Question

If √2-√3/3√2+2√3=a+b√6 where a and b are rational numbers, then find the values of a and b. Plz

Answer 1

Answer:

(√2 - √3)/(3√2 + 2√3) = a + b√6

LHS = (√2 - √3)(3√2 -2√3)/(3√2 +2√3)(3√2 - 2√3)  

       = { √2(3√2 -2√3) -√3 (3√2 -2√3)}/{(3√2)²-(2√3)² }  

       ={ 6 -2√6 -3√6+6 }/{ 18 - 12}  

       ={ 12 - 5√6 }/6  

⇒2 - (5/12)√6 = a + b√6

so, a = 2 and b = -5/12

Answer 2

Answer:

(3 + sqrt(5)) / (3 - sqrt(5)) = (a + b*sqrt(5)) /1 , we have one equation of 2 fractions.

The cross product gives: (3 + sqrt(5)) = (3 - sqrt(5)) * (a + b*sqrt(5)) 

Getting rid of the brackets: 3 + sqrt(5) = 3a + 3b*sqrt(5) - 5b - a* sqrt(5)

3 + sqrt(5) = (3a -5b) + (3b - a)*sqrt(5) Separating the Rational from the Irrational,

we get 2 linear equations: 3 = (3a -5b) and 

1 = (3b - a)

Multiplying the second equation by 3 gives: 3 = - 3a + 9b

adding the first equation: 3 = 3a -5b

we get : 3+3= - 3a + 3a + 9b - 5b 

Eliminating the variable a : 6 = 4 b this gives b = 6/4 = 3/2

Now knowing b=3/2 , we substitute it in 1 = (3b - a) and we get a = 7/2

The answer is: a = 7/2 ; and b = 3/2