Question

If \[x^4 + \frac{1}{x^4} = 623\] then \[x + \frac{1}{x} =\]

Options

• 27

• 25

• \[3\sqrt{3}\]
• \[- 3\sqrt{3}\]

Answer 1

In the given problem, we have to find the value of  `x+1/x`

Given  `x^4 + 1/x^4 = 623`

We shall use the identity `(a+b)^2 = a^2 +b^2 +2ab`

Here put`x^4 +1/x^4 = 623`,

`(x^2 +1/x^2)^2 = (x^2)^2 + 1/(x^2)^2 + 2 (x^2 xx 1/x^2)`

`(x^2 +1/x^2)^2 = x^4 + 1/x^4+ 2 (x^2 xx 1/x^2)`

`(x^2 +1/x^2)^2 = x^4+ 1/x^4+2`

`(x^2 +1/x^2)^2 = 625+2`

`(x^2 +1/x^2)^2 = 625`

`(x^2 +1/x^2) xx (x^2 +1/x^2) = 25xx25`

`(x^2 +1/x^2)  = 25`

We shall use the identity `(a+b)^2 = a^2 +b^2 +2ab` we get,

`(x+1/x)^2 = x^2 +1/x^2 +2(x xx 1/x)`

`(x+1/x)^2 = 25 +2 (x xx 1/x)`

`(x+1/x)^2 = 25 +2`

`(x+1/x)^2 = 27`

Taking square root on both sides we get,

`sqrt((x+1/x) xx (x+1/x)) = sqrt(3 xx 3xx 3)`

`(x+1/x) = 3sqrt3`

Hence the value of  `(x+1/x)`is `3sqrt3`.