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प्रश्न
The rationalisation factor of \[2 + \sqrt{3}\] is
विकल्प
\[2 - \sqrt{3}\]
\[2 + \sqrt{3}\]
\[\sqrt{2} - 3\]
\[\sqrt{3} - 2\]
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उत्तर
We know that rationalization factor for `a+sqrt b` is `a-sqrtb` . Hence rationalization factor of `2+sqrt3` is `2-sqrt3 `.
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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?
Simplify the following expressions:
`(3 + sqrt3)(3 - sqrt3)`
Express the following with rational denominator:
`30/(5sqrt3 - 3sqrt5)`
In the following determine rational numbers a and b:
`(4 + 3sqrt5)/(4 - 3sqrt5) = a + bsqrt5`
Find the value of `6/(sqrt5 - sqrt3)` it being given that `sqrt3 = 1.732` and `sqrt5 = 2.236`
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)`
Simplify: \[\frac{7 + 3\sqrt{5}}{3 + \sqrt{5}} - \frac{7 - 3\sqrt{5}}{3 - \sqrt{5}}\]
Simplify \[\sqrt{3 + 2\sqrt{2}}\].
Rationalise the denominator of the following:
`1/(sqrt5+sqrt2)`
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)`
