Mathematics 9709/23 — May/June 2018

Walkthrough

We want all values of $x$ for which the distance-like expression $|3x - 2|$ is smaller than $|x + 5|$. Because absolute values are always non-negative, squaring both sides does not change the direction of the inequality. This is the key step: it lets us replace the awkward modulus signs with ordinary algebraic expressions.

After squaring, we expand:

Subtracting the right-hand side from the left gives:

Next we solve the equality $8x^2 - 22x - 21 = 0$ to find the boundary points. The factorisation is:

so the boundary points are $x = -\frac{3}{4}$ and $x = \frac{7}{2}$.

Because the quadratic has a positive leading coefficient, its graph is a U-shape. A U-shaped quadratic is below zero only between its two roots. Therefore the solution is the open interval:

A quick check: for $x = 0$, $|3(0)-2| = 2$ and $|0+5| = 5$, and indeed $2 < 5$, so $0$ lies in the interval. For $x = 2$, $|6-2| = 4$ and $|2+5| = 7$, and $4 < 7$, so $2$ also lies in the interval. For $x = 4$, $|12-2| = 10$ and $|4+5| = 9$, and $10 < 9$ is false, so values beyond $\frac{7}{2}$ are not in the solution.

Key Takeaways

• Squaring both sides is a valid way to remove modulus signs because both sides are non-negative.
• Solving the corresponding equality gives the critical values where the two expressions are equal.
• For a quadratic inequality with a positive leading coefficient, the expression is negative between the roots.
• The final answer must be written as an open interval because the original inequality is strict.

Common Mistakes

• Forgetting to square both sides correctly, especially the middle term $2ab$ when expanding $(3x - 2)^2$.
• Reversing the inequality sign when moving terms; moving all terms to one side should keep the direction unchanged.
• Using closed endpoints $\leq$ or $\geq$ when the original inequality is strict.
• Stating the interval outside-in, for example $x < -\frac{3}{4}$ or $x > \frac{7}{2}$, instead of between the roots.

Things to Be Careful About

• The critical values must be found exactly: $-\frac{3}{4}$ and $\frac{7}{2}$.
• Check the factorisation: $(4x + 3)(2x - 7)$ expands to $8x^2 - 22x - 21$.
• Since the inequality is strict, both endpoints are excluded.
• If using the alternative linear-equation method, solve both $3x - 2 = x + 5$ and $3x - 2 = -(x + 5)$ to obtain the two critical values.