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!

AQA AS Maths question from '09 paper

Tags:
  1. Jan 9, 2010 #1
    Hi,
    I have my AS Maths exam on Monday and so I was looking at some previous papers and I couldn't do this question:

    The quadratic equation (k+1)x² + 12x + (k-4) = 0 has real roots
    (a) Show that k² - 3k - 40 ≤ 0

    I don't even understand where to start so I was hoping you could help me.

    Thanks!
     
  2. jcsd
  3. Jan 9, 2010 #2
    I think it may be, that as it has real roots, it must mean that b^2-4ac ≤ 0
    so just substitute your values of a b and c into the formula to get k² - 3k - 40

    are you sure its -40 and not -140 ?
    cause then it would work i think!
     
  4. Jan 9, 2010 #3
    oh no sorry, i forgot to multiply it out, silly me, it would work with 40.
    and also 0 ≤ b^2-4ac if it has real roots. sorry i'm a bit foggy on this stuff!
     
  5. Jan 9, 2010 #4
    Can I ask how how you got the values for a and c?
    a=
    b= 12
    c=
    Because I dont understand why the x is there?

    By the way -- thanks for the response, I really really appreciate it!
     
  6. Jan 9, 2010 #5
    well you dont have any integer values for the a and c
    so instead, you just substitue in the constants before the x^2 for a, and the last value for c

    so a would simply be k+1
    and c would be k-4

    do you remember how to find if an equation has real roots ? that is where the b^2-4ac comes in, you could google it if you don't. If the answer is less than zero it has no real roots, equal to zero it has one real root etc. But yours is telling you it has real roots. plural. so therefore your answer to b^2-4ac is, it must be bigger than 0. Your answer will include k once you multiply it out, and you should get the equation you stated before
    k² - 3k - 40 ≤ 0
     
  7. Jan 9, 2010 #6
    Yeah, I understand now. Thanks alot, I really appreciate it! :smile:
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook