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Let $$A= \sqrt{93+28\sqrt{11}}$$ if $B$ is the integer part of $A$ and $C$ is the fraction part of $C$, what is the value of $$B+C^2$$

I tried manipulating it by setting

$$ A=B+C$$

but I can't transform it into the expression, do I need approximate the integer part of A in order to solve this or is there another way?

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  • $\begingroup$ Presumably you mean $C$ is the fractional part of $A$ $\endgroup$ – Ross Millikan Mar 23 at 14:07
  • $\begingroup$ What do you mean by fractional part? That number is irrational. $\endgroup$ – Allawonder Mar 23 at 14:10
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Let's find intgers $x,y$ such that $$(x+y\sqrt{11})^2= 93+28\sqrt{11}$$

Then $$x^2+11y^2 =93 \;\;\;\wedge \;\;\; 2xy = 28$$

Since $11y^2<99\implies y^2 <9 \implies |y|\leq 2$ and $xy=14$. Playing with numbers we see that $x=7$ and $y=2$ works well, so $$\boxed{\sqrt{93+28\sqrt{11}} = 7+2\sqrt{11}}$$

Since $$13= 7+2\cdot 3<7+2\sqrt{11} <7+2\cdot \sqrt{49\over 4} = 14$$

we see $B =13$...

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Hint:

$(7+2\sqrt{11})^2=\cdots$

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