Mathematics 9709/23 — October/November 2013

Walkthrough

We want to solve $|x+1| < |3x+5|$. The modulus of a number is its distance from zero, so both sides are non-negative. When both sides of an inequality are non-negative, squaring preserves the inequality. This turns the modulus signs into ordinary brackets: $(x+1)^2 < (3x+5)^2$.

Now expand the brackets:

$$(x+1)^2 = x^2 + 2x + 1, \qquad (3x+5)^2 = 9x^2 + 30x + 25$$

So the inequality becomes:

$$x^2 + 2x + 1 < 9x^2 + 30x + 25$$

Move all terms to the right-hand side:

$$0 < 8x^2 + 28x + 24$$

Dividing by the positive number 4 keeps the inequality direction the same:

$$2x^2 + 7x + 6 > 0$$

Factorise the quadratic:

$$(2x+3)(x+2) > 0$$

The critical values are where the product is zero, $x = -2$ and $x = -\frac{3}{2}$. Because the quadratic has a positive leading coefficient, its graph is a U-shape, so it is positive outside the interval between the roots. Therefore the solution is $x < -2$ or $x > -\frac{3}{2}$.

Key Takeaways

• Squaring both sides is a reliable way to remove modulus signs when both sides are non-negative.
• The critical values of a modulus inequality are found by solving the corresponding equality.
• For a quadratic with a positive leading coefficient, the expression is positive outside the interval between its roots.
• A strict inequality means the critical values themselves are not included in the solution.

Common Mistakes

• Expanding $(3x+5)^2$ incorrectly, for example writing $3x^2 + 25$.
• Forgetting to subtract all terms from one side before factorising.
• Reversing the interval: for $2x^2 + 7x + 6 > 0$, the solution is outside the roots, not between them.
• Including $x = -2$ or $x = -\frac{3}{2}$ in the answer, even though the original inequality is strict.

Things to Be Careful About

• Squaring is valid here because both sides are absolute values, hence non-negative; the inequality direction is unchanged.
• Dividing by 4 is safe because 4 is positive.
• The critical values are ordered as $-2 < -\frac{3}{2}$, so the final answer must be written as $x < -2$ or $x > -\frac{3}{2}$.
• If using a sign table or test points, check at least one value in each interval to confirm the direction.