Advertisements
Advertisements
प्रश्न
Find the values the following correct to three places of decimals, it being given that `sqrt2 = 1.4142`, `sqrt3 = 1.732`, `sqrt5 = 2.2360`, `sqrt6 = 2.4495` and `sqrt10 = 3.162`
`(3 - sqrt5)/(3 + 2sqrt5)`
Advertisements
उत्तर
We know that rationalization factor for `3 + 2sqrt5` is `3 - sqrt5`. We will multiply numerator and denominator of the given expression `(3 - sqrt5)/(3 + 2sqrt5)` by `3 - 2sqrt5` to get
`(3 - sqrt5)/(3 + 2sqrt5) xx (3 - 2sqrt5)/(3 - 2sqrt5) = ((3)^2 - 3 xx 2 xx sqrt5 - 3 xx sqrt5 + 2 xx (sqrt5)^2)/((3)^2 - (2sqrt5)^2)`
` = (9 - 9sqrt5 + 10)/(9 - 20)`
`= (19 - 9sqrt5)/(-11)`
`= (9sqrt5 - 19)/11`
Putting the value of `sqrt5`, we get
`(9sqrt5 - 19)/11 = (9(2.236) - 19)/11`
`= (20.124 - 19)/11`
`= 1.124/11`
= 0.102
Hence the given expression is simplified to 0.102
APPEARS IN
संबंधित प्रश्न
Find the value to three places of decimals of the following. It is given that
`sqrt2 = 1.414`, `sqrt3 = 1.732`, `sqrt5 = 2.236` and `sqrt10 = 3.162`
`2/sqrt3`
Rationales the denominator and simplify:
`(3 - sqrt2)/(3 + sqrt2)`
Rationales the denominator and simplify:
`(2sqrt3 - sqrt5)/(2sqrt2 + 3sqrt3)`
Simplify
`1/(2 + sqrt3) + 2/(sqrt5 - sqrt3) + 1/(2 - sqrt5)`
If \[x = 3 + 2\sqrt{2}\],then find the value of \[\sqrt{x} - \frac{1}{\sqrt{x}}\].
Rationalise the denominator of the following:
`1/(sqrt7-2)`
Value of `root(4)((81)^-2)` is ______.
Find the value of a and b in the following:
`(5 + 2sqrt(3))/(7 + 4sqrt(3)) = a - 6sqrt(3)`
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`
Simplify:
`(8^(1/3) xx 16^(1/3))/(32^(-1/3))`
