Polynomial question help!! (1 Viewer)

[poptarts12345]

how do I do this question?
ANS : a =2, b=1/3, c=5/2

 

[SplashJuice]

Let each of the brackets equal to zero.

a-2=0

1-3b=0

5-2c=0

 

[Drdusk]

how do I do this question?View attachment 28706
ANS : a =2, b=1/3, c=5/2

Let each of the brackets equal to zero.

a-2=0

1-3b=0

5-2c=0

Is this a typo or am I hallucinating or something from sleep deprivation?

Because the equation given is a quadratic. How can a quadratic have 3 zeros??

 

[SplashJuice]

Is this a typo or am I hallucinating or something from sleep deprivation?

Because the equation given is a quadratic. How can a quadratic have 3 zeros??

You're not hallucinating, and tbh I've got no clue because it can't

 

[asharnadeem]

I think its called the vanishing condition whereby let's say if a quadratic has three roots it is then a zero polynomial.

 

[idkkdi]

Is this a typo or am I hallucinating or something from sleep deprivation?

Because the equation given is a quadratic. How can a quadratic have 3 zeros??

You're not hallucinating, and tbh I've got no clue because it can't

I think its called the vanishing condition whereby let's say if a quadratic has three roots it is then a zero polynomial.

Can we define a zero polynomial as another polynomial? Zero polynomials have undefined degrees, does this mean that its degree is undefinable or simply definable for an undefinable amount of values?

i.e. 0 = 0x^2 + 0x + 0
0 = 0x^3 + 0x^2 + 0x+ 0

We cannot define the other way around, that a quadratic is a zero polynomial, but can we say that a zero polynomial is a quadratic?

 

[fan96]

The question introduced as simply a polynomial (which it is, if we assume are constants).

It never claimed this polynomial was a quadratic and in fact it is not, because quadratic equations must be expressible in the form
where .

So there are no problems with terminology here, as long as we remember that this polynomial isnt a quadratic to begin with.

 

[Drdusk]

The question introduced as simply a polynomial (which it is, if we assume are constants).

It never claimed this polynomial was a quadratic and in fact it is not, because quadratic equations must be expressible in the form
where .

So there are no problems with terminology here, as long as we remember that this polynomial isnt a quadratic to begin with.

Well the answers have a,b and c as constants.

So clearly is it not a quadratic polynomial??

Kind of by working backwards in a sense. Since a,b and c are real it must be a quadratic polynomial as opposed to it is a quadratic polynomial if a,b and c are real.

 

[fan96]

Since a,b and c are real it must be a quadratic polynomial as opposed to it is a quadratic polynomial if a,b and c are real.

That isn't quite what I said - I said if were constants then the expression is a polynomial. What sort of polynomial it is, we can't tell without further information.
For the polynomial to be quadratic we require , which cannot be the case since the polynomial is the zero polynomial.

 

[Drdusk]

That isn't quite what I said - I said if were constants then the expression is a polynomial.
For the polynomial to be quadratic we require , which cannot be the case since the polynomial is the zero polynomial.

Ah yes mb.

 

[CM_Tutor]

how do I do this question?View attachment 28706
ANS : a =2, b=1/3, c=5/2

This is a really bad question, partly for the reasons indicated.

(1) The answer given is wrong. Taking



produces the polynomial . I can argue this has infinitely many roots as for all . I cannot argue that there are precisely three zeroes.

(2) There is an arguable way to get precisely three zeroes:

Case 1: Take , in which case we have one root:



Case 2: Take , in which case we can have two roots:



So, if , there are two zeroes for

However, there is no set of values for as constants for which has precisely three roots. I suppose one could argue that the answer given does have three roots (and infinitely many more), making it the "best" answer in some sense, but fundamentally it is an ambiguous and flawed question... that's my opinion, anyway!