<s>[INST] Finding the exact value of a radical How do I show that $\sqrt{97 +56\sqrt3}$ reduces to $7 +4\sqrt3?$. Without knowing intitially that it reduces to that value.
 [/INST] If you suspect that $\sqrt{97+56\sqrt3}$ (or similar) equals something
like $a+\sqrt b$ with $a$, $b$ rational you must have
$$97+56\sqrt 3=a^2+b+2a\sqrt b.$$
To match up the surds you'll have to have $a^2+b=97$
and $2a\sqrt b=56\sqrt3$. Thus $a^2b=28^2\times 3=2352$. Therefore
$a^2$ and $b$ are the solutions of the quadratic
$$x^2-97x+2352=0.$$
The solutions are $x=48$ and $x=49$. Then $a^2=49$ as that's the one
that's a square, and $b^2=48$. So
$$\sqrt{97+56\sqrt3}=\sqrt{49}+\sqrt{48}=7+4\sqrt3.$$
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