# Perfect Square (Quadratic function)

by Ak94
Tags: perfect square, quadratic
 P: 6 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-$$\alpha$$)(x-$$\beta$$) where $$\alpha$$ and $$\beta$$ are the values of x for which y = 0, y is a perfect square when Discriminant of quadratic = 0 (this ensures that $$\alpha$$ = $$\beta$$) 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?
Mentor
P: 21,407
 Quote by Ak94 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-$$\alpha$$)(x-$$\beta$$) where $$\alpha$$ and $$\beta$$ are the values of x for which y = 0, y is a perfect square when Discriminant of quadratic = 0 (this ensures that $$\alpha$$ = $$\beta$$) 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?
How about if y = 2x2 + 4x + 2 = 2(x + 1)(x + 1)? Is that a perfect square?
 P: 6 hm.. 2x2+4x+2=[$$\sqrt{2}$$(x+1)]2 and since $$\sqrt{2}$$ 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.. ?
Math
Emeritus
Thanks
PF Gold
P: 39,682
Perfect Square (Quadratic function)

 Quote by Ak94 hm.. 2x2+4x+2=[$$\sqrt{2}$$(x+1)]2 and since $$\sqrt{2}$$ 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.. ?
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?
 HW Helper P: 1,378 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 square.
 P: 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?
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,682 Not necessarily. I would call "$a(x- b)^2$" a "perfect square" for any numbers a and b- a does not have to be positive.
 P: 6 I see.. so D = 0 is the only condition.. to everyone who helped out, thanks :)

 Related Discussions Precalculus Mathematics Homework 4 Precalculus Mathematics Homework 3 Calculus & Beyond Homework 11 Calculus & Beyond Homework 8 Linear & Abstract Algebra 2