Individual Fractional Part of Roots!?
x 3 − x 2 − 5 x + 1 = 0
If the roots of the above equation are α , β , and γ , find the value of { α } + { β } + { γ } , where { ⋅ } denotes the fractional part function .
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3 solutions
How did you figure out that α + β + γ = 1?
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By Vieta's formula , the sum of the roots of a polynomial of degree n is the opposite of the coefficient on the x n − 1 term.
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No it is not. The sum of the roots of a polynomial of degree n is i = 1 ∑ n r i = − a n a n − 1 , where r 1 , r 2 , … , r n are its complex roots.
from Vieta's. ax^3+bx^2+cx+d=0; a(x-x1)(x-x2)(x-x3)=0; ax^3-a(x1+x2+x3)x^2+a(x1x2+x2x3+x3x1)-ax1x2x3=0; then,we can know x1+x2+x3=-b/a
I get zero.
Solve[x^3 - x^2 - 5 x + 1 == 0, x] // N
{-1.90321 + 1.11022 10^-16i}, {0.193937 - 2.22045 10^-16i}, {2.70928 - 1.4803*10^-16i}}
X=FractionalPart[Re[-1.9032119259115532
+1.1102230246251565
*^-16 i]]
-0.903212
Y=FractionalPart[Re[0.19393656647463042
-2.220446049250313
*^-16i]]
0.193937
Z=FractionalPart[Re[2.7092753594369228
- 1.4802973661668753
*^-16i]]
0.709275
X+Y+Z=0
Evidently, the theorem states:
(A+B+C) - (X+Y+Z) where,
A=Re[-1.9032119259115532
+1.1102230246251565
*^-16i]
-1.90321
B=Re[0.19393656647463042
-2.220446049250313
*^-16i]
0.193937
C=Re[2.7092753594369228
- 1.4802973661668753
*^-16i]
2.70928
So
(A+B+T)-(X+Y+Z) = 1.
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The fractional part of a real number is always in the range [ 0 , 1 ) . Therefore, the value of X you found is incorrect; in fact, it should be 1 + X ≈ 0 . 0 9 6 7 9 instead, which should give the correct value of X + Y + Z .
I recommend you learn Latex.
This problem became Level 1??
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Yeah, I notice that with the Problems of the Week, the problem levels tends to decrease dramatically due to the number of people viewing and solving the questions. They typically go back up when the next batch of problems comes out.
What's a vietta?
What kind of deduction is this?
Because there are different definitions for "the fractional part," the problem is faulty.
Before the problem was edited
Since α + β + γ = 1 by Vieta's Theorem, { α } + { β } + { γ } must be an integer. Given the five choices, the only reasonable answer is 1.
Thanx sir, its a great and concise answer, not too many steps to go. xD. Sir can u plz explain the Vieta's theorem.
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Read this wiki .
This is not texting. Spell out your words.
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Be nice, Linda. Mainak accomplished what he set out to do: ask me a question. I understood him. Communication accomplished. Pretty amazing, considering that English is not my native language, and most likely not Mainak's either. You try writing in your second or third language.
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@Arjen Vreugdenhil – No, it was not an issue for Mainak. It was laziness for that user just not bothering to spell out the words. Clear?
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@Linda Slovik – Laziness or efficiency?
Is your post pettiness or accuracy?
This is not a grammar site. Don't be a grammar police. (Even my english is not that strong) :)... So if we can explain others and express ourselves properly .. then its okay!
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@Md Zuhair – The remark was not about grammar. Can you tell the difference!?
I see you managed to spell out your other words in English. Don't be lazy. Just continue spelling out all the rest of your words.
This solution is not totally correct, because you have not considered negative solutions and the fact that {a} is always nonnegative, even if a<0. Consider this example:
1,2+1,5-1,7=1; 0,2+0,5+0,7=1,4
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No, because with the definition { x } = x − ⌊ x ⌋ , { − 1 . 7 } = ( − 1 . 7 ) − 2 = 0 . 3 . Thus, { α } + { β } + { γ } = ( α + β + γ ) − ( ⌊ α ⌋ + ⌊ β ⌋ + ⌊ γ ⌋ ) = 1 − integer = integer .
(Actually, there are conflicting definitions: see here . Perhaps the problem should be clarified.)
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Yes, you're right. Sorry, I've mistaken with the fractional part of negative numbers. Thanks for the explanation!
The problem has been edited now so that all the answer choices are all integers, so this solution isn't valid anymore.
Substituting 1 for each variable was the closest to = 0. I guessed correctly! Hooray No clue how to arrive at the correct answer. Even after reading the correct method 🤣
Since { x } = x − ⌊ x ⌋ , we have
{ α } + { β } + { γ } = ( α + β + γ ) − ( ⌊ α ⌋ + ⌊ β ⌋ + ⌊ γ ⌋ ) .
We know from Vieta's that α + β + γ = 1 , so we focus on ⌊ α ⌋ + ⌊ β ⌋ + ⌊ γ ⌋ . Let f ( x ) = x 3 − x 2 − 5 x + 1 . We want to find the consecutive integer intervals that each root of f lies in. Thus, we calculate f ( x ) at various integer values of x , which gives us
f ( − 2 ) f ( − 1 ) f ( 0 ) f ( 1 ) f ( 2 ) f ( 3 ) = − 1 = 4 = 1 = − 4 = − 5 = 4 .
From this, we can deduce that there is a root in each of the the intervals ( − 2 , − 1 ) , ( 0 , 1 ) , and ( 2 , 3 ) . This covers all three possible real roots of the polynomial, so the floor of each is just the lower bound on each interval. This gives us ⌊ α ⌋ + ⌊ β ⌋ + ⌊ γ ⌋ = − 2 + 0 + 2 = 0 , and { α } + { β } + { γ } = 1 − 0 = 1 .