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Question
If a and b are rational numbers, find the value of a and b:
`a + bsqrt(6) = (2sqrt(3) - sqrt(2))/(2sqrt(3) + sqrt(2))`
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Solution
Given: `a + bsqrt(6) = (2sqrt(3) - sqrt(2))/(2sqrt(3) + sqrt(2))` where a and b are rational numbers.
Stepwise calculation:
1. Rationalise the denominator:
`(2sqrt(3) - sqrt(2))/(2sqrt(3) + sqrt(2)) xx (2sqrt(3) - sqrt(2))/(2sqrt(3) - sqrt(2))`
= `(2sqrt(3) - sqrt(2))^2/((2sqrt(3))^2 - (sqrt(2))^2`
2. Calculate numerator:
`(2sqrt(3) - sqrt(2))^2`
= `(2sqrt(3))^2 - 2 xx 2sqrt(3) xx sqrt(2) + (sqrt(2))^2`
= `4 xx 3 - 4sqrt(6) + 2`
= `12 - 4sqrt(6) + 2`
= `14 - 4sqrt(6)`
3. Calculate denominator:
`(2sqrt(3))^2 - (sqrt(2))^2`
= 4 × 3 – 2
= 12 – 2
= 10
4. So the expression becomes:
`(14 - 4sqrt(6))/10`
= `14/10 - 4/10sqrt(6)`
= `7/5 - 2/5sqrt(6)`
5. Comparing with `a + bsqrt(6)`:
`a = 7/5, b = -2/5`
