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Perfect Square (Quadratic function)

  • Thread starter Ak94
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  • #1
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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-[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?
 

Answers and Replies

  • #2
33,648
5,318

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-[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?
 
  • #3
6
0
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.. ?
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
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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?
 
  • #5
statdad
Homework Helper
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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.
 
  • #6
6
0
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
HallsofIvy
Science Advisor
Homework Helper
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956
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
6
0
I see.. so D = 0 is the only condition..
to everyone who helped out, thanks :)
 

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