Question. Without a calculator, evaluate $64812^2 - 64822 \times 64802$.

Solution.

Observe: $64822 = 64812 + 10$ and $64802 = 64812 - 10$. Let $x = 64812$, $a = 10$.

$$64812^2 - 64822 \times 64802 = x^2 - (x + a)(x - a)$$

Expand the product: $(x + a)(x - a) = x^2 - a^2$.

$$= x^2 - (x^2 - a^2) = a^2 = 10^2 = 100$$

Answer: $100$.

Lesson: when two large numbers are $a$ above and $a$ below a third, their product is $x^2 - a^2$. Subtracting from $x^2$ leaves only $a^2$.

Question. Given $(x + y)^2 = 30$ and $xy = 5$, find $\left(\dfrac{x - y}{4}\right)^2$.

Solution.

Step 1 — Link $(x-y)^2$ to known quantities.

$$(x + y)^2 - (x - y)^2 = 4xy$$

So:

$$(x - y)^2 = (x + y)^2 - 4xy = 30 - 4(5) = 30 - 20 = 10$$

Step 2 — Apply the outer transformation:

$$\left(\frac{x - y}{4}\right)^2 = \frac{(x - y)^2}{16} = \frac{10}{16} = \frac{5}{8}$$

Answer: $\dfrac{5}{8}$.

Lesson: never try to find $x$ and $y$ separately. Use the identity $4xy = (a+b)^2 - (a-b)^2$ to jump straight from what you have to what you need.

Question. Given $x^2 + \dfrac{9}{x^2} = 30$ and $x > 0$, find $x + \dfrac{3}{x}$.

Solution.

Step 1 — Set up the identity. Squaring $x + \dfrac{3}{x}$:

$$\left(x + \frac{3}{x}\right)^2 = x^2 + 2(x)\left(\frac{3}{x}\right) + \frac{9}{x^2} = x^2 + 6 + \frac{9}{x^2}$$

Step 2 — Substitute. Given $x^2 + \dfrac{9}{x^2} = 30$:

$$\left(x + \frac{3}{x}\right)^2 = 30 + 6 = 36$$

Step 3 — Take the root: $x + \dfrac{3}{x} = \pm 6$.

Step 4 — Use the sign constraint. Since $x > 0$, $\dfrac{3}{x} > 0$, so $x + \dfrac{3}{x} > 0$. Take the positive root.

Answer: $x + \dfrac{3}{x} = 6$.

Lesson: never solve for $x$ directly. Square the target, substitute the given, then take $\pm$ root with care.

Question. Factorise $8y^2 z - 18z + 4x^2 y^2 - 9x^2$ completely.

Solution — 2-by-2 grouping.

Step 1 — Group:

• Pair 1: $8y^2 z - 18z = 2z(4y^2 - 9)$.
• Pair 2: $4x^2 y^2 - 9x^2 = x^2(4y^2 - 9)$.

Step 2 — Pull out common bracket $(4y^2 - 9)$:

$$= (4y^2 - 9)(2z + x^2)$$

Step 3 — Check each factor. $4y^2 - 9 = (2y)^2 - 3^2$ is a DOS:

$$= (2y - 3)(2y + 3)(2z + x^2)$$

Answer: $(2y - 3)(2y + 3)(2z + x^2)$.

Lesson: after grouping, always re-check whether the leftover bracket is itself a DOS or a perfect square.

Question. Show that $(7n + 2)^2 - (7n - 1)^2$ is a multiple of 3 for all integer $n$.

Solution — DOS first, then factor.

This is a difference of squares with $a = 7n + 2$, $b = 7n - 1$:

$$(7n + 2)^2 - (7n - 1)^2 = [(7n + 2) - (7n - 1)] \cdot [(7n + 2) + (7n - 1)]$$ $$= (3)(14n + 1)$$ $$= 3(14n + 1)$$

Since $n$ is an integer, $14n + 1$ is an integer, so $3(14n + 1)$ is a multiple of 3.

$$\therefore (7n + 2)^2 - (7n - 1)^2 \text{ is always a multiple of 3.} \quad \checkmark$$

Lesson: DOS turns a messy difference into a product. If one of the factors is a constant (here, 3), divisibility falls out for free.

Question. Simplify $5a^2 - (3a - 2b)^2$.

Solution.

Step 1 — Expand the square:

$$(3a - 2b)^2 = 9a^2 - 12ab + 4b^2$$

Step 2 — Substitute back:

$$5a^2 - (9a^2 - 12ab + 4b^2)$$

Step 3 — Distribute the minus (bracket the whole thing!):

$$= 5a^2 - 9a^2 + 12ab - 4b^2$$ $$= -4a^2 + 12ab - 4b^2$$

Step 4 — Common factor $-4$:

$$= -4(a^2 - 3ab + b^2)$$

Answer: $-4(a^2 - 3ab + b^2)$.

Lesson: when subtracting a binomial square, bracket the whole expanded form before distributing the minus. Every sign flips.