If \[x = 3 + 2\sqrt{2}\],then find the value of \[\sqrt{x} - \frac{1}{\sqrt{x}}\].
Solution
Given that:.`x = 3+2sqrt2` It can be written in the form `(a+b)^2 = a^2 +b^2 +2ab` as
`sqrtx = sqrt(3+2sqrt2)`
` = sqrt(2+1+2xx 1xxsqrt2)`
` = sqrt((sqrt2)^2+ (1)^2 +2 xx 1 xx sqrt2 `
` = sqrt((sqrt2+1)^2)`
` = sqrt2 +1`
Therefore,
`1/sqrtx = 1/(sqrt2+1)`
We know that rationalization factor for `sqrt2`+1 is `sqrt2`-1 . We will multiply numerator and denominator of the given expression `1/(sqrt2+1)`by, `sqrt2-1,`to get
`1/(sqrt2 +1) xx (sqrt2-1)/(sqrt2-1) = (sqrt2-1)/((sqrt2) ^2 - (1)^2)`
`=(sqrt2-1)/(2-1)`
`= sqrt2 - 1`
Hence
`sqrtx - 1/sqrtx = sqrt2 +1 - (sqrt2 - 1)`
` = sqrt 2+1 - sqrt2 +1`
` =2 `
Therefore, value of the given expression is 2.