Additional Mathematics 4037/12 — October/November 2020

Walkthrough

The curve and line meet where their equations give the same $y$ for the same $x$, so set $2x^2 + k + 4 = (k+4)x$ and collect everything on one side: $2x^2 - (k+4)x + (k+4) = 0$. This is a quadratic in $x$ whose coefficients depend on $k$.

The question says there are two distinct intersection points, which means this quadratic must have two distinct real roots. The algebraic test for that is a strictly positive discriminant: $b^2 - 4ac > 0$ with $a = 2$, $b = -(k+4)$, $c = k+4$.

Substituting gives $(k+4)^2 - 8(k+4) > 0$. Rather than expanding, factor out the common $(k+4)$: $(k+4)(k-4) > 0$. This quadratic in $k$ crosses zero at $k = -4$ and $k = 4$; since its leading coefficient is positive, it is positive outside these roots, giving $k < -4$ or $k > 4$.

Key Takeaways

• Intersections of a curve and a line are found by equating the expressions and solving.
• 'Two distinct points' translates directly into discriminant $> 0$ (tangency would be $= 0$, no contact $< 0$).
• Solving a quadratic inequality by factorising and reading off the sign pattern is faster than expanding.

Common Mistakes

• Writing the discriminant as $(k+4)^2 - 8(k+4)$ but then solving it as an equation ($= 0$) instead of an inequality — you need the strict inequality for two distinct points.
• Sign errors with $b = -(k+4)$ when squaring or computing $-4ac$.
• Giving the wrong interval direction: $(k+4)(k-4) > 0$ means outside the roots, not between them.
• Forgetting that $k = \pm 4$ themselves are excluded (at those values the line is tangent — only one point).

Things to Be Careful About

• The inequality is strict: $k < -4$ or $k > 4$, not $\leq$ or $\geq$, because 'distinct' rules out tangency.
• Show the rearranged quadratic explicitly — the mark scheme awards B1 for $2x^2 - (k+4)x + (k+4) = 0$ before any discriminant work.
• Factorising $(k+4)^2 - 8(k+4)$ as $(k+4)(k-4)$ avoids messy expansion; if you do expand, check your arithmetic carefully.