Question

If \[x - \frac{1}{x} = \frac{1}{2}\],then write the value of \[4 x^2 + \frac{4}{x^2}\]

Answer 1

We have to find the value of  `4x^2 + 4/x^2`

Given  `x- 1/x = 1/2`

Using identity  `(a-b)^2 = a^2 - 2ab +b^2`

Here  `a= x,b= 1/x`

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

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

 `(x-1/x)^2 = x^2 -2 +1/x^2`

By substituting the value of  `x - 1/x = 1/x` we get 

 `(1/2)^2 = x^2 + 1/x^2 - 2`

By transposing – 2 to left hand side we get 

 `1/4 +2 = x^2 +1/x^2`

By taking least common multiply we get 

          `1/4+2/1 = x^2 + 1/x^2 `

 `1/4 +2/1 xx 4/4 = x^2 + 1/x^2`

           `1/4 +8/4 = x^2 +1/x^2`

             `(1+8)/4 = x^2 + 1/x^2`

                `(+9)/4 = x^2 + 1/x^2`

By multiplying 4 on both sides we get 

`4 xx 9/4 = 4x^2 + 4 xx 1/x^2`

`4 xx 9/4 = 4x^2 + 4/x^2`

`9 =4x^2 + 4/x^2`

Hence the value of  `4x^2 + 4/x^2` is 9.