Advertisements
Advertisements
प्रश्न
Find the value of a and b in the following:
`(7 + sqrt(5))/(7 - sqrt(5)) - (7 - sqrt(5))/(7 + sqrt(5)) = a + 7/11 sqrt(5)b`
Advertisements
उत्तर
We have, `(7 + sqrt(5))/(7 - sqrt(5)) - (7 - sqrt(5))/(7 + sqrt(5)) = a + 7/11 sqrt(5)b`
⇒ `((7 + sqrt(5))^2 - (7 - sqrt(5))^2)/((7 - sqrt(5))(7 + sqrt(5))) = a + 7/11 sqrt(5)b`
⇒ `([7^2 + (sqrt(5))^2 + 2 xx 7 xx sqrt(5)] - [7^2 + (sqrt(5))^2 - 2 xx 7 xx sqrt(5)])/(7^2 - (sqrt(5))^2) = a + 7/11 sqrt(5)b`
⇒ `(49 + 5 + 14sqrt(5) - 49 - 5 + 14sqrt(5))/(49 - 5) = a + 7/11 sqrt(5)b` ...`[("Using identity" (a + b)^2 = a^2 + 2ab + b^2),((a - b)^2 = a^2 - 2ab - b^2),("and" (a - b)(a + b) = a^2 - b^2)]`
⇒ `(28sqrt(5))/44 = a + 7/11 sqrt(5)b`
⇒ `7/11 sqrt(5) = a + 7/11 sqrt(5)b`
⇒ `0 + 7/11 sqrt(5) = a + 7/11 sqrt(5)b`
On comparing both sides, we get
a = 0 and b = 1
APPEARS IN
संबंधित प्रश्न
Simplify the following expressions:
`(3 + sqrt3)(5 - sqrt2)`
Rationalise the denominator of the following
`(sqrt2 + sqrt5)/3`
In the following determine rational numbers a and b:
`(3 + sqrt2)/(3 - sqrt2) = a + bsqrt2`
The rationalisation factor of \[2 + \sqrt{3}\] is
Rationalise the denominator of the following:
`(2 + sqrt(3))/(2 - sqrt(3))`
Rationalise the denominator of the following:
`(sqrt(3) + sqrt(2))/(sqrt(3) - sqrt(2))`
Find the value of a and b in the following:
`(3 - sqrt(5))/(3 + 2sqrt(5)) = asqrt(5) - 19/11`
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.
`sqrt(2)/(2 + sqrt(2)`
If `a = (3 + sqrt(5))/2`, then find the value of `a^2 + 1/a^2`.
Find the value of `4/((216)^(-2/3)) + 1/((256)^(- 3/4)) + 2/((243)^(- 1/5))`
