
#1
Jun1810, 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 ax^{2} + 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 = ax^{2} + 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? 



#2
Jun1810, 02:45 PM

Mentor
P: 21,074





#3
Jun1810, 02:49 PM

P: 6

hm..
2x^{2}+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 x^{2} is a perfect square, we can conclude that the entire quadratic will be one.. ? 



#4
Jun1810, 03:03 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,904

Perfect Square (Quadratic function) 



#5
Jun1810, 03:07 PM

HW Helper
P: 1,344

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 noninteger coefficients are allowed, it is a perfect square. 



#6
Jun1810, 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? 



#7
Jun1910, 07:54 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,904

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.




#8
Jun1910, 04:42 PM

P: 6

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


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