Additional Mathematics 4037/22 — October/November 2016

Walkthrough

The modulus $|4x - 3|$ equals either $4x - 3$ (when $4x - 3 \geq 0$) or $-(4x - 3)$ (when $4x - 3 < 0$). So the equation $|4x - 3| = x$ splits into two linear equations.

Case 1: $4x - 3 = x$. Subtracting $x$ from both sides gives $3x - 3 = 0$, so $x = 1$. Checking: $|4(1) - 3| = 1 = x$. Valid.

Case 2: $4x - 3 = -x$. Adding $x$ to both sides gives $5x - 3 = 0$, so $x = 0.6$. Checking: $|4(0.6) - 3| = |2.4 - 3| = 0.6 = x$. Valid.

Both solutions check out, so the solution set is $x = 1$ and $x = 0.6$. (An alternative accepted route squares both sides: $(4x-3)^2 = x^2$ gives $15x^2 - 24x + 9 = 0$, i.e. $3(x-1)(5x-3) = 0$, yielding the same two roots.)

Key Takeaways

• $|A| = B$ splits into $A = B$ or $A = -B$.
• Each case must be solved and the solutions checked against the original equation.
• Squaring both sides is an alternative route but risks introducing extraneous roots, so checking is essential there.

Common Mistakes

• Using both $4x - 3 = -x$ and $-(4x - 3) = -x$ (the mark scheme explicitly disallows using both forms of the negative case — it is the same equation twice).
• Discarding $x = 0.6$ by assuming the modulus equation has only one solution.
• Sign errors when expanding $-(4x - 3)$.
• In the squaring route, expanding $(4x-3)^2$ incorrectly or failing to factorise the resulting quadratic correctly.

Things to Be Careful About

• The mark scheme awards the first mark for the positive case and a method mark for correctly setting up the negative case — you must show $4x - 3 = -x$ explicitly, not just quote answers.
• Answers must come from correct working (www — without wrong working); quoting both answers without the case equations scores poorly.
• Always substitute each answer back into the original equation to confirm it is not extraneous.