Where the Formulas Come From

From $ax^2 + bx + c = 0$: divide by $a$, move the constant, and complete the square: $\left(x + \dfrac{b}{2a}\right)^2 = \dfrac{b^2 - 4ac}{4a^2}$. Square-root both sides and solve: $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. $\blacksquare$

In plain terms. The formula is not magic — it is completing the square ONCE, in general, so you never have to do it again.

Example. $x^2 - 6x + 5 = 0$: $(x-3)^2 = 4$, so $x = 3 \pm 2$ — the formula gives the same $5$ and $1$.

If $r$ and $s$ are the roots, the quadratic factors as $a(x - r)(x - s) = a\left(x^2 - (r+s)x + rs\right)$. Matching coefficients with $ax^2 + bx + c$: $r + s = -\dfrac{b}{a}$ and $rs = \dfrac{c}{a}$. $\blacksquare$

In plain terms. Multiply the factored form back out — the sum and product of the roots are sitting in the coefficients.

Example. $x^2 - 7x + 12$: roots sum to $7$ and multiply to $12$ ($3$ and $4$), read off with no solving.

Write the series forwards and backwards and add: each column gives $a_1 + a_n$ (what one end gains, the other loses), so $2S_n = n(a_1 + a_n)$. $\blacksquare$

In plain terms. Same Gauss pairing as $1 + \cdots + 100$, just starting anywhere and stepping by anything.

Example. $5 + 8 + 11 + \cdots + 35$ ($11$ terms): $S = \tfrac{11}{2}(5 + 35) = 220$.

Multiply the sum by $r$ and subtract: $S_n - rS_n = a - ar^n$ — every middle term appears once in each copy and cancels. Divide by $1 - r$. $\blacksquare$ For $|r| < 1$, $r^n \to 0$ gives the infinite form $S = \dfrac{a}{1-r}$.

In plain terms. Shifting the whole series one step and subtracting wipes out everything but the two ends.

Example. $1 + \tfrac12 + \tfrac14 + \cdots = \dfrac{1}{1 - \tfrac12} = 2$.

First the case where one chord is a diameter: the triangle from the center is isosceles (two radii), so the central angle, an exterior angle of it, equals the sum of two equal base angles — twice the inscribed angle. Any other position splits (or subtracts) into two such cases. $\blacksquare$

In plain terms. An angle at the rim is always half the angle at the center over the same arc — proved with one isosceles triangle.

Example. An angle inscribed in a semicircle sits on a $180^\circ$ arc, so it is $90^\circ$ — Thales’ theorem for free.

Chords $AB$ and $CD$ meet at $P$. Angles $\angle PAC$ and $\angle PDB$ subtend the SAME arc, so they are equal; likewise $\angle PCA = \angle PBD$. Hence $\triangle PAC \sim \triangle PDB$, and matching sides gives $\dfrac{PA}{PD} = \dfrac{PC}{PB}$, i.e. $PA \cdot PB = PC \cdot PD$. $\blacksquare$

In plain terms. The inscribed angle theorem hands you two similar triangles, and the product formula is just their ratio cross-multiplied.

Example. Pieces $3, 8$ and $4, x$: $3 \cdot 8 = 4x$ gives $x = 6$.

Squares are never negative: $\left(\sqrt a - \sqrt b\right)^2 \ge 0$. Expand: $a - 2\sqrt{ab} + b \ge 0$, i.e. $a + b \ge 2\sqrt{ab}$ — with equality exactly when $\sqrt a = \sqrt b$, i.e. $a = b$. $\blacksquare$

In plain terms. The whole inequality is one squared bracket refusing to be negative. Equality means the bracket was zero: the numbers were equal.

Example. For $a + b = 20$, the product $ab$ is largest at $a = b = 10$: $ab \le \left(\tfrac{20}{2}\right)^2 = 100$.

$1^2 + 2^2 + \cdots + n^2 = \dfrac{n(n+1)(2n+1)}{6}$.

A set with $n$ elements has $2^n$ subsets, and $\sum_{k}\dbinom nk = 2^n$.