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प्रश्न
In the following determine rational numbers a and b:
`(3 + sqrt2)/(3 - sqrt2) = a + bsqrt2`
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उत्तर
We know that rationalization factor for `3 - sqrt2` is `3 + sqrt2`. We will multiply numerator and denominator of the given expression `(3 + sqrt2)/(3 - sqrt2)` by `3 + sqrt2` to get
`(3 + sqrt2)/(3 - sqrt2) xx (3 + sqrt2)/(3 + sqrt2) = ((3)^2 + (sqrt2)^2 + 2 xx 3 sqrt2)/((3)^2 - (sqrt2)^2)`
`= (9 + 2 + 6sqrt2)/(9 - 2)`
` = (11 + 6sqrt2)/7`
`= 11/7 + 6/7 sqrt2`
On equating rational and irrational terms, we get
`a + bsqrt2 = 11/7 + 6/7 sqrt2`
Hence we get a = 11/7, b = 6/7
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संबंधित प्रश्न
Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, π = `c/d`. This seems to contradict the fact that π is irrational. How will you resolve this contradiction?
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`(3 + sqrt3)(3 - sqrt3)`
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`3/sqrt5`
Rationalise the denominator of the following:
`3/(2sqrt5)`
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`16/(sqrt41 - 5)`
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