I assume you want to solve for a and b in
(√2+√3)/(3√2-2√3) = a+b√6
get rid of the radicals in the bottom by recalling that (x-y)(x+y) = x^2-y^2. So, multiply top and bottom by 3√2+2√3 and you have
(√2+√3)(3√2+2√3)/((3√2)^2-(2√3)^2)
= (3*2+2√6+3√6+2*3)/(18-12)
= (12+5√6)/6
= 2 + 5/6 √6
find a and b for the following:-
1).(root of 2+root of 3 / 3 root of 2 - 2 root of 3) = a+b root of 6
2). {(root of 5 - 2/root of 5+2)-(root of 5+2/root of 5 - 2)} = a+b root of 5
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