Quadratic equations and inequalities

Factorising: the fastest route

A quadratic equation $ax^2 + bx + c = 0$ is easiest when it factorises, because a product is zero only when one of its factors is zero. For $x^2 + bx + c$, look for two numbers that multiply to $c$ and add to $b$. For $x^2 - 5x - 14$: the pair $-7$ and $2$ multiplies to $-14$ and adds to $-5$, so

$$x^2 - 5x - 14 = (x - 7)(x + 2) = 0 \quad\Longrightarrow\quad x = 7 \text{ or } x = -2.$$

Check one root: $49 - 35 - 14 = 0$.

When the leading coefficient is not 1, try factor pairs of the first and last coefficients and check the middle term: $2x^2 - 7x + 3 = (2x - 1)(x - 3)$, since $2x \times (-3) + (-1) \times x = -7x$.

Two patterns are worth recognising on sight:

$$a^2 - b^2 = (a - b)(a + b), \qquad a^2 \pm 2ab + b^2 = (a \pm b)^2.$$

They also help with arithmetic: $49 \times 51 = (50 - 1)(50 + 1) = 2500 - 1 = 2499$.

The formula and the discriminant

If no factor pair works, use

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$$

The discriminant $D = b^2 - 4ac$ tells you how many real roots there are before you finish: two if $D > 0$, one (a repeated root) if $D = 0$, none if $D < 0$. A question that only asks "how many solutions" needs only $D$.

Two more facts save time. The roots of $ax^2 + bx + c = 0$ add to $-\frac{b}{a}$ and multiply to $\frac{c}{a}$. And never divide both sides by $x$: from $x^2 = 6x$, write $x^2 - 6x = 0$, so $x(x - 6) = 0$, giving $x = 0$ or $x = 6$. Dividing by $x$ silently throws away $x = 0$.

Linear inequalities

Solve a linear inequality like an equation, with one extra rule: multiplying or dividing both sides by a negative number reverses the sign. The reason is that negating swaps the order of numbers: $2 < 5$, but $-2 > -5$.

$$7 - 3x > 19 \;\Longrightarrow\; -3x > 12 \;\Longrightarrow\; x < -4.$$

Test a value: $x = -5$ gives $7 + 15 = 22 > 19$, which fits, while $x = 0$ gives $7 > 19$, which does not. Adding or subtracting never reverses the sign; only multiplying or dividing by a negative does.

When two conditions must hold together, solve each and keep the overlap. If an inequality is strict ($<$ or $>$), the boundary value itself is excluded; check endpoints whenever a question counts integers.

Quadratic inequalities

To solve $x^2 - 2x - 15 \le 0$:

1. Find the roots of the matching equation: $(x - 5)(x + 3) = 0$, so $x = -3$ and $x = 5$.
2. Picture the graph. With a positive $x^2$ coefficient, the parabola opens upwards: it is below zero between the roots and above zero outside them.
3. Read off the answer: $\le 0$ means between, so $-3 \le x \le 5$.

Form (leading coefficient positive, roots $r_1 < r_2$)	Solution
$(x - r_1)(x - r_2) < 0$	$r_1 < x < r_2$
$(x - r_1)(x - r_2) > 0$	$x < r_1$ or $x > r_2$

Test one value to confirm: $x = 0$ gives $-15 \le 0$, true, and 0 is between the roots. If the leading coefficient is negative, multiply the whole inequality by $-1$ first, reversing the sign.

Squares and square roots

$x^2 = 9$ has two solutions, $x = 3$ and $x = -3$, and $x^2 < 9$ means $-3 < x < 3$, not just $x < 3$. A square is never negative, so $x^2 \ge 0$ for every $x$ and an equation like $x^2 = -4$ has no real solution.