Solving Equations and Inequalities

A linear equation contains the unknown to the power 1 only. The golden rule is: whatever you do to one side, do to the other.

One-step and two-step equations:

$3x + 7 = 22 \implies 3x = 15 \implies x = 5$

Unknown on both sides — collect all $x$ terms on one side first:

$5x - 3 = 2x + 9 \implies 3x = 12 \implies x = 4$

Equations with brackets — expand first:

$4(2x - 1) = 3(x + 5) \implies 8x - 4 = 3x + 15 \implies 5x = 19 \implies x = 3.8$

Equations with fractions — multiply through by the LCM of the denominators:

$\frac{x + 1}{3} = \frac{2x - 1}{5}$

Multiply both sides by 15: $5(x + 1) = 3(2x - 1) \implies 5x + 5 = 6x - 3 \implies x = 8$

A quadratic equation has the form $ax^2 + bx + c = 0$. AQA expects three algebraic methods and one graphical method.

Method 1 — Factorising (fastest when it works):

$x^2 + 5x + 6 = 0 \implies (x + 2)(x + 3) = 0 \implies x = -2 \text{ or } x = -3$

For $ax^2 + bx + c$ with $a \neq 1$, find two numbers that multiply to $ac$ and add to $b$, then split the middle term.

Method 2 — Completing the square:

$x^2 + 6x + 7 = 0$ $(x + 3)^2 - 9 + 7 = 0$ $(x + 3)^2 = 2$ $x = -3 \pm \sqrt{2}$

For $ax^2 + bx + c$: first divide through by $a$, complete the square, then solve.

Method 3 — Quadratic formula (always works):

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

The discriminant $\Delta = b^2 - 4ac$ tells you the number of roots:

• $\Delta > 0$: two distinct real roots
• $\Delta = 0$: one repeated root (the curve just touches the $x$-axis)
• $\Delta < 0$: no real roots

Graphical method: draw $y = ax^2 + bx + c$ and read off where it crosses the $x$-axis. Results are approximate.

Simultaneous equations are two equations that must both be satisfied at the same time.

Elimination method (two linear equations):

$2x + 3y = 12 \quad (1)$ $5x - 3y = 9 \quad (2)$

Add (1) and (2): $7x = 21 \implies x = 3$. Substitute back: $y = 2$.

If coefficients don't match, multiply one or both equations to make them equal before adding or subtracting.

Substitution method (especially for one linear, one quadratic):

$y = 2x - 1 \quad (1)$ $x^2 + y^2 = 10 \quad (2)$

Substitute (1) into (2): $x^2 + (2x-1)^2 = 10 \implies 5x^2 - 4x - 9 = 0 \implies (x+1)(5x-9) = 0$

So $x = -1, y = -3$ or $x = \frac{9}{5}, y = \frac{13}{5}$.

Graphical method: plot both equations; the coordinates of the intersection point(s) are the solution(s). This gives approximate answers only.

Iteration is a numerical method for finding approximate roots of equations that are difficult to solve algebraically. An iterative formula has the form:

$x_{n+1} = f(x_n)$

Starting from an initial estimate $x_0$, you substitute repeatedly to obtain a sequence $x_1, x_2, x_3, \ldots$ that converges to a root.

Example: show that $x^3 + 2x - 5 = 0$ has a root between 1 and 2, then use the iterative formula $x_{n+1} = \sqrt[3]{5 - 2x_n}$ to find the root to 3 decimal places.

$n$	$x_n$
0	1.5
1	$\sqrt[3]{5 - 3} = \sqrt[3]{2} \approx 1.2599$
2	$\sqrt[3]{5 - 2(1.2599)} \approx 1.3200$
3	$\approx 1.3016$
4	$\approx 1.3076$
5	$\approx 1.3056$

The root converges to $x \approx 1.306$ (3 d.p.).

Showing a root exists in an interval: evaluate $f(a)$ and $f(b)$. If the signs differ (one positive, one negative), there is a root between $a$ and $b$ by the change-of-sign rule. State: "Since $f(1) < 0$ and $f(2) > 0$ there is a root in the interval $(1, 2)$."

An inequality is solved like an equation with one crucial difference: multiplying or dividing by a negative number reverses the inequality sign.

Single inequality in one variable:

$3x - 4 < 11 \implies 3x < 15 \implies x < 5$

$-2x + 1 \geq 7 \implies -2x \geq 6 \implies x \leq -3 \quad (\text{sign flips when dividing by } -2)$

Compound inequality — solve both parts simultaneously:

$-3 \leq 2x + 1 < 9 \implies -4 \leq 2x < 8 \implies -2 \leq x < 4$

Number line representation:

• Open circle $\circ$ for strict inequality ($<$ or $>$)
• Closed/filled circle $\bullet$ for inclusive inequality ($\leq$ or $\geq$)

Set notation:

• $\{x : x < 5\}$ — the set of all $x$ such that $x < 5$
• $\{x : -2 \leq x < 4\}$ — the set of $x$ in the compound inequality above

Integer solutions: if asked for integer values satisfying $-2 \leq x < 4$, list: $-2, -1, 0, 1, 2, 3$.

To solve a quadratic inequality, first solve the corresponding equation to find the critical values (roots), then use the shape of the parabola to determine which intervals satisfy the inequality.

Steps:

1. Rearrange so one side is 0.
2. Factorise (or use the formula) to find the roots $\alpha$ and $\beta$ ($\alpha < \beta$).
3. Sketch (or visualise) the parabola.
4. Read off the solution:
   • If $a > 0$ (opens upward): $ax^2 + bx + c < 0 \implies \alpha < x < \beta$ (between the roots)
   • If $a > 0$: $ax^2 + bx + c > 0 \implies x < \alpha$ or $x > \beta$ (outside the roots)

Example:

$x^2 - 5x + 6 < 0$

Roots: $(x-2)(x-3) = 0 \implies x = 2$ or $x = 3$.

The parabola opens upward, so it is below the $x$-axis between the roots:

$\text{Solution: } 2 < x < 3 \quad \text{i.e. } \{x : 2 < x < 3\}$

Example 2:

$x^2 - x - 6 \geq 0 \implies (x-3)(x+2) \geq 0$

Roots $x = -2, x = 3$; parabola opens upward ⟹ at or above $x$-axis outside the roots:

$\text{Solution: } x \leq -2 \text{ or } x \geq 3 \quad \text{i.e. } \{x : x \leq -2\} \cup \{x : x \geq 3\}$