1. Solve the equation `1 (x^2 + 1/x^2) - 8 ( x + 1/x ) + 14 = 0`

Solution:
`1(x^2+1/x^2)-8(x+1/x)+14=0`

Let `x+1/x=m`

`=>(x+1/x)^2=m^2`

`=>x^2+1/x^2+2=m^2`

`=>x^2+1/x^2=m^2 - 2`

Substituting this values in the given equation, we get
`(m^2-2)-8m+14=0`

`=>m^2-8m+12=0`

`m^2-8m+12=0`

`=>m^2-8m+12 = 0`

`=>m^2-2m-6m+12 = 0`

`=>m(m-2)-6(m-2) = 0`

`=>(m-2)(m-6) = 0`

`=>(m-2) = 0" or "(m-6) = 0`

`=>m = 2" or "m = 6`

Now, `x+1/x=2`

`=>x^2+1=2x`

`=>x^2-2x+1=0`

`x^2-2x+1=0`

`=>x^2-2x+1 = 0`

`=>(x)^2 - 2(x)(1) + (1)^2 = 0`

`=>(x-1)^2 = 0`

`=>(x-1) = 0`

`=>x = 1`

Now, `x+1/x=6`

`=>x^2+1=6x`

`=>x^2-6x+1=0`

`x^2-6x+1=0`

`=>x^2-6x+1 = 0`

factor is not possible for equation `x^2-6x+1=0`

Solution is possible using the method of perfect square.
Comparing the given equation with the standard quadratic equation `ax^2+bx+c=0,`

we get, `a=1, b=-6, c=1.`

`:. Delta=b^2-4ac`

`=(-6)^2-4 (1) (1)`

`=36-4`

`=32`

`:. sqrt(Delta)=sqrt(32)=4*sqrt(2)`



Now, `alpha=(-b+sqrt(Delta))/(2a)`

`=(-(-6)+4*sqrt(2))/(2*1)`

`=(6+4*sqrt(2))/2`

`=3+2*sqrt(2)`



and, `beta=(-b-sqrt(Delta))/(2a)`

`=(-(-6)-4*sqrt(2))/(2*1)`

`=(6-4*sqrt(2))/2`

`=3-2*sqrt(2)`

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