Mathematics 9709/31 — October/November 2010

Walkthrough

We need to solve $2|x-3| > |3x+1|$. Since absolute values are always non-negative, squaring both sides is valid and removes the modulus signs. This turns the modulus inequality into a standard quadratic inequality. After expanding and simplifying, we get $(x+7)(x-1)<0$. A quadratic with positive leading coefficient is negative between its two roots, so the solution is $-7<x<1$. The critical values are where the two sides are equal.

Key Takeaways

• Squaring both sides is a standard way to remove modulus signs when both sides are non-negative.
• Modulus inequalities often reduce to quadratic inequalities.
• The sign of a factorised quadratic determines the interval: positive outside the roots, negative between the roots for a positive leading coefficient.

Common Mistakes

• Forgetting to square the coefficient: $2|x-3|$ must become $4(x-3)^2$, not $2(x-3)^2$.
• Multiplying by $-1$ without reversing the inequality sign.
• Using $\leq$ instead of $<$. The original inequality is strict, so endpoints are excluded.
• Giving only critical values without the final interval.

Things to Be Careful About

• When squaring, both sides must be non-negative; here they are because moduli are non-negative.
• Do not condone $x \leq -7$ or $x \leq 1$; the answer must be strict.
• If using the alternative linear-equation method, solve $2(x-3) = \pm(3x+1)$ to get the critical values, then test an interval.