If x=3+2√2,then find the value of √x-√1/x

Answer 1

method of undetermined coefficients:

[A+B*sqrt(2) ]^2 = 3 + 2*sqrt(2)

A^2 + 2ab*sqrt(2) + 2B^2 = 3 + sqrt(2)

(A^2 + 3b^2) + 2ab*sqrt(2) = 3 + sqrt(2)

A^2 + 3b^2 = 3 and 2ab = 2

A^2 + 2b^2 = 3 and ab=1

2nd equation says: b = 1/a

plugging into first equation:

A^2 + 2(1/a)^2 = 3

A^4 + 2 = 3a^2

A^4 - 3a^2 + 2 = 0

(A^2 - 2)(A^2 - 1) = 0

A^2 - 2 = 0 or A^2 = 1

A = +- sqrt(2) or A=+-1

In the former case:

B = 1/ +-sqrt(2) = +-sqrt(2)/2

In the latter case:

B = 1/+-1 = +-1

So the square root of X is:

A+B*sqrt(2) = +-sqrt(2) + sqrt(2)/2 * sqrt(2) = +-sqrt(2)+-1

or

A+B*sqrt(2) = +-1 +- sqrt(2)

Squaring the four possibilities:

(1+sqrt(2))^2 = 2 + 2*sqrt(2) + 2 = 3+sqrt(2) <--- there is the first square root

(1 - sqrt(2))^2 = 1 - 2*sqrt(2) + 2 = 3 - sqrt(2) <--- NO, that is not correct

(-1 + sqrt(2))^2 = 1 - 2*sqrt(2) + 2 = 3 - 2*sqrt(2) <--- No, that is not correct

(-1 - sqrt(2))^2 = 1 + 2*Sqrt(2) + 2 = 3 + sqrt(2) <---- yes, that is the second square root

1/sqrt(x) = sqrt(x)/x = +-1+-sqrt(2) / [3 + 2*sqrt(2)]

= [+-1+- sqrt(2)][ 3 - 2*sqrt(2)] / [ 3 - 8]

= [+-2 -+ 2*sqrt(2) +-3*sqrt(2) -+ 4 ]/5

= [-+2 +- sqrt(2)]/5

Changing the signs:

[+-2 -+ sqrt(2)]/5

finally adding x:

[3+2*sqrt(2)]/1 [+-2 -+ sqrt(2)]/5

[15 + 10*sqrt(2) +-2 -+ sqrt(2) ]/5

(15 +-2) + (10 +-1)*Sqrt(2)]/5

Answer 2

Let's go with algebraic here, not sure what you wanted (numeric or algebraic)

First notice that root(x)-root(1/x) = (x-1)/root(x) by finding common denominator of course.

now we plug in the value of x = 3+2root(2)

we have then root(x)-root(1/x) = [2+2root(2)]/[3+2root(2)] you can simplify this a bit further to get

$2\sqrt{2}-2$ by rationalizing the denominator