Advertisements
Advertisements
Question
The value of \[\sqrt{5 + 2\sqrt{6}}\] is
Options
\[\sqrt{3} - \sqrt{2}\]
\[\sqrt{3} + \sqrt{2}\]
\[\sqrt{5} + \sqrt{6}\]
none of these
Advertisements
Solution
Given that:`sqrt(5+2sqrt6)`.It can be written in the form `(a-b )^2 = a^2 +b^2 - 2 ab` as
`sqrt(5+2sqrt6) = sqrt(3+2+2xxsqrt3 xxsqrt2)`
` =sqrt((sqrt3)^2 + (sqrt2)^2+ 2 xx sqrt3 xxsqrt2)`
`= sqrt((sqrt3+sqrt2)^2)`
` = sqrt3 +sqrt2.`
APPEARS IN
RELATED QUESTIONS
Simplify:-
`2^(2/3). 2^(1/5)`
Given `4725=3^a5^b7^c,` find
(i) the integral values of a, b and c
(ii) the value of `2^-a3^b7^c`
Assuming that x, y, z are positive real numbers, simplify the following:
`(sqrt2/sqrt3)^5(6/7)^2`
Prove that:
`(3^-3xx6^2xxsqrt98)/(5^2xxroot3(1/25)xx(15)^(-4/3)xx3^(1/3))=28sqrt2`
Show that:
`{(x^(a-a^-1))^(1/(a-1))}^(a/(a+1))=x`
Show that:
`(3^a/3^b)^(a+b)(3^b/3^c)^(b+c)(3^c/3^a)^(c+a)=1`
Find the value of x in the following:
`(3/5)^x(5/3)^(2x)=125/27`
Write \[\left( \frac{1}{9} \right)^{- 1/2} \times (64 )^{- 1/3}\] as a rational number.
If x = \[\sqrt[3]{2 + \sqrt{3}}\] , then \[x^3 + \frac{1}{x^3} =\]
The value of \[\sqrt{3 - 2\sqrt{2}}\] is
