Express $x^2-6x-2$ in the form $(x-a)^2 + b$ where $a$ and $b$ are constants. Answer ___ Hence, state the minimum value of $y = x^2-6x-2$. Answer ___ Simplify $\displaystyle \frac{2x^2-18}{3x^2+11x+6}$. Answer ___ Hence, or otherwise, express $1 - \displaystyle \frac{2x^2-18}{3x^2+11x+6}$ as a single fraction in its simplest form. Answer ___ Given that $R^2 = h(p-q)^2$. Find the value of $R$ when $h = 20$, $p = 10$ and $q = 5. Answer ___ Express $q$ in terms of $h$, $p$ and $R$. Answer ___

1. Form the completed square using half of the coefficient of $x$.

Half of $-6$ is $-3$, so use $(x-3)^2$.

$(x-3)^2=x^2-6x+9$

2. Adjust the constant term to make the expression equal to the original expression.

Since $(x-3)^2=x^2-6x+9$,

$x^2-6x-2=(x-3)^2-9-2=(x-3)^2-11$

1. Identify the minimum value from the completed square form.

From 1(a)(i),
$y=(x-3)^2-11.$
Since $(x-3)^2\geq 0$, the minimum value of $y$ is
$0-11=-11.$

1. Factorise the numerator by taking out the common factor and using difference of squares.

$2x^2-18=2(x^2-9)=2(x-3)(x+3)$

2. Factorise the quadratic denominator.

$3x^2+11x+6=(3x+2)(x+3)$

3. Cancel the common factor to simplify the algebraic fraction.

$\frac{2x^2-18}{3x^2+11x+6}=\frac{2(x-3)(x+3)}{(3x+2)(x+3)}=\frac{2(x-3)}{3x+2}$

1. Use the result from part 1(b)(i) and express $1$ with the same denominator.

From part (b)(i),
$\frac{2x^2-18}{3x^2+11x+6}=\frac{2(x-3)}{3x+2}.$
Hence,
$1-\frac{2x^2-18}{3x^2+11x+6}=1-\frac{2(x-3)}{3x+2}$
$=\frac{3x+2}{3x+2}-\frac{2(x-3)}{3x+2}.$

2. Subtract the numerators and simplify the resulting fraction.

$=\frac{3x+2-2x+6}{3x+2}$
$=\frac{x+8}{3x+2}.$

1. Substitute the given values of $h$, $p$ and $q$ into the formula.

Given $R^2=h(p-q)^2$, with $h=20$, $p=10$ and $q=5$,

$R^2=20(10-5)^2$

2. Evaluate and take the square root to find $R$.

$R^2=20(5)^2=20\times25=500$

$R=\sqrt{500}=10\sqrt{5}$

1. Isolate the squared expression involving $q$.

$R^2=h(p-q)^2$
$\frac{R^2}{h}=(p-q)^2$

2. Take square roots and rearrange to make $q$ the subject.

$p-q=\pm\sqrt{\frac{R^2}{h}}=\pm\frac{R}{\sqrt{h}}$
Hence,
$q=p\mp\frac{R}{\sqrt{h}}$
Equivalently,
$q=p\pm\frac{R}{\sqrt{h}}$