Perfect Square (Quadratic function)

AI Thread Summary
A quadratic function of the form ax² + bx + c is a perfect square when its discriminant is zero, ensuring that the roots are equal. Additionally, the coefficient 'a' must be a perfect square for the entire expression to be classified as a perfect square. The discussion highlights that the definition of "perfect square" can vary, particularly regarding whether coefficients must be integers. It concludes that while D = 0 is a necessary condition, the sign of 'a' does not need to be positive for the expression to be a perfect square. Overall, the key conditions for a quadratic to be a perfect square are D = 0 and 'a' being a perfect square.
Ak94
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Homework Statement


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...

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?
 
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Ak94 said:

Homework Statement


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...

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?
 
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.. ?
 
Ak94 said:
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?
 
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

<br /> 2x^2 + 4x + 2 = \left( \sqrt 2 (x+1) \right)^2<br />

is not a perfect square is correct. If non-integer coefficients are allowed, it is a perfect
square.
 
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?
 
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.
 
I see.. so D = 0 is the only condition..
to everyone who helped out, thanks :)
 
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