Solve $\sqrt{4x+12} - \sqrt{x+3} = 2$. The volume of right circular cone is $4\pi\text{ cm}^3$. The radius of its base is $(1+\sqrt{2})\text{ cm}$. Find, without the use of a calculator, the height of the cone in the form $(a+b\sqrt{2})\text{ cm}$, where $a$ and $b$ are integers.

1. Simplify the first surd by factoring out a perfect square

$\sqrt{4(x+3)} - \sqrt{x+3} = 2$
$2\sqrt{x+3} - \sqrt{x+3} = 2$

2. Combine the like surds

$\sqrt{x+3} = 2$

3. Square both sides to solve for $x$

$x + 3 = 2^2$
$x + 3 = 4$
$x = 1$

1. Set up the equation for the volume of the cone

$\frac{1}{3}\pi r^2 h = V$
$\frac{1}{3}\pi (1+\sqrt{2})^2 h = 4\pi$

2. Expand the squared term and isolate $h$

$(1+2\sqrt{2}+2) h = 12$
$(3+2\sqrt{2}) h = 12$
$h = \frac{12}{3+2\sqrt{2}}$

3. Rationalise the denominator to find the height in the required form

$h = \frac{12(3-2\sqrt{2})}{(3+2\sqrt{2})(3-2\sqrt{2})}$
$h = \frac{36 - 24\sqrt{2}}{3^2 - (2\sqrt{2})^2}$
$h = \frac{36 - 24\sqrt{2}}{9 - 8}$
$h = 36 - 24\sqrt{2}$