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प्रश्न
Find the value of a and b in the following:
`(3 - sqrt(5))/(3 + 2sqrt(5)) = asqrt(5) - 19/11`
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उत्तर
We have, `(3 - sqrt(5))/(3 + 2sqrt(5)) = asqrt(5) - 19/11`
For rationalising the above equation, we multiply numerator and denominator of LHS by `3 - 2sqrt(5)`, we get
⇒ `((3 - sqrt(5)))/(3 + 2sqrt(5)) xx (3 - 2sqrt(5))/(3 - 2sqrt(5)) = asqrt(5) - 19/11`
⇒ `(3(3 - 2sqrt(5)) - sqrt(5)(3 - 2sqrt(5)))/((3)^2 - (2sqrt(5))^2) = asqrt(5) - 19/11` ...[Using identity, (a – b)(a + b) = a2 – b2]
⇒ `(9 - 6sqrt(5) - 3sqrt(5) + 10)/(9 - 4 xx 5) = asqrt(5) - 19/11`
⇒ `(19 - 9sqrt(5))/(9 - 20) = asqrt(5) - 19/11`
⇒ `(19 - 9sqrt(5))/(-11) = asqrt(5) - 19/11`
⇒ `(9sqrt(5))/11 - 19/11 = asqrt(5) - 19/11`
⇒ `(9sqrt(5))/11 = asqrt(5)`
⇒ `a = 9/11`
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संबंधित प्रश्न
Simplify the following expression:
`(sqrt5 - sqrt2)(sqrt5 + sqrt2)`
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`(2sqrt5 + 3sqrt2)^2`
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If x= \[\sqrt{2} - 1\], then write the value of \[\frac{1}{x} . \]
Classify the following number as rational or irrational:
`1/sqrt2`
Rationalise the denominator of the following:
`(sqrt(3) + sqrt(2))/(sqrt(3) - sqrt(2))`
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`(3sqrt(5) + sqrt(3))/(sqrt(5) - sqrt(3))`
Rationalise the denominator in the following and hence evaluate by taking `sqrt(2) = 1.414, sqrt(3) = 1.732` and `sqrt(5) = 2.236`, upto three places of decimal.
`1/(sqrt(3) + sqrt(2))`
Simplify:
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