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Homework Help: Find α+β+αβ

  1. Apr 11, 2017 #1
    1. The problem statement, all variables and given/known data
    P(x) =ax2+bx+c where a, b and c are in arithmetic progression and are positive. α and β are the roots of the equation and are integers. Find the value of α+β+αβ. (Answer is 7)
    2. Relevant equations
    x = {−b ± √(b2 − 4ac)} /2a
    3. The attempt at a solution

    Since a, b and c are in arithmetic progression, b-a=c-b.
    α+β+αβ = -b/a+c/a = c-b/a = b-a/a
     
  2. jcsd
  3. Apr 11, 2017 #2

    scottdave

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    You know that α and β are roots, so k*(x - α)(x - β) is equal to the polynomial. Expand this out with FOIL k*(x² - (α+β)x + αβ), set this equal to ax² + bx + c. Since they are roots, you do not know what the coefficient of x squared is, yet. So you can represent that k*(x² - (α+β)x + αβ) = ax² + bx + c. Now set the coefficients of the x² terms equal, the x terms, and the constant terms. This should give you a system of equations (along with your knowledge of the arithmetic progression). See if that gets you the answer.
     
  4. Apr 11, 2017 #3
    @scottdave, What is k?
     
    Last edited: Apr 11, 2017
  5. Apr 11, 2017 #4

    scottdave

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    k is a constant which makes the two polynomials equal. You can have several distinct polynomials with the same roots. The difference is a proportionality constant. You will find that k is equal to a.
     
  6. Apr 11, 2017 #5
    @scottdave Do you mean that equations which are related by a multiplicative factor have equal roots?
     
  7. Apr 11, 2017 #6
    Sorry, but i am not getting the answer. I am getting some weird equation.
     
  8. Apr 11, 2017 #7

    mfb

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    Add brackets around the numerator and simplify.

    You can also use the condition that the roots are integers.
     
  9. Apr 11, 2017 #8
    @mfb, i cannot understand how to simplify it further.
     
  10. Apr 11, 2017 #9

    mfb

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    $$\frac{b-a}{a} = \frac b a - \frac a a$$
    And the second term can be simplified.

    Afterwards you can have a look at the condition that α and β are integers.
     
  11. Apr 11, 2017 #10
    @mfb, I thought you meant some other simplification. I could do that simplification by myself but i cannot simplify it further. I am getting -(α+β )-1=α+β+αβ. I cannot recall any related property of integer which would help me simplify it further.
     
  12. Apr 11, 2017 #11

    mfb

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    α+β+αβ = b/a - 1. Keep that result for now.

    α and β are integers. What does that imply for α+β+αβ?
    What does it imply if you look at the formula for α (or β)?
     
  13. Apr 11, 2017 #12
    I found out that if α or β are integers, then [{−b ± √(b2 − 4ac)} /2a] must also be integers. ('s' in 'integers' is due to the ± sign).

    For them to be integers {−b ± √(b2 − 4ac)} and 2a must also be integers and {−b ± √(b2 − 4ac)} must be a multiple of 2a.

    For {−b ± √(b2 − 4ac)} to be integer and an multiple of 2a, -b and √(b2 − 4ac) should be integers and should leave the same remainder on dividing by 2a.
    For -b to be an integer b should be an integer (positive, given in the question) and for √(b2 − 4ac) to be an integer (b2 − 4ac) has to be a positive integer as well as a perfect square.

    The most important conclusion is (b2 − 4ac) is a perfect square.
     
  14. Apr 11, 2017 #13

    mfb

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    That directly follows from the definition, yes.
    Why? You can show that, but it is not trivial.
    Sure.
    Right so far.
     
