Express $2x^2 - 6x - 12$ in the form $k(x - a)^2 - b$, where $k$ is an integer. Answer: ... Hence, solve $2x^2 - 6x - 12 = 0$. Answer: x = ... or ... Simplify $\frac{3x+1}{2x^2+11x+12} - \frac{1}{x+4}$. Answer: ...

1. Complete the square and write the expression in the required form

1. Substitute the completed-square form and isolate the square term

From part (a)(i),
$2x^2-6x-12=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{33}{2}.$
So
$2\left(x-\dfrac{3}{2}\right)^2-\dfrac{33}{2}=0$
$2\left(x-\dfrac{3}{2}\right)^2=\dfrac{33}{2}$
$\left(x-\dfrac{3}{2}\right)^2=\dfrac{33}{4}.$

2. Take square roots and solve for the two values of $x$

$x-\dfrac{3}{2}=\pm\sqrt{\dfrac{33}{4}}$
$x-\dfrac{3}{2}=\pm\dfrac{\sqrt{33}}{2}$
$x=\dfrac{3}{2}\pm\dfrac{\sqrt{33}}{2}$
$x=\dfrac{3\pm\sqrt{33}}{2}.$

1. Factorise the quadratic denominator to identify the common denominator

$2x^2+11x+12=(2x+3)(x+4)$

So
$\frac{3x+1}{2x^2+11x+12}-\frac{1}{x+4}=\frac{3x+1}{(2x+3)(x+4)}-\frac{1}{x+4}.$

2. Write both fractions over the same denominator

The common denominator is $(2x+3)(x+4)$.

$\frac{3x+1}{(2x+3)(x+4)}-\frac{1}{x+4}$
$=\frac{3x+1}{(2x+3)(x+4)}-\frac{2x+3}{(2x+3)(x+4)}.$

3. Subtract the numerators and simplify

$\frac{3x+1-(2x+3)}{(2x+3)(x+4)}$
$=\frac{3x+1-2x-3}{(2x+3)(x+4)}$
$=\frac{x-2}{(2x+3)(x+4)}.$