Question: Solve $2x^2 - 5x + 1 = 0$ by completing the square, giving answers to 3 significant figures.

Step 1. The method needs the $x^2$ coefficient to be $1$ — divide everything by $2$:

$$x^2 - \dfrac{5}{2}x + \dfrac{1}{2} = 0$$

Step 2. Complete the square with half the $x$-coefficient, $\left(\dfrac{5}{4}\right)$, added and subtracted:

$$\left(x - \dfrac{5}{4}\right)^2 - \left(\dfrac{5}{4}\right)^2 + \dfrac{1}{2} = 0$$ $$\left(x - \dfrac{5}{4}\right)^2 = \dfrac{25}{16} - \dfrac{1}{2} = \dfrac{17}{16}$$

Step 3. Square-root both signs:

$$x - \dfrac{5}{4} = \pm\sqrt{\dfrac{17}{16}}$$

Step 4.

$$x = \dfrac{5}{4} \pm \dfrac{\sqrt{17}}{4} = 2.28 \;\text{ or }\; 0.219 \quad (3 \text{ s.f.})$$

Check. Sum of roots $\approx 2.50 = \frac{5}{2}$ ✓ (matches $-\frac{b}{a}$).

⚠ Watch out: Divide through by $a$ first — completing the square on $2x^2$ directly derails. The added term is $\left(\frac{\text{coefficient of }x}{2}\right)^2$, and the $\pm$ at the square-root step is where the second solution lives; dropping it halves the answer.

Question: Solve $\dfrac{x}{x-3} - \dfrac{5}{x+2} = 4$, giving your answers to 2 decimal places.

Step 1. Combine the left side over the common denominator $(x-3)(x+2)$:

$$\dfrac{x(x+2) - 5(x-3)}{(x-3)(x+2)} = 4 \;\Rightarrow\; \dfrac{x^2 - 3x + 15}{x^2 - x - 6} = 4$$

Step 2. Cross-multiply and bring everything to one side:

$$x^2 - 3x + 15 = 4x^2 - 4x - 24$$ $$3x^2 - x - 39 = 0$$

Step 3. "2 decimal places" signals the formula, with $a = 3$, $b = -1$, $c = -39$:

$$x = \dfrac{1 \pm \sqrt{1 + 468}}{6} = \dfrac{1 \pm \sqrt{469}}{6}$$

Step 4.

$$x = 3.78 \;\text{ or }\; x = -3.44 \quad (2 \text{ d.p.})$$

Check. Neither answer is $3$ or $-2$ (which would zero a denominator), so both are valid.

Every fractional equation follows this exact pipeline: combine, cross-multiply, gather to zero, solve, then screen the roots against the denominators.

⚠ Watch out: The bracket expansion $-5(x-3) = -5x + 15$ — the $+15$ sign flip is the step this question exists to test. And an answer request in decimals is the paper telling you the quadratic won't factorise: don't burn time hunting factors.

Question: A motorist drove $360$ km at an average speed of $x$ km/h, and returned by the same route at a speed $10$ km/h less. The difference between the two times was $30$ minutes. Show that $x^2 - 10x - 7200 = 0$, solve it, and find the return speed.

Step 1. Times, from $\text{time} = \dfrac{\text{distance}}{\text{speed}}$: outward $\dfrac{360}{x}$ h, return $\dfrac{360}{x-10}$ h.

Step 2. The slower trip takes longer, and $30$ min $= \dfrac{1}{2}$ h:

$$\dfrac{360}{x-10} - \dfrac{360}{x} = \dfrac{1}{2}$$

Step 3. Combine and cross-multiply:

$$\dfrac{360x - 360(x-10)}{x(x-10)} = \dfrac{1}{2} \;\Rightarrow\; \dfrac{3600}{x^2 - 10x} = \dfrac{1}{2}$$ $$x^2 - 10x = 7200 \;\Rightarrow\; x^2 - 10x - 7200 = 0 \; \text{(shown)}$$

Step 4. Factorise and interpret:

$$(x - 90)(x + 80) = 0 \;\Rightarrow\; x = 90 \;\text{ or }\; x = -80 \;(\text{rejected — speed can't be negative})$$

Return speed $= x - 10 = 80$ km/h.

⚠ Watch out: Subtract the smaller time from the larger ($\frac{360}{x-10}$ first) and convert the $30$ minutes to hours before equating. Always reject the impossible root with a reason, and answer the actual question — the return speed, $80$, not $90$.

Question: Solve $x^2 - 3x + 2 = 0$ (a) by factorisation, (b) using the quadratic formula.

Step 1. (a) Factorise — two numbers multiplying to $+2$ and adding to $-3$: they are $-1$ and $-2$:

$$(x-1)(x-2) = 0$$

Step 2. A product is zero only when a factor is zero:

$$x - 1 = 0 \;\text{ or }\; x - 2 = 0 \;\Rightarrow\; x = 1 \;\text{ or }\; x = 2$$

Step 3. (b) Formula, with $a = 1$, $b = -3$, $c = 2$:

$$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \dfrac{3 \pm \sqrt{9 - 8}}{2} = \dfrac{3 \pm 1}{2}$$

Step 4.

$$x = 2 \;\text{ or }\; x = 1$$

— the same pair, confirming both methods.

Choosing: try factorisation first (fastest when it works); reach for the formula when the factors won't appear or the question demands decimal answers.

⚠ Watch out: In the formula, $-b$ with $b = -3$ becomes $+3$ — the double negative is the top scoring leak. And the zero-product step needs the equation equal to zero first: factorising $x^2 - 3x = -2$ as it stands proves nothing.