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प्रश्न
If `x = (3 + sqrt(2))/(3 - sqrt(2)), y = (3 - sqrt(2))/(3 + sqrt(2))`, find y2.
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उत्तर
Given: `x = (3 + sqrt(2))/(3 - sqrt(2)), y = (3 - sqrt(2))/(3 + sqrt(2))`
We need to find y2.
Step-wise calculation:
1. Rationalise the denominator of y:
`y = (3 - sqrt(2))/(3 + sqrt(2)) xx (3 - sqrt(2))/(3 - sqrt(2))`
= `(3 - sqrt(2))^2/((3)^2 - (sqrt(2))^2`
= `(3 - sqrt(2))^2/(9 - 2)`
= `(3 - sqrt(2))^2/7`
2. Expand the numerator:
`(3 - sqrt(2))^2`
= `3^2 + (sqrt(2))^2 - 2 xx 3 xx sqrt(2)`
= `9 + 2 - 6sqrt(2)`
= `11 - 6sqrt(2)`
3. So, `y = (11 - 6sqrt(2))/7`.
4. Now, find y2:
`y^2 = ((11 - 6sqrt(2))/7)^2`
`y^2 = (11 - 6sqrt(2))^2/7^2`
5. Expand the numerator:
`(11 - 6sqrt(2))^2`
= `11^2 + (6sqrt(2))^2 - 2 xx 11 xx 6sqrt(2)`
= `121 + 72 - 132sqrt(2)`
= `193 - 132sqrt(2)`
6. Thus, `y^2 = (193 - 132sqrt(2))/49`.
