Express the equation $\dfrac{x-2}{x-3} + \dfrac{x-4}{x-5} = \dfrac{10}{3}$; $(x \neq 3, 5)$ as a quadratic equation in standard form. Hence, find the roots of the equation so formed.
Generated by claude-sonnet-4-6 · 2026-06-15 10:32 · grounding rag
Model Answer
Step 1: Simplify the given equation
$$\frac{x-2}{x-3} + \frac{x-4}{x-5} = \frac{10}{3}$$
Taking LCM on LHS:
$$\frac{(x-2)(x-5)+(x-4)(x-3)}{(x-3)(x-5)} = \frac{10}{3}$$
$$\frac{(x^2-7x+10)+(x^2-7x+12)}{x^2-8x+15} = \frac{10}{3}$$
$$\frac{2x^2-14x+22}{x^2-8x+15} = \frac{10}{3}$$
Step 2: Cross-multiply
$$3(2x^2-14x+22) = 10(x^2-8x+15)$$
$$6x^2-42x+66 = 10x^2-80x+150$$
$$0 = 4x^2-38x+84$$
$$\boxed{2x^2-19x+42 = 0}$$
This is the required standard form.
Step 3: Find the roots by factorisation
Split $-19x$ as $-12x - 7x$:
$$2x^2-12x-7x+42 = 0$$
$$2x(x-6)-7(x-6) = 0$$
$$(2x-7)(x-6) = 0$$
$$x = \frac{7}{2} \quad \text{or} \quad x = 6$$
The roots are $x = \dfrac{7}{2}$ and $x = 6$.
Source: Chapter 4, Sections 4.2 and 4.3
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Explanation
- What examiners look for: Correct LCM step, proper expansion, cross-multiplication leading to the standard quadratic, then clean factorisation with both roots stated. Each algebraic step earns marks.
- Neither root is 3 or 5, so the domain restriction $x \neq 3, 5$ is satisfied — worth a quick mental check but not always required to write.
- Splitting the middle term is the expected method here since the discriminant is a perfect square ($b^2-4ac = 361-336 = 25$). The quadratic formula also works and is equally acceptable.