Mathematics 9709/23 — May/June 2017

Walkthrough

We need to solve $|x + a| = |2x - 5a|$. Since both sides are non-negative, squaring both sides is a valid first step: it removes the modulus signs without losing solutions.

After squaring, we get

$$(x + a)^2 = (2x - 5a)^2.$$

Expanding gives

$$x^2 + 2ax + a^2 = 4x^2 - 20ax + 25a^2.$$

Bringing everything to one side:

$$3x^2 - 22ax + 24a^2 = 0.$$

This quadratic factorises as

$$(3x - 4a)(x - 6a) = 0.$$

Setting each factor to zero gives the two solutions

$$x = \frac{4a}{3} \quad \text{and} \quad x = 6a.$$

Alternatively, we could have split into the two linear cases:

$$x + a = 2x - 5a \quad \Rightarrow \quad x = 6a,$$

and

$$x + a = -(2x - 5a) \quad \Rightarrow \quad 3x = 4a \quad \Rightarrow \quad x = \frac{4a}{3}.$$

Both methods are acceptable; the mark scheme gives the first mark for stating a non-modulus equation, the second for attempting to solve it, and the third for the two correct final answers.

Key Takeaways

• To solve $|A| = |B|$, you can square both sides because both sides are non-negative.
• Squaring leads to a quadratic equation, which can be solved by factorisation.
• Alternatively, split into the two cases $A = B$ and $A = -B$.
• When a constant parameter such as $a$ appears, the answers are expressed in terms of that parameter.

Common Mistakes

• Forgetting the second case when using the linear-equation method: only writing $x + a = 2x - 5a$ and missing $x + a = -(2x - 5a)$.
• Making sign errors when expanding $(2x - 5a)^2$.
• Incorrectly factorising the quadratic, especially the coefficient of $x^2$.
• Giving only one solution instead of two.

Things to Be Careful About

• Since $a$ is stated to be positive, both $\frac{4a}{3}$ and $6a$ are valid and distinct. No extra checking is required.
• When squaring, no extraneous solutions arise here because both original sides are non-negative.
• Be careful not to divide by $a$ prematurely; factorising the quadratic is safer and keeps both solutions visible.