Roots of this equation are,`(x^3+1/x^3)+(x^2+1/x^2)-6(x+1/x)-7=0`

Options

A. `(3 + sqrt(5))/(2)`
B. `(-3 - sqrt(5))/(2)`
C. `(3 - sqrt(5))/(2)`
D. `(-3 + sqrt(5))/(2)`

Answer

1,2,3,4

Explanation

Let ` x + (1)/(x) = t`
Then given equation is
`(t^(3) - 3t) + (t^(2) - 2) - 6t - 7 = 1`
or ` t^(3) + t^(2) - 9t-9 = 0 `
`therefore (t + 1) (t^(2) - 9) = 0`
`therefore t = 1, 3, - 3`
If ` x + (1)/(x) = - 1`, then `x^(2) + x + 1 = 0` . This equation has no real
roots.
If ` x + (1)/(x) = - 3` , then ` x^(2) + 3x + 1 = 0 `. Roots are ` (-3pm sqrt(5))/(2)`
` If x + (1)/(x) = 3 `, then ` x^(2) - 3x + 1 = 0 ` . Roots are ` (3pm sqrt(5))/(2)`