Perfect Square (Quadratic function)

[Ak94]

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?

 

[Mark44]

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?

 

[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.. ?

 

[HallsofIvy]

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?

 

[statdad]

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.

 

[Ak94]

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?

 

[HallsofIvy]

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.

 

[Ak94]

I see.. so D = 0 is the only condition..
to everyone who helped out, thanks :)