Asked by Kaur
                If x + (1/x)=6
Find
x - (1/x)
            
        Find
x - (1/x)
Answers
                    Answered by
            mathhelper
            
    x + 1/x = 6
multiply by x -1/x
(x+1/x)(x - 1/x) = 6(x-1/x)
x^2 - 1/x^2 = 6x - 6/x
times x^2
x^4 - 1 = 6x^3 - 6x
x^4 - 6x^3 + 6x - 1 = 0
by observation x = 1 and x = -1
by long division after that :
(x+1)(x-1)(x^2 - 6x + 1) = 0
x= -1, +1, 3+2√2, 3-2√2
Since we squared all roots must be verified,
by observation x = 1 and x = -1 do not work in the original x + 1/x = 6
but x = 3 ± 2√2 do work
so x - 1/x , if x = 3+2√2
= 3 + 2√2 - 1/(3 + 2√2)
= 3 + 2√2 - 1/(3 + 2√2)*(3 - 2√2)/(3 - 2√2)
= 3 + 2√2 - (3 - 2√2)/(9 - 8)
= 4√2
or , if x = 3 - 2√2
x - 1/x
= 3 - 2√2 - 1/(3 - 2√2)
= ...
= 3 - 2√2 - (3 + 2√2)/1
= -4√2
    
multiply by x -1/x
(x+1/x)(x - 1/x) = 6(x-1/x)
x^2 - 1/x^2 = 6x - 6/x
times x^2
x^4 - 1 = 6x^3 - 6x
x^4 - 6x^3 + 6x - 1 = 0
by observation x = 1 and x = -1
by long division after that :
(x+1)(x-1)(x^2 - 6x + 1) = 0
x= -1, +1, 3+2√2, 3-2√2
Since we squared all roots must be verified,
by observation x = 1 and x = -1 do not work in the original x + 1/x = 6
but x = 3 ± 2√2 do work
so x - 1/x , if x = 3+2√2
= 3 + 2√2 - 1/(3 + 2√2)
= 3 + 2√2 - 1/(3 + 2√2)*(3 - 2√2)/(3 - 2√2)
= 3 + 2√2 - (3 - 2√2)/(9 - 8)
= 4√2
or , if x = 3 - 2√2
x - 1/x
= 3 - 2√2 - 1/(3 - 2√2)
= ...
= 3 - 2√2 - (3 + 2√2)/1
= -4√2
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