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Math Help - Bi-square equation

  1. #1
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    Question Bi-square equation

    Hey everyone!
    So there is this equation that's giving me a big headache. The task goes like this:

    Solve the following equations:

    (t+1)^4 + (t+5)^4 = 82
    Book result: t1= -4, t2= -2, t3= -3+5i, t4= -3-5i

    I tried to do it in many ways but I can't solve it.


    Here is a similar task, if you think it could help you:

    (k-7)^4 - 13(k-7)^2 + 36 = 0

    z = (k-7)

    z^2 - 13z + 36 = 0 [from this we get a=1, b=-13, c=36 and we are going to use the quadratic formula]
    z = [ -b +/- √( b^2 - 4ac ) ] / 2a
    z = [ 13 +/- √( 169 - 144 ) ] / 2
    z = [ 13 +/- 5 ] / 2
    z1 = 9
    z2 = 4

    (k-7)^2 = 9 |√
    |k-7|= 3
    k1-7 = 3
    k1 = 10
    k2-7 = -3
    k2= 4

    (k-7)^2 = 4 |√
    |k-7|= 2
    k3 = 9
    k4 = 5

    Results: k1=10, k2=4, k3=9, k4=5.
    Last edited by IoNForce; October 26th 2010 at 08:50 AM.
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  2. #2
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    I don't know where this leads but you can try substituting t + 1 = u to see if this helps simplify.
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  3. #3
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    Quote Originally Posted by IoNForce View Post
    Hey everyone!
    So there is this equation that's giving me a big headache. The task goes like this:

    Solve the following equations:

    (t+1)^4 + (t+5)^2 = 82
    Book result: t1= -4, t2= -2, t3= -3+5i, t4= -3-5i

    I tried to do it in many ways but I can't solve it.
    When you expand this you get a quartic with integer coefficients of the form:

    t^4+...-56=0

    So the rational root theorm tell you that the rational roots (if any) of this are amoung \pm7, \pm 14, \pm 28, \pm 56. \pm 2, \pm 4, \pm 8


    Check these to find the rational roots, and what you find is that there is a typo: -2 is not a root.

    CB
    Last edited by CaptainBlack; October 26th 2010 at 09:14 AM.
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  4. #4
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    OK, just to let you know, I managed to solve it!

    Here it is:

    (t+1)^4 + [t + 5]^4 = 82
    t+1 = p
    p^4 + [p + 4]^4 = 82
    p^4 + [p ^2+ 16 + 8p]^2 = 82
    p^4 + p^4 + 64p^2 + 256 + 16^3 + 32 p^2 + 256p - 82 = 0
    2p^4 + 16p^3 + 96 p^2 + 256 p + 174 = 0
    p = -1 => p + 1 = 0 {The left side MUST b equal to the right side of the equation}
    2p^3 (p+1) + 14p^2 (p+1) + 82 p (p+1) + 174 (p+1) = 0
    (p+1)[2p^3 + 14p^2+ 82 p +174] = 0
    p = -3 => p + 3 = 0 {The left side MUST b equal to the right side of the equation}
    (p+1)[2p^2 (p+3) + 8p (p+3) + 58 (p+3) ] = 0
    (p+1)(p+3) [2p^2 + 8p + 58] = 0

    t1 + 1 = p1
    t1 + 1 = -1
    t1 = -2
    t2 + 1 = p2
    t2 + 1 = -3
    t2 = -4

    2p^2 + 8p + 58 = 0 |:2
    p^2 + 4p + 29 = 0 => a = 1, b = 4, c = 29
    p3,4 = [ -b +/- √( b^2 - 4ac ) ] / 2a
    p3,4 = [ -4 +/- √( 16 - 116 ) ] / 2
    p3,4 = 2(-2 +/- 5i) / 2
    p3 = -2 + 5i
    p4 = -2 - 5i
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by IoNForce View Post
    OK, just to let you know, I managed to solve it!

    Here it is:

    (t+1)^4 + [t + 5]^4 = 82
    t+1 = p
    p^4 + [p + 4]^4 = 82
    p^4 + [p ^2+ 16 + 8p]^2 = 82
    p^4 + p^4 + 64p^2 + 256 + 16^3 + 32 p^2 + 256p - 82 = 0
    2p^4 + 16p^3 + 96 p^2 + 256 p + 174 = 0
    p = -1 => p + 1 = 0 {The left side MUST b equal to the right side of the equation}
    2p^3 (p+1) + 14p^2 (p+1) + 82 p (p+1) + 174 (p+1) = 0
    (p+1)[2p^3 + 14p^2+ 82 p +174] = 0
    p = -3 => p + 3 = 0 {The left side MUST b equal to the right side of the equation}
    (p+1)[2p^2 (p+3) + 8p (p+3) + 58 (p+3) ] = 0
    (p+1)(p+3) [2p^2 + 8p + 58] = 0

    t1 + 1 = p1
    t1 + 1 = -1
    t1 = -2
    t2 + 1 = p2
    t2 + 1 = -3
    t2 = -4

    2p^2 + 8p + 58 = 0 |:2
    p^2 + 4p + 29 = 0 => a = 1, b = 4, c = 29
    p3,4 = [ -b +/- √( b^2 - 4ac ) ] / 2a
    p3,4 = [ -4 +/- √( 16 - 116 ) ] / 2
    p3,4 = 2(-2 +/- 5i) / 2
    p3 = -2 + 5i
    p4 = -2 - 5i
    Makes it easier for us if you post the right question from the start, or give a new post pointing out you have changed the question.

    CB
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  6. #6
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    Yeah, sorry I had a typo and I corrected it as soon as I noticed...
    I'll keep it in mind if it happens again.
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