Secondary 3 A Math · Quadratic Functions

Discriminant: the range of k for two distinct real roots

The question

The equation

x^2 + (k + 2)x + (2k + 1) = 0

has two distinct real roots. Find the range of values of $k$ .

(4 marks)

The solution, line by line

01
Comparing with $ax^2 + bx + c = 0$ : $\; a = 1$ , $\; b = k + 2$ , $\; c = 2k + 1$ .
Why this earns the mark
Write the coefficients down before touching the discriminant. Nearly every lost mark here starts with $b = k$ instead of $b = k + 2$ — the bracket is part of the coefficient.
02M1
Two distinct real roots $\;\Rightarrow\; b^2 - 4ac > 0$
Why this earns the mark
This is the mark for translating the words into a condition. "Two distinct real roots" is strictly greater than zero — $\geq$ would allow the repeated root the question has excluded. State the condition explicitly; an examiner cannot award this from working alone.
03M1
$(k + 2)^2 - 4(1)(2k + 1) > 0$
Why this earns the mark
Substitution done in one clean line, brackets intact. Keep the $4(1)(2k+1)$ unexpanded for a moment — expanding and substituting in the same breath is where sign slips happen.
04
$k^2 + 4k + 4 - 8k - 4 > 0$ $k^2 - 4k > 0$
Why this earns the mark
Expand and simplify. Note $-4(2k + 1) = -8k - 4$ : the $4$ must be distributed across both terms. The constants cancel, which is a good sign you have expanded $(k+2)^2$ correctly.
05M1
$k(k - 4) > 0$
Why this earns the mark
Factorise rather than reaching for the quadratic formula. A quadratic inequality is solved from its roots, so getting to factored form is the method — here the critical values are $k = 0$ and $k = 4$ .
06A1
The curve $y = k(k - 4)$ is a $\cup$ -shaped parabola cutting the axis at $0$ and $4$ , so it is above the axis outside the roots:
$k < 0 \quad \text{or} \quad k > 4$
Why this earns the mark
The accuracy mark. Sketch the parabola — positive $k^2$ coefficient means $\cup$ , and $> 0$ means the parts above the axis, which are the two outer branches. Writing $0 < k < 4$ (the inside) is the single most common error in this question, and $k < 0$ or $k > 4$ must be joined by or, never by a chain like $4 < k < 0$ , which describes nothing.

Final answer

k < 0 \quad \text{or} \quad k > 4

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