Quadratics and exponents

Quadratics: get to zero, then factorise

A quadratic equation has a squared term. The reliable method is to move everything to one side so it equals zero, then factorise:

$$x^2 - 3x = 28 \;\Rightarrow\; x^2 - 3x - 28 = 0 \;\Rightarrow\; (x - 7)(x + 4) = 0$$

A product is zero only when one of its factors is zero, so $x = 7$ or $x = -4$. That zero-product rule is why you must reach zero first: $(x - 7)(x + 4) = 10$ tells you nothing about either factor on its own.

To factorise $x^2 + bx + c$, look for two numbers that multiply to $c$ and add to $b$. For $x^2 - 3x - 28$ you want a product of $-28$ and a sum of $-3$: that is $-7$ and $4$. When the leading coefficient is not 1, divide it out if you can, or test factor pairs.

Sum and product of the roots

For $x^2 + bx + c = 0$, the two roots add to $-b$ and multiply to $c$. With $x^2 - 3x - 28 = 0$ the roots add to 3 and multiply to $-28$, which matches $7 + (-4)$ and $7 \times (-4)$. Questions that ask for the sum or product of the solutions can be answered without finding them.

Never divide by a variable

From $x^2 = 5x$ it is tempting to divide by $x$ and write $x = 5$. That loses a solution, because $x$ might be zero. Instead write $x^2 - 5x = 0$, factorise as $x(x - 5) = 0$, and keep both $x = 0$ and $x = 5$.

Three special products

These come up constantly, in both directions:

Product	Expanded
$(a + b)^2$	$a^2 + 2ab + b^2$
$(a - b)^2$	$a^2 - 2ab + b^2$
$(a + b)(a - b)$	$a^2 - b^2$

The middle term $2ab$ is the one people drop. Used backwards, the difference of squares turns arithmetic into a shortcut: $51 \times 49 = (50 + 1)(50 - 1) = 2500 - 1 = 2499$. The squares also connect sums and products: $a^2 + b^2 = (a + b)^2 - 2ab$.

Exponent rules

Rule	Example
$a^m \cdot a^n = a^{m+n}$	$2^3 \cdot 2^4 = 2^7$
$\frac{a^m}{a^n} = a^{m-n}$	$\frac{5^6}{5^2} = 5^4$
$(a^m)^n = a^{mn}$	$(3^2)^3 = 3^6$
$(ab)^n = a^n b^n$	$6^3 = 2^3 \cdot 3^3$
$a^0 = 1$ for $a \neq 0$	$7^0 = 1$
$a^{-n} = \frac{1}{a^n}$	$2^{-3} = \frac{1}{8}$
$a^{1/n} = \sqrt[n]{a}$	$27^{1/3} = 3$

The rules work only for the same base (to multiply) or the same exponent (to combine bases). There is no rule for $a^m + a^n$: you cannot add exponents across a plus sign.

Equations with powers

When both sides can be written as powers of one base, set the exponents equal. Rewrite 4, 8, 16 and 32 as powers of 2, and 9, 27 and 81 as powers of 3:

$$4^{x} = 8^{x-1} \;\Rightarrow\; 2^{2x} = 2^{3x - 3} \;\Rightarrow\; 2x = 3x - 3 \;\Rightarrow\; x = 3$$

When powers of the same base are added or subtracted, factor out the smallest one: $3^{x+2} - 3^{x} = 3^{x}(3^2 - 1) = 8 \cdot 3^{x}$. That turns a sum you cannot simplify into a product you can.

Signs and even powers

An even power is never negative, and it hides the sign: $(-3)^2 = 3^2 = 9$. So $x^2 = 9$ has two solutions, $3$ and $-3$, while $x^3 = -27$ has one, $-3$. The symbol $\sqrt{9}$ means the non-negative root, 3, which is why $\sqrt{x^2} = |x|$ rather than $x$.