Simplify $\displaystyle\frac{16a^6 b^2}{5} - \displaystyle\frac{8a^3}{25ab^3}$. Answer ___ (i) Express $-4x + x^2 - 6$ in the form $a + (x + b)^2$, where $a$ and $b$ are integers. Answer ___ (ii) Write down the coordinates of the minimum point of the graph of y = -4x + x^2 - 6. Answer (,) (c) $l = \frac{1}{2}(m^2 - n)$ (i) Evaluate $l$ when $m = 4$ and $n = -5$. Answer $l$ = ___ (ii) Express $n$ in terms of $l$ and $m$. Answer ___ (d) Solve $\displaystyle\frac{x+1}{2x+3} + \displaystyle\frac{3x}{4x^2-9} = 2$. Answer $x$ = ___ or ___

1. Write both terms with a common denominator.

$\frac{16a^6b^2}{5}-\frac{8a^3}{25ab^3}$
First simplify the second fraction:
$\frac{8a^3}{25ab^3}=\frac{8a^2}{25b^3}$
The common denominator is $25b^3$.
$\frac{16a^6b^2}{5}=\frac{80a^6b^5}{25b^3}$

2. Subtract the fractions and factorise the numerator.

$=\frac{8a^2(10a^4b^5-1)}{25b^3}$

1. Rewrite the quadratic in standard order and prepare to complete the square.

$-4x+x^2-6=x^2-4x-6$

2. Complete the square for the $x^2-4x$ part.

$x^2-4x=(x-2)^2-4$

3. Combine the constant terms to express in the required form.

$x^2-4x-6=(x-2)^2-4-6=(x-2)^2-10$

So in the form $a+(x+b)^2$, $a=-10$ and $b=-2$.

1. Read the minimum point from the completed square form.

From part (i),
$y=-4x+x^2-6=(x-2)^2-10.$
Since $(x-2)^2\geq 0$, the minimum occurs when $x-2=0$, so $x=2$.
Then $y=-10$.

Minimum point: $(2,-10)$.

1. Substitute $m=4$ and $n=-5$ into the formula and evaluate.

$l=\frac{1}{2}(m^2-n)$

$=\frac{1}{2}\left(4^2-(-5)\right)$

$=\frac{1}{2}(16+5)$

$=\frac{21}{2}$

1. Clear the fraction by multiplying both sides by 2.

$l=\dfrac{1}{2}(m^2-n)$

$2l=m^2-n$

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

$2l=m^2-n$

$n=m^2-2l$

1. Factorise the denominator and state the common denominator.

$4x^2-9=(2x+3)(2x-3)$
The common denominator is $(2x+3)(2x-3)$, with $x\neq \pm\dfrac{3}{2}$.

2. Clear the fractions by multiplying throughout by the common denominator.

$\frac{x+1}{2x+3}+\frac{3x}{(2x+3)(2x-3)}=2$
Multiplying throughout by $(2x+3)(2x-3)$,
$(x+1)(2x-3)+3x=2(2x+3)(2x-3).$

3. Expand and simplify to obtain a quadratic equation.

$(x+1)(2x-3)+3x=2(4x^2-9)$
$2x^2-x-3+3x=8x^2-18$
$2x^2+2x-3=8x^2-18$
$6x^2-2x-15=0$

4. Solve the quadratic equation and check the solutions are valid.

For $6x^2-2x-15=0$,
$x=\frac{-(-2)\pm\sqrt{(-2)^2-4(6)(-15)}}{2(6)}$
$x=\frac{2\pm\sqrt{364}}{12}$
$x=\frac{2\pm2\sqrt{91}}{12}$
$x=\frac{1\pm\sqrt{91}}{6}$
These values are not $\pm\dfrac{3}{2}$, so both are valid.