Prove that condition is rational

Question

I tried to solve this about hour, but I can't...

$$\begin{align} \sqrt{7+4\sqrt{3}} - \sqrt{3} \end{align}$$

Answer should be 2. I don't need to solve this for me, I just need explanation how to solve this condition. Thanks.

Answer 1

Starting with the identity $$7+4\sqrt{3}=7+4\sqrt{3}$$ $$7+4\sqrt{3}=4+4\sqrt{3}+3$$ $$7+4\sqrt{3}=\left(2+\sqrt{3}\right)^2$$ $$\sqrt{7+4\sqrt{3}}=2+\sqrt{3}$$ $$\sqrt{7+4\sqrt{3}}-\sqrt{3}=2$$

Answer 2

$$\begin{align}\sqrt{7+4\sqrt 3}-\sqrt 3&=\sqrt{7+2\sqrt{12}}-\sqrt 3\\&=\sqrt{\left(3+4+2\sqrt{3\times 4}\right)}-\sqrt 3\\&=\sqrt{\left(\sqrt 3+\sqrt 4\right)^2}-\sqrt 3\\&=\left(\sqrt 3+\sqrt 4\right)-\sqrt 3\\&=2.\end{align}$$

Answer 3

More generally, $$(x + y \sqrt{c})^2 = x^2 + y^2 c + 2 x y \sqrt{c}$$ So if you have a quantity $a + b \sqrt{c}$ that you suspect is the square of something of the form $x + y \sqrt{c}$ (where $a$, $b$, $x$ and $y$ are rational), you need to solve $$ \eqalign{x^2 + y^2 c &= a\cr 2 x y &= b\cr}$$ Assuming $b \ne 0$, you'll have $y = b/(2x)$, and then the first equation will give you a quadratic to solve for $x^2$. Remember you want a solution that is rational.

Answer 4

The solutions above all rely in one way or another on knowing ahead of time that the value is $2$. But what if you don't know that? And how can you find solutions to other problems that are like this?

Set $\sqrt{7 + 4\sqrt{3}} - \sqrt{3} = x$. Then $x + \sqrt{3} = \sqrt{7 + 4\sqrt{3}}$. Squaring both sides of this we get $$ x^2 + 2x\sqrt{3} + 3 = 7 + 4 \sqrt{3}$$ or $$x^2 + 2x\sqrt{3} = 4 + 4 \sqrt{3}$$ $$x^2-4 = (4-2x)\sqrt{3}$$

Now we want to know if the above equation has a rational solution. Notice that if $x$ is rational, then the left-hand-side of the above equation is rational, and the right-hand-side of the equation is of the form $(\mathrm{rational})\sqrt{3}$. The only way for those to be equal is if they are both zero. So we can see that we will have a solution if and only if we can find $x$ such that $x^2=4$ and $2x=4$. Obviously $x=2$ does it.