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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`
`(1 + sqrt2)/(3 - 2sqrt2)`
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उत्तर
We know that rationalization factor for `3 - 2sqrt2` is `3 + 2sqrt2`. We will multiply numerator and denominator of the given expression `(1 + sqrt2)/(3 - 2sqrt2)` by `3 + 2sqrt2` to get
`(1 + sqrt2)/(3 - 2sqrt2) xx (3 + 2sqrt2)/(3 + 2sqrt2) = (3 + 2 xx sqrt2 + 3 xx sqrt2 + 2 xx (sqrt2)^2)/((3)^2 - (2sqrt2)^2)`
`= (3 + 2sqrt2 + 3sqrt2 + 4)/(9 - 8)`
`= (7 + 5sqrt2)/1`
Putting te value of `sqrt2` we get
`7 + 5sqrt2 = 7 + 5(1.4142)`
= 7 + 7.071
= 14.071
Hence the given expression is simplified to 14.071
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संबंधित प्रश्न
Classify the following numbers as rational or irrational:
`2-sqrt5`
Rationalise the denominator of the following
`(sqrt2 + sqrt5)/3`
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`
`(sqrt5 + 1)/sqrt2`
In the following determine rational numbers a and b:
`(3 + sqrt2)/(3 - sqrt2) = a + bsqrt2`
If x= \[\sqrt{2} - 1\], then write the value of \[\frac{1}{x} . \]
If \[x = 3 + 2\sqrt{2}\],then find the value of \[\sqrt{x} - \frac{1}{\sqrt{x}}\].
\[\sqrt{10} \times \sqrt{15}\] is equal to
\[\sqrt[5]{6} \times \sqrt[5]{6}\] is equal to
If \[\frac{\sqrt{3 - 1}}{\sqrt{3} + 1}\] =\[a - b\sqrt{3}\] then
Simplify:
`64^(-1/3)[64^(1/3) - 64^(2/3)]`
