wanted to say thanks!

    I'm really excited to keep working at it now... (Although, sometimes it still gives me almost too much to ponder!)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook