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Perfect square

  1. Feb 5, 2007 #1
    1. The problem statement, all variables and given/known data
    [tex]\left(\frac{\dot{x}}{a}\right)^2 = K\left[b\frac{a}{x} + c\left(\frac{a}{x}\right)^2 + (1-b-c)\right][/tex]

    Show that, for [itex]b<1[/itex], there is a value of [itex]c[/itex] that makes the right hand side a perfect square of a function of x.

    2. The attempt at a solution
    I guess that a perfect square is a square of a function that can be written as

    [tex](x \pm d)^2 = x^2 \pm 2xd + d^2[/tex]

    but I have not been successful in this problem.
  2. jcsd
  3. Feb 5, 2007 #2


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    That's the right general idea, but for your equation you want something like ((a/x) + d)^2.

    If you have an expression like ax^2 + bx + c and it is a perfect square, what is special about the two roots of the quadratic equation ax^2 + bx + c = 0? What condition can you use to decide what kind of roots a quadratic equation has?
  4. Feb 5, 2007 #3
    If [itex]ax^2+bx+c[/itex] is a perfect square, then [itex]b=2 \sqrt{ac}[/itex] and the roots of [itex]ax^2+bx+c=0[/itex] is a double root [itex]x_{1,2}=-\sqrt{c/a}[/itex]. I'm not sure what you mean with the conditions..
  5. Feb 5, 2007 #4


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    First let u= a/x. Now, what value of c makes cu2+ bu+ 1-b-c a perfect square? Also recall that Ax2+Bx+ C is a perfect square if and only if B2-4AC= 0. However, it seems to me that b would have to be less than or equal to 1/2 for this to be true, not just less than 1.
  6. Feb 5, 2007 #5
    and if [itex]b^2=4ac[/itex] then [itex]ax^2+bx+c[/itex] is a perfect square. So you know the condition already.

    Look again at your quadratic (shame it uses the same letters as those in the general form of the quadratic you have given)

    In your quadratic what is equivalent to 'x' 'a', 'b', 'c' in the general form?
  7. Feb 5, 2007 #6

    [tex]c = \frac{b-1 \pm \sqrt{1-2b}}{2}[/tex]

    but that's two values. I think I also want to have [itex]-1 \leq c \leq 1[/itex] for my problem.
  8. Feb 6, 2007 #7
    Ok, this is correct. But now I need to find an exact solution for [itex]x(t)[/itex] in this case. Help..? =)
  9. Feb 6, 2007 #8


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    Now write that right hand side as a perfect square, take the square root of both sides and solve the resulting separable differential equation.
  10. Feb 6, 2007 #9
    Then I get, using [itex]u=a/x[/itex],

    [tex]u=\sqrt{\frac{1-b-c}{c}} \left[ \frac{(exp[K(1-b-c)^{1/4}t]+1)^2}{(exp[K(1-b-c)^{1/4}t]-1)^2} - 1 \right][/tex]

    and when I then try to solve for x and show that this is a solution I get these infinite calculations so this doesn't seem right.
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