Question

Prove algebraically that 2(√(2+√3)) = √6 + √2?

Answer 1

Topic :-

Square roots

Given to prove :-

$2( \sqrt{2 + \sqrt{3} } ) = \sqrt{6} + \sqrt{2}$

Proof !

Take L.H.S

$2( \sqrt{ {2} + \sqrt{3} )}$

$\sqrt{(2 + \sqrt{3} )2 {}^{2} }$

$\sqrt{2 + \sqrt{3} (4)}$

$\sqrt{8 + 4 \sqrt{3} }$

$\sqrt{8 + \sqrt{3 \times 4 {}^{2} } }$

$\sqrt{8 + \sqrt{16 \times 3} }$

$\sqrt{8 + \sqrt{48} }$

So, here 48 can be written as 12 × 4

$\sqrt{8 + \sqrt{4 \times 12} }$

$\sqrt{8 + \sqrt{4} \times \sqrt{12} }$

$\sqrt{8 + 2 \sqrt{12} }$

12 can be written as 2× 6

8 can be written as 2 + 6

$\sqrt{2 + 6 + 2 \sqrt{2 \times 6} \ }$

2 can be written as (√2)² 6 can be written as (√6)²

$\sqrt{( \sqrt{2}) {}^{2} + ( \sqrt{6}) {}^{2} + 2 \sqrt{2} \sqrt{6} }$

It is in form of a² + b² + 2ab = (a+b)² So,

$\sqrt{( \sqrt{2} + \sqrt{6}) {}^{2} }$

$\sqrt{2} + \sqrt{6}$

So,

$2( \sqrt{2 + \sqrt{3} } ) = \sqrt{6} + \sqrt{2}$

Hence proved !

Know more some algebraic identities:-

(a+ b)² = a² + b² + 2ab

( a - b )² = a² + b² - 2ab

( a + b )² + ( a - b)² = 2a² + 2b²

( a + b )² - ( a - b)² = 4ab

( a + b + c )² = a² + b² + c² + 2ab + 2bc + 2ca

a² + b² = ( a + b)² - 2ab

(a + b )³ = a³ + b³ + 3ab ( a + b)

( a - b)³ = a³ - b³ - 3ab ( a - b)

If a + b + c = 0 then a³ + b³ + c³ = 3abc

Answer 2

$\large \tt 2( \sqrt{2 + \sqrt{3} }) = \sqrt{6} + \sqrt{2}$

$\sf ( \cancel{ \sqrt{2} )( \sqrt{2}} )( \sqrt{2} + \sqrt{3} ) = \sqrt{6} + \cancel{ \sqrt{2} }$

$\sf as \: \: per \: \: \: binomial \: \: expansion \: \:$

$\large \tt \: \sqrt{6} = \sqrt{6}$

• Hence Proved