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Homework Help: Perfect Square (Quadratic function)

  1. Jun 18, 2010 #1
    1. The problem statement, all variables and given/known data
    What is the condition, for a quadratic function of the form
    ax2 + bx + c = y
    to be a perfect square? (x, y are real here)

    There's a question of this type in a book I'm working with, and I'd just like to have some general conditions for any quadratic...

    3. The attempt at a solution

    Since y = ax2 + bx + c = a(x-[tex]\alpha[/tex])(x-[tex]\beta[/tex])
    where [tex]\alpha[/tex] and [tex]\beta[/tex] are the values of x for which y = 0,
    y is a perfect square when Discriminant of quadratic = 0 (this ensures that [tex]\alpha[/tex] = [tex]\beta[/tex]) and when a is a perect square..

    Are these the required conditions for any quadratic function (of the given form) to be a perfect square? Any condition I may have missed?
  2. jcsd
  3. Jun 18, 2010 #2


    Staff: Mentor

    How about if y = 2x2 + 4x + 2 = 2(x + 1)(x + 1)? Is that a perfect square?
  4. Jun 18, 2010 #3
    and since [tex]\sqrt{2}[/tex] isn't an integer, the expression isn't a perfect square...
    so that does imply that when D = 0, and coefficient of x2 is a perfect square, we can conclude that the entire quadratic will be one.. ?
  5. Jun 18, 2010 #4


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    Science Advisor

    So I guess it depends on what your definition of "perfect square" is! And I've never seen one that required constants to be integers. Is that from your textbook?
  6. Jun 18, 2010 #5


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    Homework Helper

    Are you required to use integers as coefficients? I know some introductory algebra books specify that you factor only if you have integer coefficients, but you have not specified this. If so, your comment that

    2x^2 + 4x + 2 = \left( \sqrt 2 (x+1) \right)^2

    is not a perfect square is correct. If non-integer coefficients are allowed, it is a perfect
  7. Jun 18, 2010 #6
    it's kind of ambigously mentioned in the book, so I'll talk to my teacher about it.., thanks for pointing that out :)

    So a perfect square number would have to be the square of an integer, but that is not needed for an expression.. ?

    Also.. I guess D = 0 is still a necessary condition
    Since this perfect square would be greater than or equal to 0, the parabola representing the function would have to open upwards, so it would be greater than 0

    So, the necessary (and sufficient) conditions are D = 0, a > 0? (assuming that any real coefficient is allowed)
    Is that right?
  8. Jun 19, 2010 #7


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    Science Advisor

    Not necessarily. I would call "[itex]a(x- b)^2[/itex]" a "perfect square" for any numbers a and b- a does not have to be positive.
  9. Jun 19, 2010 #8
    I see.. so D = 0 is the only condition..
    to everyone who helped out, thanks :)
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