Question

Arrange in ascending order
( root 8 + root 2)
( root 6 + root 4)
( root 1 + root 9)
( root 3 + root 7)
( root 5 + root 5)

Answer 1

heya friend here's ur answer :-)

$( \sqrt{8 }+ \sqrt{2} ) = (2 \sqrt{2} + \sqrt{2} ) = 3 \sqrt{2} \\ (\sqrt{6} + \sqrt{4} ) = \sqrt{6} + 2 = 3 \sqrt{6} \\ (\sqrt{1} + \sqrt{9} ) = ( \sqrt{1} + 3) = 4 \sqrt{1} \\ ( \sqrt{3} + \sqrt{7} ) \\ ( \sqrt{5} + \sqrt{5} ) = 2 \sqrt{5}$
now squaring all of them

$(3 \sqrt{2} ) {}^{2} = 18...(iii) \\ (3 \sqrt{6} ) {}^{2} = 54 ...(v)\\ (4 \sqrt{1} ) {}^{2} = 16 ...(ii)\\ ( \sqrt{3} + \sqrt{7} ) {}^{2} = 10....(i )\\ (2 \sqrt{5} ) {}^{2} = 20...(iv)$



so the ascending order is :-

$\sqrt{3} + \sqrt{7} < \sqrt{1} + \sqrt{9} < \sqrt{8} + \sqrt{2} < \sqrt{5} + \sqrt{5} < \sqrt{6} + \sqrt{4}$


hope it helps