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.
 A: Wikipedia describes a simple method to write
$$
\sqrt{a+b \sqrt{c}\ } = \sqrt{d}+\sqrt{e}
$$
with
$$
d=\frac{a + \sqrt {a^2-b^2c}}{2},
\qquad
e=\frac{a - \sqrt {a^2-b^2c}}{2}
$$
This works iff $a^2 - b^2c$ is a square.
For $\sqrt{97 +56\sqrt3}$ we have $a^2 - b^2c=1$ and so
$$
\sqrt{97 +56\sqrt3} = \sqrt{49}+\sqrt{48} = 7 + 4 \sqrt 3
$$
A: $$\sqrt{97 +56\sqrt3}=\sqrt{49+48 +56\sqrt3} =\sqrt{7^2+(4\sqrt 3)^2 +2 \times 7 \times 4\sqrt3} = \sqrt{(7+4\sqrt3)^2 }= 7 +4\sqrt3 $$
A: While the accepted solution is fine if you're willing to solve a quadratic equation, it's worth developing some theory to avoid this for two reasons:


*

*To save effort spent solving quadratic equations, replacing it by effort spent doing these different and exciting calculations instead :)

*So that we can also figure out something like $\sqrt[3]{1351 + 780 \sqrt{3}}$, where we'd have to solve a cubic equation instead, and who knows how to do that?


The theory is that if $(a + b \sqrt 3)^2 = 97 + 56 \sqrt 3$, then we also have $(a - b \sqrt{3})^2 = 97 - 56 \sqrt 3$. (You can check this by comparing the expansions. It ultimately comes from the fact that $\sqrt3$ and $-\sqrt3$ are "symmetric" as far as the rational numbers are concerned: you can't tell them apart by their algebraic properties.)
Multiplying these two equations together and applying difference of squares, we get
$$(a + b\sqrt 3)^2 (a - b\sqrt 3)^2 = (97 + 56 \sqrt3)(97 - 56 \sqrt 3) \implies (a^2 - 3b^2)^2 = 97^2 - 3 \cdot 56^2 = 1.$$
So now we have three equations for $a$ and $b$, not two:
\begin{align}
a^2 + 3b^2 &= 97 \\
2ab &= 56 \\
a^2 - 3b^2 &= \pm1.
\end{align}
The first two come from comparing coefficients in $(a + b \sqrt 3)^2 = 97 + 56 \sqrt 3$, as before. The third comes from the fact that if $(a^2 - 3b^2)^2 = 1$, then either $a^2 - 3b^2 = 1$ or $a^2 - 3b^2 = -1$.
If we consider $a^2 - 3b^2 = 1$ as the first possibility, then together with $a^2 + 3b^2 = 97$ we have two linear equations in $a^2$ and $b^2$. Adding them together, we get $2a^2 = 98$, so $a^2 = 49$, and $a = \pm 7$. Subtracting them, we get $6b^2 = 96$, so $b^2 = 16$, and $b = \pm 4$. Since $2ab = 56$, $a$ and $b$ are both positive or both negative, so we get $7 + 4\sqrt3$ and $-7 - 4\sqrt3$ as our two solutions (both valid).
The second case, when $a^2 - 3b^2 = -1$, tells us $2a^2 = 96$ and $6b^2 = 98$, which does not give us a rational solution. (Actually, it tells us that $a = 4 \sqrt 3$ and $b = \frac{7}{\sqrt 3}$, giving us the same answer in a "backwards" way.) But we can ignore it: we already got an answer.
A: 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
$$\sqrt{97+56\sqrt3}=\sqrt{49}+\sqrt{48}=7+4\sqrt3.$$
