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!

Homework Help: Expressing c in terms of b

  1. Nov 1, 2009 #1
    1. The problem statement, all variables and given/known data

    The roots of [tex]x^2 + bx + c = 0[/tex] are [tex]r_1[/tex] and [tex]r_2[/tex],
    where [tex]|r_1 - r_2| = 1[/tex].

    Express c in terms of b.

    2. Relevant equations

    [tex]x^2 + bx + c = 0[/tex]

    [tex]|r_1 - r_2| = 1[/tex]

    3. The attempt at a solution
    Im not even sure where to start with this problem, and i dont see how the roots are relevant. I just dont know what the question is asking.
  2. jcsd
  3. Nov 1, 2009 #2


    User Avatar
    Gold Member

    Can you write r1 and r2 in terms of b and c? i.e. actually find the roots.

    The condition then says that the difference between these roots must be 1. Using the roots that you found above, set their difference equal to one. When does this equation hold true? How must b and c be related to satisfy this equation?
  4. Nov 1, 2009 #3


    Staff: Mentor

    The roots are very relevant. If r1 and r2 are roots of your quadratic equation, it can be written as (x - r1)(x - r2) = 0.

    Multiply the product above and equate the coefficient of x with b, and the constant term above with c.

    The other condition, |r1 - r2| = 1, says that either r1 = r2 + 1 or that r2 = r1 + 1.
  5. Nov 1, 2009 #4
    (x - r1)(x - r2) =

    [tex]x^2 -r_1x - r_2x -r_1r_2[/tex]

    so how do i combine the middle terms to get a single coefficient for b.
  6. Nov 1, 2009 #5


    User Avatar
    Gold Member

    Take out a factor of x:

    (x - r_1)(x - r_2) =

    x^2 -x(r_1+r_2) +r_1r_2[/tex]

    Also, be careful with your signs. The constant term should be +r1r2
  7. Nov 1, 2009 #6
    okay, im seeing it a little more now, but i still dont understand how to relate b and c. I guess you can say that c is the product and b is the sum (of the values r1 and r2), but thats not relating the two to each other. And i still dont understand why its important that the absolute value is equal to 1.
  8. Nov 1, 2009 #7


    User Avatar
    Gold Member

    It is important because that is the condition that the question requires. Think about when you normally find the solutions to a quadratic equation; you will (normally) get two unique solutions? What the question is saying is that when you solve this particular one, the difference between these two roots is 1.

    If you use the quadratic formula to solve the given equation, what two roots do you get?
  9. Nov 1, 2009 #8
    So i have to use the quadratic equation on [tex]x^2 -x(r_1+r_2) +r_1r_2[/tex]

    x = -(r_1+r_2) +- sqrt{(r_1+r_2)^2 - 4(r_1r_2)}/2

    that hardly seems solvable, i know im doing something wrong.
  10. Nov 1, 2009 #9


    User Avatar
    Gold Member

    Apply it to the original equation to find x in terms of b and c.
  11. Nov 1, 2009 #10
    sorry, i dont understand what you mean by apply it to the original equation.
  12. Nov 1, 2009 #11


    User Avatar
    Gold Member

    x^2 + bx + c = 0

    Solving this gives:



    The condition says that the difference between these is 1.
  13. Nov 1, 2009 #12
    i understand that part, but i dont understand how i solve it. Can i use a systems of equations?
  14. Nov 1, 2009 #13


    User Avatar
    Gold Member

    Well the [tex]|r_1 - r_2| = 1 [/tex] part says that the difference between these two solutions must be 1.

    How about you actually find the difference between the two solutions above and then determine what the relationship between b and c must be for this difference to equal 1.
  15. Nov 1, 2009 #14

    [tex]\frac{-b+\sqrt{b^2-4c}}{2} - \frac{-b-\sqrt{b^2-4c}}{2}=1[/tex]

    multiply each side by 2

    [tex]-b+\sqrt{b^2-4c} - {-b-\sqrt{b^2-4c}=2[/tex]


    [tex]-b+\sqrt{b^2-4c} + b +\sqrt{b^2-4c}=2[/tex]

    -b and b cancel


    square both sides




    divide everything by 2


    add 4c to both sides and subtract 2 from both sides


    divide by 4 on both sides


    is that all i can do? did i do it right?

    or am i supposed to solve for b?


    add 4c to both sides


    square root both sides

    Last edited: Nov 1, 2009
  16. Nov 1, 2009 #15
    Sorry for the bump, but i am worried i did this wrong, can someone tell me if i did this correctly?
  17. Nov 1, 2009 #16


    User Avatar
    Gold Member

    Have another look at that step where you squared both sides.

    In general, it is NOT true that [tex](a+b)^2=a^2+b^2[/tex]
  18. Nov 2, 2009 #17
    in that case i dont know what to do....
  19. Nov 2, 2009 #18


    User Avatar
    Gold Member

    You had the right idea, just a small error in the algebra.

    \sqrt{b^2-4c}+\sqrt{b^2-4c}= 2\sqrt{b^2-4c}=2

    How about divide both sides by 2?


    Can you see what to do from there? For that equation to hold, how must b and c be related?
  20. Nov 2, 2009 #19
    omg, thank you so much!!!!!!
  21. Nov 2, 2009 #20


    User Avatar
    Gold Member

    So you see it now? :smile:

    Glad to be of assistance :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook