Qustion aout a quadratic function: "...whose zero values are 11" ?

[Javier77]

Hi there.
I was doing an exercise that read:
"Find the quadratic function for f(x) whose zero values are 11"
So I thought, "ok, they are asking me a function where if you plug in the number 11 you get a resut of zero".
So I thougth: (x-11)x= 0
Which gives: X squared minus 11x.
When you plug in 11 you get 11*11 minus 11*11 and the result is zero.
But they regarded this answer as incorrect. They said the right answer was:
(x-11)(x-11) which means x squared-22x + 20
In this case plugging 11 gives a result of zero.
But I wonder, what's wrong with (x-11)x?
Isnt that a quadratic function?
Thanks in advance.

 

[Deleted member 4993]

Hi there.
I was doing an exercise that read:
"Find the quadratic function for f(x) whose zero values are 11"
So I thought, "ok, they are asking me a function where if you plug in the number 11 you get a resut of zero".
So I thougth: (x-11)x= 0
Which gives: X squared minus 11x.
When you plug in 11 you get 11*11 minus 11*11 and the result is zero.
But they regarded this answer as incorrect. They said the right answer was:
(x-11)(x-11) which means x squared-22x + 20
In this case plugging 11 gives a result of zero.
But I wonder, what's wrong with (x-11)x?
Isnt that a quadratic function?
Thanks in advance.

The word value was in plural number!

So both the values must be 11 → hence f(x) = (x-1)^2

I personally would call this a trick question!

 

[JeffM]

The word value was in plural number!

So both the values must be 11 → hence f(x) = (x-1)^2 You mean (x - 11)^2.

I personally would call this a trick question!

I agree that it is a trick question as worded. Moreover, the question falsely implies that there is a unique quadratic that satisfies the given condition. So it is not even a good trick question. It could, however, be worded so as not to be a trick and actually to teach something.

Find a quadratic such that both its roots are 11.

That would lead fairly obviously to

\(\displaystyle p(x) = a(x - 11)(x - 11) = a(x - 11)^2, \text { where } a \ne 0.\)

This would remind students that the roots do not identify a unique quadratic and that the assertion that every quadratic has two roots does not imply that the two roots are different, contrary to standard English usage.

EDIT: I would prefer to teach that every polynomial with real coefficients and degree n has at most n real roots and, if n is odd, at least one real root. It is not as succinct as saying that a polynomial of degree n has n roots, but the qualifications necessary to make the succinct formulation understandable are not explicit.

 

[mmm4444bot]

But I wonder, what's wrong with (x-11)x?

Isnt that a quadratic function?

Yes, it is, but it has two different roots.

x = 11 and x = 0

As others have already said, they want a quadratic polynomial with roots:

x = 11 and x = 11

Has your class talked about "multiplicity" of roots?

 

[Javier77]

Yes, it is, but it has two different roots.

x = 11 and x = 0

As others have already said, they want a quadratic polynomial with roots:

x = 11 and x = 11

Has your class talked about "multiplicity" of roots?

Thanks everybody for your answers.

 

[Javier77]

Thanks everybody for your answer.

I see it now.

x(11-x) gives 11 and 0 as answers, and that problem only accepted 11, so (x-11)(x-11) is the only funtion which gives only 11 as correct answer.

Thanks.