  15. Apr 11, 2017 #14
    (i)If α and β are integers then α+β+αβ is also an integer.
    (ii) b2>=4ac
    =>(b/a)(b/c)>=4
    =>-(α+β)(b/c)>=4
    =>-(α+β){(b*a)/(c*a)}>=4
    =>-(α+β)*[{-(α+β)}/αβ]>=4
    =>(α+β)2/αβ>=4
    =>(α+β)2/>=4αβ
    =>α22+2αβ-4αβ>=0
    =>(α-β)2>=0
    =>α>=β
     
  16. Apr 11, 2017 #15

    mfb

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    A squared real number is always >=0, there is no need to derive this over 9 lines.
    That does not follow from the previous line.
    Correct. And you know that α+β+αβ = b/a - 1. That tells you if b/a has to be an integer.

    If b/a is an integer, you can rewrite your arithmetic progression and express c in terms of a and b. That should lead to some interesting results in the square root - which needs a perfect square in it.
     
  17. Apr 12, 2017 #16
    Did you mean this
    b-a=c-b
    =>b-a+b=c
    =>2b-a=c
    =>2ak-a=c
    =>a(2k-1)=c
    or this
    b-a=ak-b
    =>2b-a-ak=0
    =>2b-a(1+k)=0
     
  18. Apr 12, 2017 #17
    If a, b, and c are the coefficients of a quadratic equation, what is the sum of the roots and what is the product of the roots?
     
  19. Apr 12, 2017 #18
  20. Apr 12, 2017 #19
    I'm not sure you used this, so $$\alpha + \beta +\alpha\beta=\frac{c-b}{a}$$Certainly, this isn't generally equal to 7, even for an arithmatic progression.
     
  21. Apr 12, 2017 #20

    mfb

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    We also know ##\alpha## and ##\beta## are integers.
    The problem is solvable, but it needs a few steps to get there.
    Both will work, but the first definition of k is much more intuitive.
     
  22. Apr 12, 2017 #21
    I am unable to understand how to proceed. Should i solve for k and then use it's value for determining a,b and c and thus α+β+αβ?
     
  23. Apr 12, 2017 #22
    Ok. I think i got it.
    Since α, β are integers, α+β+αβ is also an integer.
    Since α+β+αβ is an integer, (b/a-1) is also an integer.
    Since (b/a-1) is an integer, b/a is also an integer.
    Since b/a is an integer, b is a multiple of a.
    Therefore b=ak
    Since a, b and c are in A.P. in the same order, b-a=c-b
    =>2b-a=c
    =>2ak-a=c
    =>a(2k-1)=c
    Therefore, P(x)=ax2+bx+c=ax2+akx+a(2k-1)
    =>ax2+akx+a(2k-1)=0
    Dividing both sides by a, we get,
    x2+kx+(2k-1)=0
    Since this equation is the same as the original one, it also has integer roots and since it has integer roots, the discriminant is a perfect square.
    Therefore, k2-4.1.(2k-1)=q2 (say)
    =>k2-8k+4=q2
    =>(k-4)2-12=q2
    =>(k-4)2-q2=12
    =>(k-4+q)(k-4-q)=12
    =>k-4+q=6 or k-4+q=4 ---(1)
    and
    k-4-q=2 or k-4-q=3 ---(2)
    Adding equations(1) and (2) we get,
    2k-8=8 or 2k-8=7
    =>k=8 or k=15/2
    For k=8, b=8a, c=15a and (c-b)/a=(15a-8a)/a=7a/a =7.
    Therefore α+β+αβ=7.

    Why did we get k=15/2?(The value of k for which an invalid value for α+β+αβ is obtained)
     
    Last edited: Apr 12, 2017
  24. Apr 12, 2017 #23

    mfb

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    Looks great!
    Just some details:
    4.2/2.1 = 2 is an integer, but 4.2 and 2.1 are not. You don't need it to write b=ak with an integer k.
    It leads to 12 (with a non-integer q), but it violates the earlier condition that k has to be an integer.
     
  25. Apr 12, 2017 #24
    4.2 and 2.1 are 8 and 2 respectively which are integers. Aren't they?
     
  26. Apr 12, 2017 #25

    mfb

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    That is a decimal dot, not a multiplication sign ("*").
     
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