How do I show that $\sqrt{97 +56\sqrt3}$ reduces to $7 +4\sqrt3?$. Without knowing intitially that it reduces to that value.
[Math] Finding the exact value of a radical
algebra-precalculus
algebra-precalculus
How do I show that $\sqrt{97 +56\sqrt3}$ reduces to $7 +4\sqrt3?$. Without knowing intitially that it reduces to that value.
Best Answer
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 $$\sqrt{97+56\sqrt3}=\sqrt{49}+\sqrt{48}=7+4\sqrt3.$$