Algebra problem (x+(1/x) problem)

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In summary, using the complex solution of ##\frac{\sqrt{3}-i}{2}##, we can find that ##a^{2016} + (\frac{1}{a})^{2016}## is equal to 2.
  • #1
terryds
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Homework Statement



If ##a+\frac{1}{a}=\sqrt{3}##, then ##a^{2016} + (\frac{1}{a})^{2016} ## equals

A. 0
B. 1
C. √2
D. √3
E. 2

Homework Equations

The Attempt at a Solution


[/B]
I find no real solution for a.
What I can do is just find a^2 + (1/a)^2 which equals 1, a^4 + (1/a)^4 = -1, a^8 + (1/a)^8 = -1, and so on...
Since 2016 is not a square number, I don't know how to determine the value

Is there any theory about problem like this?
Please help
 
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  • #2
terryds said:

Homework Statement



If ##a+\frac{1}{a}=\sqrt{3}##, then ##a^{2016} + (\frac{1}{a})^{2016} ## equals

A. 0
B. 1
C. √2
D. √3
E. 2

Homework Equations

The Attempt at a Solution


[/B]
I find no real solution for a.
What I can do is just find a^2 + (1/a)^2 which equals 1, a^4 + (1/a)^4 = -1, a^8 + (1/a)^8 = -1, and so on...
Since 2016 is not a square number, I don't know how to determine the value

Is there any theory about problem like this?
Please help
Start from ##a²+1/{a²}=1##.
Multiply by ##a²## to get rid of the denominator, you get ##a^4 +1 =a²##.
Now notice that 2016 is a multiple of 6. So multiply ##a^4 +1 =a²## again by ##a²## to get ##a^6## in the equation.
 
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  • #3
terryds said:

The Attempt at a Solution


[/B]
I find no real solution for a.
Find the complex solution. Write it in trigonometric of exponential form, which makes it easy to rise to power 2016. (2016=25*32 *7)
 
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  • #4
Samy_A said:
Start from ##a²+1/{a²}=1##.
Multiply by ##a²## to get rid of the denominator, you get ##a^4 +1 =a²##.
Now notice that 2016 is a multiple of 6. So multiply ##a^4 +1 =a²## again by ##a²## to get ##a^6## in the equation.

a^6 + a^2 = a^4
a^6 = a^4 - a^2 = (a^2 - 1) - a^2 = -1
a^12 = 1
a^24 = 1
...
a^2016 = 1

So, 1/a^2016 = 1
a^2016 + (1/a)^2016 = 1 + 1 = 2
Thank you
 
  • #5
ehild said:
Find the complex solution. Write it in trigonometric of exponential form, which makes it easy to rise to power 2016. (2016=25*32 *7)

I'm interested to try this approach..

The solution for a is ##\frac{\sqrt{3}\pm i}{2}##
Should I use the plus or the minus as the root?
If it's plus, ##e^{i\frac{1}{6}\pi}##
If it's minus, ##e^{i\frac{11}{6}\pi}##

And, then what should I do??
 
  • #6
terryds said:
I'm interested to try this approach..

The solution for a is ##\frac{\sqrt{3}\pm i}{2}##
Should I use the plus or the minus as the root?
If it's plus, ##e^{i\frac{1}{6}\pi}##
If it's minus, ##e^{i\frac{11}{6}\pi}##

And, then what should I do??
What are their 12th power?
 
  • #7
ehild said:
What are their 12th power?

The 'plus' root^12 = ##e^{i2\pi}= 1##
The 'minus' root^12 = ##e^{i22\pi} = 1##

Okay, since 2016 can be divided by 12, a^2016 = 1
And 1/a^2016 = 1
So, a^2016 + 1/a^2016 = 1+1 = 2

Thanks a lot!
 
  • #8
terryds said:
The 'plus' root^12 = ##e^{i2\pi}= 1##
The 'minus' root^12 = ##e^{i22\pi} = 1##

Okay, since 2016 can be divided by 12, a^2016 = 1
And 1/a^2016 = 1
So, a^2016 + 1/a^2016 = 1+1 = 2

Thanks a lot!
The 'minus' root^12 = ##e^{-i2\pi} = 1## :smile:
 
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1. What is the "x+(1/x) problem" in Algebra?

The "x+(1/x) problem" is a type of algebraic expression where a variable, x, is added to the reciprocal of x, which is 1/x. It can also be written as x+1/x.

2. How do I solve an "x+(1/x) problem"?

To solve an "x+(1/x) problem", you need to first simplify the expression by finding a common denominator. Then, combine the terms using the rules of addition and subtraction. Finally, solve for the value of x by isolating it on one side of the equation.

3. Can an "x+(1/x) problem" have multiple solutions?

Yes, an "x+(1/x) problem" can have multiple solutions. This is because there can be more than one value of x that satisfies the equation. It is important to check your solution by plugging it back into the original equation to ensure it is valid.

4. Are there any restrictions on the variable x in an "x+(1/x) problem"?

Yes, there are restrictions on the variable x in an "x+(1/x) problem". Since division by zero is undefined, x cannot be equal to 0. Additionally, if the expression is in the form of x+1/x, x cannot be equal to -1 as this would result in division by zero.

5. What real-world applications use the "x+(1/x) problem"?

The "x+(1/x) problem" can be used to solve various real-world problems, such as calculating the total resistance in an electrical circuit or determining the minimum distance between two points on a coordinate plane. It can also be applied in physics, finance, and engineering.

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