Can the No $4\sqrt{4-2\sqrt {3}}+\sqrt{97-56\sqrt 3}$ be an Integer?

[solakis1]

Can the No :$4\sqrt{4-2\sqrt {3}}+\sqrt{97-56\sqrt 3}$ be an iteger ,if yes prove it if no then prove it again

 

[skeeter]

integer ...

Spoiler

$4-2\sqrt{3} = 3 - 2\sqrt{3} +1 = (\sqrt{3} -1)^2$

$97-56\sqrt{3} = 49 - 2(28\sqrt{3}) + 48 = (7 - 4\sqrt{3})^2$

$4\sqrt{(\sqrt{3}-1)^2} + \sqrt{(7-4\sqrt{3})^2} = 4\sqrt{3}-4 + 7 -4\sqrt{3} = 3$

 

[solakis1]

very good ,excellent