Additional Mathematics 4037/21 — May/June 2019

Walkthrough

First expand and move everything to one side: $x(6x+7) \geq 20$ becomes $6x^2 + 7x - 20 \geq 0$. This puts the inequality in standard form with a positive leading coefficient, which tells us the parabola opens upwards, so it is non-negative outside its two roots.

Next we find those roots (the critical values) by factorising. We need two numbers multiplying to $6 \times (-20) = -120$ and adding to $7$: these are $15$ and $-8$. Splitting the middle term gives $6x^2 + 15x - 8x - 20 = 3x(2x+5) - 4(2x+5) = (3x-4)(2x+5)$.

Setting each factor to zero gives the critical values $x = \frac{4}{3}$ and $x = -\frac{5}{2}$. Because the parabola opens upwards, $y \geq 0$ for $x$ at or beyond either root — the two 'outside' regions. The inequality is $\geq$, so the endpoints themselves are included.

Key Takeaways

• Always rearrange a quadratic inequality to have zero on one side before solving.
• For a positive-coefficient quadratic, solutions of $ax^2+bx+c \geq 0$ lie in the two outside regions of the critical values; solutions of $\leq 0$ lie between them.
• Include endpoints when the inequality is strict in the inclusive sense ($\geq$ or $\leq$).

Common Mistakes

• Writing the answer as $-\frac{5}{2} \leq x \leq \frac{4}{3}$ — that is the region where the quadratic is negative, the opposite of what is required.
• Omitting the equality signs: since the original inequality is $\geq$, the critical values must be included.
• Sign errors when expanding $x(6x+7)$ or moving the 20 across.
• Factorising incorrectly (e.g. $(6x-5)(x+4)$ does not give the right middle term).

Things to Be Careful About

The mark scheme awards M1 for any correct rearrangement (any inequality sign or even an equation), A1 for both critical values $-\frac{5}{2}$ and $\frac{4}{3}$, and A1 for the final answer using the correct outside regions — follow-through applies only if your critical values are right. Give exact fractions rather than decimals, and remember both branches joined by 'or' are needed for full marks.