Register to reply

Perfect Square (Quadratic function)

by Ak94
Tags: perfect square, quadratic
Share this thread:
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?
Phys.Org News Partner Science news on Phys.org
Scientists develop 'electronic nose' for rapid detection of C. diff infection
Why plants in the office make us more productive
Tesla Motors dealing as states play factory poker
Mark44
#2
Jun18-10, 02:45 PM
Mentor
P: 21,304
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: 39,556
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: 1,372
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: 39,556
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 :)


Register to reply

Related Discussions
Converting square root to perfect square Precalculus Mathematics Homework 4
Perfect square Precalculus Mathematics Homework 3
Perfect square Calculus & Beyond Homework 11
Perfect square Calculus & Beyond Homework 8
Perfect square? Linear & Abstract Algebra 2