Sum and Product of the Roots (Quadratic Equations)

  1. 1. The problem statement, all variables and given/known data

    For the quad equation x^2 - px + 9 = 0

    1. Write down the sum of roots and product of roots
    2. Find p IF twice the sum of the roots EQUALS the product
    3. Find p IF the roots are unequal

    2. Relevant equations

    Sum = (a+b) = -b/a Product = (ab) c/a

    3. The attempt at a solution

    1. Using the formula Sum = -p Product = 9
    2. -2p = 9 -9/2 = 2p/2 = 4 1/2
    3. Totally lost

    Can someone provide guidence. Cheers
     
    Last edited: Oct 13, 2009
  2. jcsd
  3. Mark44

    Staff: Mentor

    Let r1 and r2 be the roots of the given quadratic.
    1) Sum of roots = r1 + r2 = -p
    Product of roots = r1 * r2 = 9

    You have two equations in two unknowns. Can you solve for r1 and r2 in terms of p?

    2) What's the question in this part? You have
    Part of the sentence is missing.
     
    Last edited: Oct 13, 2009
  4. 1 if the roots are a,b then the equation is x^2 - (a+b)x + ab = 0, so the coefficient of x is -(sum of the roots), and you should have p instead of -p.

    2. I have no idea what is meant here.

    3. Find p when the roots are equal first. Can you use comples numbers? if not there are more values of p where the quadratic doesn't have a solution
     
  5. Hi there

    I have edited the original question - my apologies there.

    Cheers
     
  6. I have amended my original post.

    For the quad equation x^2 - px + 9 = 0

    1. Write down the sum of roots and product of roots
    2. Find p IF twice the sum of the roots EQUALS the product
    3. Find p IF the roots are unequal

    2. Relevant equations

    Sum = (a+b) = -b/a Product = (ab) c/a

    3. The attempt at a solution

    1. Using the formula Sum = -p Product = 9
    2. -2p = 9 -9/2 = 2p/2 = 4 1/2
    3. Totally lost

    Can someone provide guidence. Cheers
     
  7. Mentallic

    Mentallic 3,763
    Homework Helper

    1) No the sum is [itex]-b/a=-(-p/1)=p[/itex] and the product is right.

    2) You're right except for taking the sum as -p rather than p.

    3) If we need p when the roots are unequal, how about we find the value(s) of p when the roots are equal, then take all other values?
     
  8. I think that I have nutted out part 3, of this question

    x^2 - px + 9 = 0

    a = 1 b = -p and c = 9

    Delta = b^2 - 4ac
    = (-p)^2 - 4(1) (9)
    = p - 36

    So if plugged into the following:

    Equal roots Delta = 0
    p - 36 = 0
    p = 36

    For real roots Delta = >(Equal to) 0
    p - 36 >(Equal to) 0

    Unreal Delta < 0
    p - 36 < 0
    p < 36

    For real and different Delta > 0
    p - 36 > 0
    p > 36

    Guidence on this would be great
     
  9. Mentallic

    Mentallic 3,763
    Homework Helper

    Yes you were very close. You had the right approach.
    You just forgot about the squaring p in the [itex]\Delta=(-p)^2-4.1.9[/itex]

    However, there were no other restrictions on the problem. It just said find p when the roots are unequal. It never said anything about the roots being real/imaginary.
    Basically, taking [itex]\Delta<0[/itex] is fine too. It just means for those values of p, the quadratic will be entirely above the x-axis.

    So finally, for roots unequal, p is all reals except [itex]p^2\neq 36[/itex] thus, [itex]p\neq \pm 6[/itex] (Note: do not forget about the plus/minus)
     
  10. Many thanks

    I understand - many thanks for taking the time to respond so thoughtfully.

    Cheers
     
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