Factorise completely $m^2 - 2mn + n^2 - p^2$. Answer ... Express as a single fraction in its simplest form $\dfrac{7}{2x-3} + \dfrac{x+1}{6-4x}$. Answer ... It is given that $p = \sqrt{1 + \dfrac{p^2}{r}}$. Find the values of $p$ when $r = 1.125$. Answer $p =$ ... or $p =$ ... Express $r$ in terms of $p$. Answer $r =$ ...

1. Group the first three terms as a perfect square.

$m^2-2mn+n^2-p^2=(m-n)^2-p^2$

2. Factorise using the difference of two squares.

$(m-n)^2-p^2=(m-n-p)(m-n+p)$

1. Recognise the relationship between the two denominators.

$6-4x=-2(2x-3)$

So
$\frac{x+1}{6-4x}=\frac{x+1}{-2(2x-3)}=-\frac{x+1}{2(2x-3)}.$

2. Express both fractions with the same denominator.

$\frac{7}{2x-3}=\frac{14}{2(2x-3)}$

Hence
$\frac{7}{2x-3}+\frac{x+1}{6-4x}=\frac{14}{2(2x-3)}-\frac{x+1}{2(2x-3)}.$

3. Combine the numerators and simplify the single fraction.

$\frac{14}{2(2x-3)}-\frac{x+1}{2(2x-3)}=\frac{14-(x+1)}{2(2x-3)}$

$=\frac{14-x-1}{2(2x-3)}=\frac{13-x}{2(2x-3)}.$

1. Substitute $r=1.125$ and convert the decimal to a fraction.

$1.125=\dfrac{9}{8}$

Given
$p=\sqrt{1+\dfrac{p^2}{r}},$
substitute $r=\dfrac{9}{8}$:
$p=\sqrt{1+\dfrac{p^2}{9/8}}=\sqrt{1+\dfrac{8p^2}{9}}.$

2. Square both sides to remove the square root and form a quadratic equation.

$p^2=1+\dfrac{8p^2}{9}$
$9p^2=9+8p^2$
$p^2=9$

3. Solve for $p$ and check against the original square-root equation.

$p^2=9$
$p=\pm 3$

But $\sqrt{1+\dfrac{p^2}{r}}$ is non-negative, so $p$ must be non-negative.

Hence $p=3.$

1. Square both sides to remove the square root.

$p=\sqrt{1+\dfrac{p^2}{r}}$

$p^2=1+\dfrac{p^2}{r}$

2. Rearrange to isolate the term containing $r$.

$p^2-1=\dfrac{p^2}{r}$

3. Make $r$ the subject.

$r(p^2-1)=p^2$

$r=\dfrac{p^2}{p^2-1}$