Mathematics 9709/22 — October/November 2018

Walkthrough

The key observation is that both $|3x - 5|$ and $2|x|$ are always non-negative, so we may square both sides of the inequality without changing its truth. Squaring a modulus expression removes the absolute value bars. We then expand:

and compare it with $(2x)^2 = 4x^2$. After bringing all terms to one side, we obtain

Dividing by 5 gives the simpler quadratic

which factorises as

The roots $x = 1$ and $x = 5$ split the number line into three intervals. Since the quadratic opens upwards, its value is negative only between the two roots. Therefore the original inequality holds for $1 < x < 5$. The endpoints are not included because the original inequality is strict.

An equally valid alternative method is to consider the two linear equations $3x - 5 = 2x$ and $3x - 5 = -2x$ to locate the boundary points $x = 5$ and $x = 1$, and then check the intervals between them. The final interval is the same.

Key Takeaways

• A modulus inequality of the form $|A| < |B|$ can be squared to give $A^2 < B^2$, provided both sides are non-negative.
• A quadratic inequality of the form $a(x - p)(x - q) < 0$ with $a > 0$ is satisfied for $p < x < q$.
• The critical values are found by solving the corresponding quadratic equation $(x - 1)(x - 5) = 0$.

Common Mistakes

• A common error shown in the mark scheme is squaring the right-hand side incorrectly to get $(3x - 5)^2 < 2x^2$; the correct square is $(2x)^2 = 4x^2$.
• Forgetting to divide the quadratic by 5, which can make factorisation harder and lead to arithmetic mistakes.
• Using $\leq$ instead of $<$: the original inequality is strict, so the endpoints must be excluded.
• Solving the quadratic equation and stopping without determining the sign of the quadratic on each interval.

Things to Be Careful About

• Squaring an inequality is only valid when both sides are non-negative; here that is guaranteed by the modulus signs.
• The direction of the inequality is preserved when squaring both non-negative sides.
• The coefficient of $x^2$ in $x^2 - 6x + 5$ is positive, so the graph is an upward-opening parabola and the inequality holds between the roots, not outside them.
• If using the alternative linear-equation method, you must test values from each interval to determine which intervals satisfy the original modulus inequality.
• The final answer is an open interval; substituting $x = 1$ or $x = 5$ gives equality, which does not satisfy the strict inequality.