Solve $3x = \sqrt{1-x} - 1$. A triangle has a base of length $(1+\sqrt{3})\text{ cm}$ and an area of $(14+8\sqrt{3})\text{ cm}^2$. Find, without using a calculator, the perpendicular height of the triangle, in cm, in the form $(a+b\sqrt{3})$, where $a$ and $b$ are integers.

1. Isolate the surd and square both sides

$3x + 1 = \sqrt{1-x}$
$(3x + 1)^2 = 1 - x$

2. Expand and simplify into a quadratic equation

$9x^2 + 6x + 1 = 1 - x$
$9x^2 + 7x = 0$

3. Solve the quadratic equation

$x(9x + 7) = 0$
$x = 0 \quad \text{or} \quad x = -\frac{7}{9}$

4. Check for extraneous roots and state the final answer

Substitute $x = 0$ into the original equation:
LHS $= 3(0) = 0$
RHS $= \sqrt{1-0} - 1 = 1 - 1 = 0$
Since LHS = RHS, $x = 0$ is a valid solution.

Substitute $x = -\frac{7}{9}$ into the original equation:
LHS $= 3\left(-\frac{7}{9}\right) = -\frac{7}{3}$
RHS $= \sqrt{1 - \left(-\frac{7}{9}\right)} - 1 = \sqrt{\frac{16}{9}} - 1 = \frac{4}{3} - 1 = \frac{1}{3}$
Since LHS $\neq$ RHS, $x = -\frac{7}{9}$ is an extraneous root.

Therefore, $x = 0$.

1. Set up the equation relating the area, base, and height of the triangle

$\frac{1}{2} \times \text{base} \times \text{height} = \text{Area}$
$\frac{1}{2}(1+\sqrt{3})h = 14+8\sqrt{3}$

2. Rearrange the equation to make $h$ the subject

$h = \frac{2(14+8\sqrt{3})}{1+\sqrt{3}}$
$h = \frac{28+16\sqrt{3}}{1+\sqrt{3}}$

3. Rationalise the denominator by multiplying the numerator and denominator by the conjugate

$h = \frac{28+16\sqrt{3}}{1+\sqrt{3}} \times \frac{1-\sqrt{3}}{1-\sqrt{3}}$
$h = \frac{28 - 28\sqrt{3} + 16\sqrt{3} - 16(3)}{1^2 - (\sqrt{3})^2}$

4. Simplify the expression to find the final answer

$h = \frac{28 - 12\sqrt{3} - 48}{1 - 3}$
$h = \frac{-20 - 12\sqrt{3}}{-2}$
$h = 10 + 6\sqrt{3}$