Question

Find 3+root 2/3-root 2=a+b root2

Answer 1

`(11)/(7)` +
`(6)/(7)`
`√(2)`

Explanation:

Given

`(3+√(2) )/(3-√(2) )`

Multiply the numerator/ denominator by the conjugate of the denominator.

The conjugate of 3 -
`√(2)` is 3 +
`√(2)` , thus

=
`((3+√(2))(3+√(2)) )/((3-√(2))(3+√(2)) )` ← expand numerator/ denominator using FOIL

=
`(9+6√(2)+2 )/(9-2)`

=
`(11+6√(2) )/(7)`

=
`(11)/(7)` +
`(6)/(7)`
`√(2)` ← in the form a + b
`√(2)`

with a =
`(11)/(7)` and b =
`(6)/(7)`

Answer 2

a =
`(11)/(7)` ; b =
`(6)/(7)`

Explanation:

`(3 + √(2))/(3 - √(2) ) = a + b√(2) \\\\`

Rationalising
`(3 + √(2))/(3 - √(2) )` gives :-

`(3 + √(2))/(3 - √(2) ) = ((3 + √(2))(3 + √(2)))/((3 - √(2))(3 + √(2)) ) = ((3 + √(2))^2)/(3^2 - (√(2))^2 ) = (11 +6√(2) )/(7)`

Comparing
`(11 + 6√(2) )/(7)` with
`a + b√(2)` gives

a =
`(11)/(7)` & b =
`(6)/(7)`