Advertisements
Advertisements
प्रश्न
The positive square root of \[7 + \sqrt{48}\] is
पर्याय
\[7 + 2\sqrt{3}\]
\[7 + \sqrt{3}\]
\[ \sqrt{3}+2\]
\[3 + \sqrt{2}\]
Advertisements
उत्तर
Given that:`7 +sqrt48`.To find square root of the given expression we need to rewrite the expression in the form `a^2 +b^2 +2ab = (a+b)^2`
`7 +sqrt48 = 3+4+2xx2xxsqrt3`
` = (sqrt3)^2 + (2)^2 +2 xx 2xx xxsqrt3`
`= (sqrt3 + 2 )^2`
Hence the square root of the given expression is `sqrt3+2`
APPEARS IN
संबंधित प्रश्न
If a = 3 and b = -2, find the values of :
aa + bb
Prove that:
`(x^a/x^b)^(a^2+ab+b^2)xx(x^b/x^c)^(b^2+bc+c^2)xx(x^c/x^a)^(c^2+ca+a^2)=1`
Solve the following equation for x:
`2^(3x-7)=256`
If `1176=2^a3^b7^c,` find a, b and c.
Assuming that x, y, z are positive real numbers, simplify the following:
`(x^((-2)/3)y^((-1)/2))^2`
If `3^(x+1)=9^(x-2),` find the value of `2^(1+x)`
Write the value of \[\sqrt[3]{7} \times \sqrt[3]{49} .\]
The square root of 64 divided by the cube root of 64 is
Which of the following is (are) not equal to \[\left\{ \left( \frac{5}{6} \right)^{1/5} \right\}^{- 1/6}\] ?
Find:-
`32^(2/5)`
