Question

Root2+ root3/ root5+root6 rationalized
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Answer 1

`2√(3)+3√(2)-√(10)-√(15)`

Explanation:

`$\frac{√(2) +√(3)} {√(5) +√(6)} $`

In order to rationalize it, we multiply the denominator by its conjugate

This way we have a difference of squares

`\boxed{(a+b)(a-b) = a^2-b^2}`

`$\frac{√(2) +√(3)} {√(5) +√(6)} \cdot \frac{√(5) -√(6)} {√(5) -√(6)} = ((√(2) +√(3))(√(5) -√(6)))/(5-6) $`

`$((√(2) +√(3))(√(5) -√(6)))/(5-6)=(\left(√(10)-2√(3)+√(15)-3√(2)\right))/(-1) $`

`-(√(10)-2√(3)+√(15)-3√(2)) = 2√(3)+3√(2)-√(10)-√(15)`