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Tricky math problem if you are bored.?
Hello,
NOTE: if my question is illegible, use a browser that displays Unicode characters like Firefox or Edge...
DISCLAIMER:
I am not trying to get my homework done. This quiz is provided as it is, and I do have the answer. So it is a challenge for anyone. Best answer will be awarded in approximatively two days. Have fun gals and guys!
= = = = = = = =
Let π and π be two real values.
Let π be a polynomial such that:
Β Β Β { Β Β π(π) = 0
Β Β Β { Β Β π(π) = 0
A. What is the value of π'(π)+π'(π) ?
B. What is the minimal degree πππ of π?
C. If π has a degree of πππ+1, what can you conclude?
Regards,
Dragon.Jade :-)
Mmm... After consideration. Question A is faulty and is hereby cancelled. It will not count toward Best Answer.
Questions B and C stand though. And they seem so trivial that no one wants to answer them correctly...?
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
That was the trick in the question.
There's no doubt the Zero polynomial defined by:
Β Β Β βπ₯ββ, Β Zero(π₯) = 0
fits the given definition of π.
Now what is the degree of this polynomial?
It is indeed a convention that some think it should be undefined, 0, -1 or -β.
If πππ were undefined or πππ=-β, then πππ+1 would have no meaning. Which means question C would be meaningless.
So those conventions need to be discarded in order to answer question C.
If πππ=-1, then πππ+1=0 and such π would be a nonzero constant polynomial then π(π)β 0 and π(π)β 0 which is contrary to the problem.
The conclusion would then be there is no π with degree πππ+1=0.
If πππ=0, then πππ+1=1 and such π would be a linear polynomial which can only have an unique zero.
The conclusion would then be that π=π.
βΊ Thanks for participating. The BA is given to Barry G for being the only one seeing that the Zero polynomial did fit the definition of π.
4 Answers
- Barry GLv 75 years agoFavorite Answer
B. The minimum degree is zero :
P(x) = x^0 - 1
which is zero for all real values of x.
C. If P has degree 1 then
P(x) = Ax+B with A <>0.
P(a) = Aa+B = 0 --> B = -Aa
P(b) = Ab+B = 0 --> Ab-Aa = A(b-a) = 0
so b=a.
- Jeff AaronLv 75 years ago
A. Insufficient information given. There are infinity possible answers.
B. Assuming a and b are different, then P must have at least degree 2. But if a and b can be the same, then P can have degree 1.
C. There are min+1 roots (but not necessarily distinct roots; some or all may be equal), but some of those may be non-real roots.
- husoskiLv 75 years ago
Seems underspecified to me.
p(x) = x^2 - x and q(x) = x^3 - 2x^2 + x are zero at a=0 and b=1, and nowhere else.
p'(x) = 2x - 1 --> p'(0) = -1, p'(1) = 1, p'(0) + p'(1) = -1 + 1 = 0
q'(x) = 3x^2 - 2x + 1 --> q'(0) = 1, p'(1) = 2, p'(0) + p'(1) = -1 + 1 = 3
So, question A has no definite answer. The others two have pretty obvious answers.
- SqdancefanLv 75 years ago
B. Assuming a and b are distinct, P must be of degree 2 or greater. The minimum degree is 2.
C. A degree 3 polynomial with 2 real roots also has a third real root. Conclusion: there must be one more real root, though it isn't necessarily distinct from both a and b.
