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Math Help - solving and equation

  1. #1
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    solving and equation

    Hello hope you're having a good day
    I need help in solving these problems

    Find a set of real numbers (R) such as
    sqrt(ab)+sqrt(cd) >= sqrt(a+b) + sqrt (c+d)

    the Second is solving this

    (a^3+b)(a+b^3)= (a+b)^4
    a and b are integers
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  2. #2
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    Can you show some of your attempts at these problems?
    Last edited by mr fantastic; March 6th 2011 at 02:49 PM.
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  3. #3
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    ok
    the farest I could reach in the first one is that a,b,c and d are real positive numbers
    in the second
    I found that if a=0 than b can be equal to anything same goes if b=0
    I know itsn't much that's what I found
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  4. #4
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    Have you tried expanding both the left hand side and the right hand side in the second problem? I suspect that it might be a good idea to do so.
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  5. #5
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    yep I did still can't be solved
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  6. #6
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    Quote Originally Posted by tmn50 View Post
    (a^3+b)(a+b^3)= (a+b)^4
    a and b are integers
    I presume that the solutions a = 0, b = anything, and a = anything, b = 0 are both already known to you, so we will go ahead and assume that neither a nor b are 0. As scounged suggested, expand both sides. After some amount of simplifying I get
    \displaystyle (b^2 - 4)a^2 - a(6b) + (1-4b^2) = 0

    This is a quadratic in a. Now take a look at the discriminant:
    \displaystyle \Delta = 36b^2 - 4(b^2 - 4)(1 - 4b^2)

    \Delta = 32b^2(b^2 - 1)

    For an integer solution for a we must require that \Delta = 32b^2(b^2 - 1) is a perfect square. Are there any integer b that do this? If you find solutions to b, use the quadratic equation in a, \displaystyle (b^2 - 4)a^2 - a(6b) + (1-4b^2) = 0
    to find if this gives an integer value for a.

    -Dan
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  7. #7
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    Quote Originally Posted by topsquark View Post
    I presume that the solutions a = 0, b = anything, and a = anything, b = 0 are both already known to you, so we will go ahead and assume that neither a nor b are 0. As scounged suggested, expand both sides. After some amount of simplifying I get
    \displaystyle (b^2 - 4)a^2 - a(6b) + (1-4b^2) = 0

    This is a quadratic in a. Now take a look at the discriminant:
    \displaystyle \Delta = 36b^2 - 4(b^2 - 4)(1 - 4b^2)

    \Delta = 32b^2(b^2 - 1)

    For an integer solution for a we must require that \Delta = 32b^2(b^2 - 1) is a perfect square. Are there any integer b that do this? If you find solutions to b, use the quadratic equation in a, \displaystyle (b^2 - 4)a^2 - a(6b) + (1-4b^2) = 0
    to find if this gives an integer value for a.

    -Dan
    thanks man but how can you find b so delta is a perfect square
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by tmn50 View Post
    thanks man but how can you find b so delta is a perfect square
    \displaystyle \Delta = 32 b^2(b^2 - 1)

    Factor all perfect squares from the RHS:
    \displaystyle \Delta = (16b^2) \cdot 2(b^2 - 1)

    That means that we need a value of an integer b such that \displaystyle 2(b^2 - 1) is a perfect square. If you don't see it, start picking values for b and see what you get. Note that, since b is squared all you need to do is check all the non-negative values for b. (And, of course once you do this, you still need to verify that these solutions for b give an integer value for a.)

    Hint: There are only two values of b that make \Delta a perfect square. Can you prove this?

    -Dan
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  9. #9
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    nope you can't
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  10. #10
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by tmn50 View Post
    nope you can't
    Can't prove there are only two b values or that there is no integer a for either of these two b values?

    First remove any perfect square factors from \Delta = 32b^2(b^2 - 1) = (16b^2) \cdot 2(b^2 - 1). So we require that 2(b^2 - 1) must be a perfect square.

    There are only two possibilities for b: b is even or it is odd.

    If b is even then let b = 2n. This implies that 2((2n)^2 - 1) = 2(4n^2 - 1). We need 4n^2 - 1 to be an even number divisible by 2 to make this a square number. But 4n^2 - 1 is odd. Thus b cannot be an even number.

    So let b be an odd number: b = 2n + 1. Then 2((2n + 1)^2 - 1) = 2(4n^2 + 4n) = 4 \cdot 2n(n + 1). Now, 4 is a perfect square, so again I am going to factor that out of the problem. This leaves the statement that 2n(n + 1) must be a square. By inspection n = 0 makes this a perfect square. So n = 0 implies b = 1. Further, the only other way that 2n(n + 1) can be a perfect square is to let 2n = n + 1. This has the solution n = 1, which says that b = 3.

    So we have only two possible solutions for an integer b: b = 1 or b = 3. Both of these give integer solutions for a.

    -Dan
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