Question

6 ÷ by 3√2-2√3=3√2-a√3 find the value of a

Answer 1

Given  (√3 - 1) / (√3 + 1)         ( rationalize numarator and denominator(√3 - 1) )
⇒ (√3 - 1)(√3 - 1) / (√3 + 1)(√3 - 1)
⇒ (√3 - 1)2 / (3 - 1)
⇒ (3 + 1-2√3 ) / 2
⇒ (2 -√3 )  = a+b√3  ⇒ a = 2 and b = -1
Sol2) :           
Given  (√2 + √3) / (3√2-2√3)     ( rationalize numarator and denominator (3√2 + 2√3 ) )
⇒ (√2 + √3)(3√2 + 2√3 ) / (3√2-2√3)(3√2 + 2√3 )
⇒ (12 + 5√6) / (18-12)
⇒ (2 + (5√6) /6 )  = a-b√6
a = 2 and b = -5/6.

Answer 2

rationalise it by multiplying dividing by
3root2+2root3

u will get 6(3root2+2root3)/18-12
as 6 cancels out

ans 3root2+2root3=3root2-aroor3

=>a=-2