Question

find the value of a and b if 3+√2/3-√2-3-√2/3+√2=a+12√2b

Answer 1

Hi friend, Here's the required answer:-

First solve the LHS:-

3+√2/3-√2-3-√2/3+√2= 3-3+√2/3-√2/3+√2-√2= 0.

Then the RHS:-

a+ 12√2b

Since it is given LHS=RHS

So, a+12√2b= 0

By hit and trial method, if we put the values of a and b as 0

Then, 0+12√2×0= 0

so a=0 and b=0.

Hope this helps you....