# If  `x=5-2sqrt6` ,  find the value of  `sqrt(x)+1/sqrt(x)` .

llltkl | College Teacher | (Level 3) Valedictorian

Posted on

`x=5-2sqrt(6)`  (given)

`=3+2-2sqrt(6)`

`=(sqrt(3))^2+(sqrt(2))^2` `-2*sqrt(3)*sqrt(2)`

`=(sqrt(3)-sqrt(2))^2` ` `

Therefore, `sqrt(x)+(1)/sqrt(x) = sqrt(3)-sqrt(2) + (1)/(sqrt(3)-sqrt(2))`

Rationalizing the second term by multiplying with `(sqrt(3)+sqrt(2))`

The R.H.S. becomes = `sqrt(3)-sqrt(2)+sqrt(3)+sqrt(2)`

Therefore, the solution to the above problem is:

`sqrt(x)+(1)/sqrt(x) =2sqrt(3)`

oldnick | (Level 1) Valedictorian

Posted on

`x=5-2sqrt(6)`     `sqrt(x)+1/sqrt(x)```

Now we have to concern about `sqrt(x)= sqrt(5-2sqrt(6))`

Note that:  `(5-2sqrt(6))=((5-2sqrt(6))(5+sqrt(6)))/(5+2sqrt(6))=` `1/(5+2sqrt(6))`

That is:   `sqrt(x) +1/sqrt(x)=sqrt(x)+ sqrt(1/x)=sqrt(5-2sqrt(6))+sqrt(5+2sqrt(5))`

So from formula:

`sqrt(a+sqrt(b))=sqrt((a+sqrt(a^2-b))/2)+sqrt((a-sqrt(a^2-b))/2)`

and:

`sqrt(a-sqrt(b))=sqrt((a+sqrt(a^2-b))/2)-sqrt((a-sqrt(a^2-b))/2)`

we get:

`sqrt(a+sqrt(b))+sqrt(a-sqrt(b))=` `2sqrt((a+sqrt(a^2-b))/2)`

then appling at this case:

`sqrt(5+2sqrt(6)) +sqrt(5-2sqrt(6))=2sqrt((5+sqrt(25-24))/2)` `=2sqrt(3)`