Solve the Tricky Equation: Finding x in a Complex Equation

[aaaa202]

Homework Statement

find x in the equation:

a = x + 1/x

Homework Equations

The Attempt at a Solution

I sat down and thought it was easy to do, but was terribly schocked at the fact, that I didn't know how to solve this equation. What is the general approach?

 

[SammyS]

Homework Statement

find x in the equation:

a = x + 1/x

Homework Equations

The Attempt at a Solution

I sat down and thought it was easy to do, but was terribly schocked at the fact, that I didn't know how to solve this equation. What is the general approach?

Multiplying both sides by x results in a quadratic equation.

 

[HallsofIvy]

If that is 1= x+ (1/x) then multiplying both sides by x gives the quadratic equation x= x^2+ 1 which is equivalent to x^2- x+ 1= 0. You will find that it has no real roots. If it is, rather, 1= (x+ 1)/x (in which case you should have used parentheses) you can again multiply both sides by x to get x= x+ 1 which is not true for any value of x.

 

[SammyS]

If that is 1= x+ (1/x) then multiplying both sides by x gives the quadratic equation x= x^2+ 1 which is equivalent to x^2- x+ 1= 0. You will find that it has no real roots.

If it is, rather, 1= (x+ 1)/x (in which case you should have used parentheses) you can again multiply both sides by x to get x= x+ 1 which is not true for any value of x.

If a ≠ 1, then \displaystyle a=\frac{x+1}{x} does have one real root.

 

[dimension10]

Homework Statement

find x in the equation:

a = x + 1/x

Homework Equations

The Attempt at a Solution

I sat down and thought it was easy to do, but was terribly schocked at the fact, that I didn't know how to solve this equation. What is the general approach?

[Post cleared by myself]

 

[dimension10]

IMO: I think that since aaaa202 has not come back to post a response, we should probably not be giving a full-blown solution .

Ya, I think you are right, I cleared my response.

 

[SammyS]

I think this is right:

a = x + 1/x
...

Please, read the rules for Homework Help this forum. Link

Your solution is far too complete.