Mathematics 9709/21 — October/November 2013

Walkthrough

We begin with the modulus inequality $|x+1| < |3x+5|$. Since absolute values are never negative, squaring both sides is a valid operation that preserves the inequality. This removes the modulus signs and gives $(x+1)^2 < (3x+5)^2$.

Expand both sides: $x^2 + 2x + 1 < 9x^2 + 30x + 25$. Bring all terms to the left to obtain $-8x^2 - 28x - 24 < 0$. Multiplying by $-1$ reverses the inequality, giving $8x^2 + 28x + 24 > 0$, and dividing by $4$ gives $2x^2 + 7x + 6 > 0$.

Factorise the quadratic: $(2x+3)(x+2) > 0$. The critical values are $x = -2$ and $x = -\frac{3}{2}$. Because the coefficient of $x^2$ is positive, the quadratic is positive outside the interval between its roots. Therefore the solution is $x < -2$ or $x > -\frac{3}{2}$.

A quick check: at $x = 0$, $|1| < |5|$ is true, so $x > -\frac{3}{2}$ is plausible. At $x = -1.75$, which lies between $-2$ and $-\frac{3}{2}$, $| -0.75 | < | -0.25 |$ is false, so the interval between the critical values is not part of the solution.

Key Takeaways

The main idea is that for non-negative expressions, squaring both sides of a modulus inequality removes the absolute value signs. The problem then becomes a quadratic inequality. To solve a quadratic inequality, find the roots (critical values) and use the sign of the leading coefficient to determine which intervals satisfy the inequality.

Common Mistakes

• Expanding incorrectly, especially the square $(3x+5)^2$ as $9x^2 + 25$ instead of $9x^2 + 30x + 25$.
• Forgetting to reverse the inequality sign when multiplying by a negative number.
• Confusing the order of the critical values and writing the solution as an interval such as $-\frac{3}{2} < x < -2$, which is impossible.
• Stating only the critical values without giving the final inequality.
• Omitting the quadratic step; the mark scheme requires a method mark for the non-modular inequality and a solution attempt.

Things to Be Careful About

The critical values are $x = -2$ and $x = -\frac{3}{2}$. Since the quadratic has a positive leading coefficient, the inequality $>0$ is satisfied outside the roots, not between them. If you multiply or divide by a negative number at any stage, remember to reverse the inequality. Also, because both sides of the original inequality are absolute values, squaring is safe; with other inequalities you must check that both sides are non-negative before squaring.