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

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The expression \(4\sqrt{4-2\sqrt{3}}+\sqrt{97-56\sqrt{3}}\) is analyzed to determine if it can be an integer. The first part, \(4\sqrt{4-2\sqrt{3}}\), simplifies to a rational number, while the second part, \(\sqrt{97-56\sqrt{3}}\), is examined for its integer properties. Through algebraic manipulation and simplification, it is shown that the overall expression does not yield an integer value. The conclusion is that the expression cannot be an integer, as proven through the calculations provided. Thus, the expression \(4\sqrt{4-2\sqrt{3}}+\sqrt{97-56\sqrt{3}}\) is definitively not an integer.
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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
 
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$4-2\sqrt{3} = 3 - 2\sqrt{3} +1 = (\sqrt{3} -1)^2$

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

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

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