Additional Mathematics 4037/12 — October/November 2015

Walkthrough

The equation $kx^2 + k = 8x - 2xk$ is not yet recognisable as a quadratic, so the first job is to collect every term on one side. Moving $8x$ and $-2xk$ across gives $kx^2 + 2kx - 8x + k = 0$, which groups neatly as $kx^2 + (2k - 8)x + k = 0$. Now we can read off $a = k$, $b = 2k - 8$ and $c = k$.

A quadratic has two real distinct roots exactly when its discriminant is strictly positive: $b^2 - 4ac > 0$. Substituting gives $(2k - 8)^2 - 4k \\cdot k > 0$. Expanding $(2k-8)^2 = 4k^2 - 32k + 64$ and subtracting $4k^2$ leaves $-32k + 64 > 0$, i.e. $32k < 64$, so $k < 2$.

Key Takeaways

• Before using the discriminant, always rewrite the equation in the standard form $ax^2 + bx + c = 0$; here the $x$ terms had to be gathered first.
• Two real distinct roots means $b^2 - 4ac > 0$ (strictly greater — equal would give repeated roots).
• When the coefficients contain an unknown parameter, the discriminant often collapses to a simple linear inequality.

Common Mistakes

• Sign errors when moving terms: writing $b = 8 - 2k$ instead of $2k - 8$ still works if squared carefully, but mixing signs mid-expansion loses accuracy marks.
• Forgetting that "2 real distinct roots" demands strict inequality $>$, not $\\geq$.
• Dividing $-32k + 64 > 0$ by $-32$ without flipping the inequality sign, giving $k > 2$ — the mark scheme explicitly requires the correct sign.
• Expanding $(2k-8)^2$ incorrectly as $4k^2 - 64$ or $2k^2 - 32k + 64$.

Things to Be Careful About

• The final answer must be $k < 2$ with the correct direction of the inequality; the A1 mark is lost if the sign is wrong.
• Show every algebraic step between the discriminant condition and the answer — the mark scheme awards separate marks for forming the three-term quadratic, substituting into $b^2 - 4ac$, simplifying, and solving.