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Perfect Square (Quadratic function)

by Ak94
Tags: perfect square, quadratic
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Ak94
#1
Jun18-10, 02:31 PM
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-[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?
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Mark44
#2
Jun18-10, 02:45 PM
Mentor
P: 21,253
Quote Quote by Ak94 View Post
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?
How about if y = 2x2 + 4x + 2 = 2(x + 1)(x + 1)? Is that a perfect square?
Ak94
#3
Jun18-10, 02:49 PM
P: 6
hm..
2x2+4x+2=[[tex]\sqrt{2}[/tex](x+1)]2
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.. ?

HallsofIvy
#4
Jun18-10, 03:03 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,491
Perfect Square (Quadratic function)

Quote Quote by Ak94 View Post
hm..
2x2+4x+2=[[tex]\sqrt{2}[/tex](x+1)]2
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.. ?
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
#5
Jun18-10, 03:07 PM
HW Helper
P: 1,371
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

[tex]
2x^2 + 4x + 2 = \left( \sqrt 2 (x+1) \right)^2
[/tex]

is not a perfect square is correct. If non-integer coefficients are allowed, it is a perfect
square.
Ak94
#6
Jun18-10, 03:27 PM
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?
HallsofIvy
#7
Jun19-10, 07:54 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,491
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.
Ak94
#8
Jun19-10, 04:42 PM
P: 6
I see.. so D = 0 is the only condition..
to everyone who helped out, thanks :)


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