Solve x + 1/x = 2{resolved}

[Dillinger4]

Logically x = 1 but i can't figure this out mathematically.
x + 1/x = 2 --> x^2 + 1 = 2x --> x^2 = 2x - 1 --> x = +/- sqr(2x -1)
What am i doing wrong.

[manavo11]

(x+1)/x=2 => (x+1)=2*x => 2*x-x=1 => x=1

The first step is to multiply both "sides" (I don't know how to say it) by x...

[da_silvy]

that maths is wrong

x + 1/x = 2

(x^2 + 1)/x = 2

so x^2 + 1 = 2x

x^2 - 2x + 1 = 0

(x-1)^2 = 0

.'. x = 1

[beros87]

this equation has 2 answers
the first one is x= 1
the second one is x= -2

[Something Else]

Originally posted by beros87
this equation has 2 answers

You are correct!
The right one: the first one is x= 1
      1 + 1/1 = 1 + 1 = 2
and the Wrong One: the second one is x= -2
&nbsp; &nbsp; &nbsp; -2 + (1/(-2)) = -2 - 1/2 = -2.5 <> 2



da_silvy's the way to go.


-Lou

[manavo11]

Originally posted by da_silvy
that maths is wrong

x + 1/x = 2

(x^2 + 1)/x = 2

so x^2 + 1 = 2x

x^2 - 2x + 1 = 0

(x-1)^2 = 0

.'. x = 1

Can you explain to me how you get from step one (x + 1/x = 2) to step two ((x^2 + 1)/x = 2)?

[sql_lall]

Firstly, that step in full is:
x + 1/x = (x^2)/x + 1/x = (x^2 + 1) / x
just putting over common denominator.

Secondly, you error is:
"x = +/- sqr(2x -1)"

That is actually right, just you have x on both sides.
Try putting in x = 1, you get 1 = sqrt(1). If you say 'x = -1', then you no longer have sqrt(1), you have sqrt(-3)!

So, in short, with x^2 = 2x - 1, you put it all on one side and make a quadratic...

OR: you can do this:

2x - x^2 = 1. x(2 - x) = 1.
No, you can also see 1 is the only answer.

[Ecniv]

Is that :
(x+1)/x = 2

or

x+ (1/x) = 2

??

I assumed the top one and couldn't see where you were getting the x^2 from...




Vince

[manavo11]

Originally posted by Ecniv
Is that :
(x+1)/x = 2

or

x+ (1/x) = 2

??

I assumed the top one and couldn't see where you were getting the x^2 from...




Vince

Same here... But now that I think of it, I was wrong assuming the parenthesis were there...

[Dillinger4]

Yeah sorry. I did this x^2 + 1 = 2x --> x^2 = 2x - 1 when i should have did this x^2 + 1 = 2x --> x^2 + 1 - 2x = 0

[TheRobster]

It's a quadratic though isn't it? I'm sure you can re-arrange it to get it in the form of: ax^2 + bx + c

x + 1/x = 2
(* all terms by x)

x^2 + 1 = 2x
(-2x to get all terms on LHS)

x^2 - 2x + 1 = 0
(factorise to give)

(x-1) (x-1) = 0

So the solution is x = 1.

this equation has 2 answers
the first one is x= 1
the second one is x= -2

I don't see how x= -2 can be right.

-2 + 1/-2 = -2.5

So that doesn't equal 2!!!

[beros87]

sorry
the equation is x^2-2x+1=0
it has the form of a+b+c=0
then the first answer is x=1
the second answer is x=c/a=1
i made a mistake and i took the second answer as b/a which is wrong
that's what happens when you have little time

[Something Else]

Originally posted by beros87
it has the form of a+b+c=0

It looks more like aX²+bX+c=0.

[TheRobster]

it has the form of a+b+c=0

It's a quadratic.

Dunno where you're getting: a+b+c = 0 from.

[beros87]

i know it has the form of ax^2 +bx+c=0 but i assumed that you already know that
but like in this case if a+b+c=0 then, it is known globally that x=1 or x=c/a
if a-b+c=0 then x=-1 or x=-c/a
these are global exceptions.

[Ecniv]

Firstly apologies for adding to a resolved post, but I think you may (just may) find the code below useful...

Public Function QuadraticEq(ByVal dblA As Double, dblB As Double, dblC As Double) As String
'---- ax^2 + bx + c = 0
 
'---- x= (-b +/- sqr(b^2-4ac))/2a
 
 
 
    Dim dblAns1 As Double, dblAns2 As Double, dblSqr As Double
    
    If (dblB ^ 2) - (4 * (dblA * dblC)) >= 0 Then dblSqr = Sqr((dblB ^ 2) - (4 * dblA * dblC))
    
    dblAns1 = ((0 - dblB) + dblSqr) / (2 * dblA)
    dblAns2 = ((0 - dblB) - dblSqr) / (2 * dblA)
    QuadraticEq = CStr(dblAns1) & "  :  " & CStr(dblAns2)
End Function

Vince

[alkatran]

Originally posted by beros87
i know it has the form of ax^2 +bx+c=0 but i assumed that you already know that
but like in this case if a+b+c=0 then, it is known globally that x=1 or x=c/a
if a-b+c=0 then x=-1 or x=-c/a
these are global exceptions.

I wouldn't call that an exception, just a special case. One that is easier to calculate. If it was an exception it would break the rule.