Question

If \[x - \frac{1}{x} = \frac{15}{4}\], then \[x + \frac{1}{x}\] = 

पर्याय

• 4

• \[\frac{17}{4}\]
• \[\frac{13}{4}\]
• \[\frac{1}{4}\]

Answer 1

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

Given  `x- 1/x = 15/4`

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

Here putting`x-1/x =15/4`,

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

              `(15 /4)^2 = x^2 +1/x^2 -2 (x xx 1/x)`

                    `225/16 = x^2 +1/x^2`

              `225/16 +2 = x^2 +1/x^2`

         `225/16 + (2 xx 16) /(1 xx 16) = x^2 +1/x^2`

                   `(225+32)/16 = x^2 +1/x^2`

                             `257/16 = x^2 +1/x^2`

Substitute  `257/16 = x^2 +1/x^2` in  `(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 = (x)^2 + (1/x)^2 + 2 xx x xx 1/x`

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

`(x+1/x)^2 = 257/16+2`

                       `(x+1/x)^2 = 257/16 + (2 xx 16)/(1 xx 16)`

                      `(x+1/x^2 )^2= (257+32)/16`

                         `(x+1/x)^2 = 289/16`

   `(x+1/x) xx (x+1/x) = (17 xx 17)/(4 xx 4)`

                          `(x+1/x) = 17/4`

Hence the value of  `x+1/x` is `17/4`.