Mathematics 9709/21 — October/November 2011

Walkthrough

The inequality is $|4 - 5x| < 3$. Because both $|4 - 5x|$ and $3$ are non-negative, we may square both sides without changing the set of solutions. This is a standard way to remove the modulus sign.

Squaring gives $(4 - 5x)^2 < 9$. Expanding the left-hand side gives $25x^2 - 40x + 16$, so the inequality becomes $25x^2 - 40x + 7 < 0$.

Next we factorise: $25x^2 - 40x + 7 = (5x - 1)(5x - 7)$. The critical values are the values of $x$ that make each factor zero, namely $x = \frac{1}{5}$ and $x = \frac{7}{5}$. For a quadratic with a positive leading coefficient, the expression is negative between its roots. Therefore the solution is $\frac{1}{5} < x < \frac{7}{5}$.

An equivalent way to see the same result is to solve the two linear equations $4 - 5x = 3$ and $4 - 5x = -3$, which give the same critical values.

Key Takeaways

This question tests the ability to solve a modulus inequality. The key idea is that squaring both sides is valid only when both sides are non-negative, and it converts the modulus inequality into a quadratic inequality. It also tests factorising a quadratic and interpreting the sign of a quadratic expression between its roots.

Common Mistakes

• Expanding $(4 - 5x)^2$ incorrectly. The correct expansion is $16 - 40x + 25x^2$; a common error is forgetting the middle term $-40x$.
• Writing the answer as two separate inequalities such as $x > \frac{1}{5}$ and $x < \frac{7}{5}$ without linking them correctly, or choosing the region outside the roots instead of between them.
• Forgetting that the inequality is strict, so the endpoints $x = \frac{1}{5}$ and $x = \frac{7}{5}$ are not included.
• Giving an unsupported final answer. The mark scheme requires the non-modular inequality or critical values to be shown.

Things to Be Careful About

The modulus inequality $|4 - 5x| < 3$ means the distance of $4 - 5x$ from $0$ is less than $3$. After squaring, the inequality sign does not change because both sides are non-negative. When factorising, check the signs carefully: $(5x - 1)(5x - 7)$ expands to $25x^2 - 40x + 7$. Since the leading coefficient is positive, the quadratic is negative between the two roots. Also, because the original inequality is strict, the critical values themselves must not be included in the final interval.