When a Line Misses a Curve: Getting the Range of k the Right Way Round

The question

The straight line $y = kx - 2k + 3$ , where $k$ is a constant, does not meet the curve $y = x^2$ .

Find the range of values of $k$ . [4 marks]

The common wrong answer

Most students set up the algebra correctly. Where the line and curve meet, the $y$ -values are equal:

x^2 = kx - 2k + 3 \quad\Rightarrow\quad x^2 - kx + 2k - 3 = 0

The line doesn’t meet the curve, so this equation has no real roots, and $b^2 - 4ac < 0$ with $a = 1$ , $b = -k$ , $c = 2k - 3$ :

\begin{aligned} (-k)^2 - 4(1)(2k - 3) &< 0 \\ k^2 - 8k + 12 &< 0 \\ (k - 2)(k - 6) &< 0 \end{aligned}

So far, so good. The critical values are $k = 2$ and $k = 6$ . Then the final line goes wrong:

k < 2 \quad\text{or}\quad k > 6

This is the region where $(k - 2)(k - 6)$ is positive, the opposite of what the inequality asks for. A quick test shows it can’t be right: $k = 7$ is in this “answer”, but with $k = 7$ the equation becomes $x^2 - 7x + 11 = 0$ , whose discriminant is $49 - 44 = 5 > 0$ . The line $y = 7x - 11$ cuts the curve twice.

It usually happens when the region is chosen from memory instead of from a sketch. Cambridge’s examiners have noted that candidates who link a quadratic inequality to a sketch of its graph are more often successful.

The correct working

Equate the line and the curve and rearrange so one side is zero:

x^2 - kx + 2k - 3 = 0

The line does not meet the curve, so this quadratic has no real roots:

b^2 - 4ac < 0

Substitute $a = 1$ , $b = -k$ , $c = 2k - 3$ :

\begin{aligned} k^2 - 4(2k - 3) &< 0 \\ k^2 - 8k + 12 &< 0 \\ (k - 2)(k - 6) &< 0 \end{aligned}

The critical values are $k = 2$ and $k = 6$ . Sketch $y = (k - 2)(k - 6)$ against $k$ : a U-shaped curve crossing the axis at $2$ and $6$ .

Sketch of the discriminant k² − 8k + 12 = (k − 2)(k − 6) against k. A U-shaped curve crosses the k-axis at k = 2 and k = 6, with its lowest point at (4, −4). The part of the curve below the axis, between k = 2 and k = 6, is highlighted as the region where the discriminant is negative. — The discriminant (k − 2)(k − 6) is negative only between the critical values 2 and 6.

We need the part below the axis, which is between the critical values, so $2 < k < 6$ .

Check with a value inside: $k = 4$ gives $x^2 - 4x + 5 = 0$ , with discriminant $16 - 20 = -4 < 0$ , so no real roots and the line misses the curve. ✓

Why it works

The discriminant $b^2 - 4ac$ counts the roots of the quadratic, and each real root is an $x$ -coordinate where the line meets the curve. Two roots means the line cuts the curve twice, zero discriminant means it touches once (a tangent), and a negative discriminant means it never meets the curve at all.

The inequality $(k - 2)(k - 6) < 0$ says a product of two brackets is negative, so one bracket must be positive and the other negative. That only happens when $k$ is bigger than $2$ but smaller than $6$ . Outside that interval both brackets have the same sign and the product is positive.

There’s a nice picture behind the numbers, too. Every line $y = k(x - 2) + 3$ passes through the point $(2, 3)$ , which is just below the curve $y = x^2$ . When $k = 2$ the line touches the curve at $(1, 1)$ , and when $k = 6$ it touches at $(3, 9)$ . Any gradient in between swings the line through the gap without touching the curve.

How to spot it next time

Always sketch the quadratic in $k$ through its critical values. For " $< 0$ " take the part below the axis; for " $> 0$ " take the parts above.
Test one value of $k$ from your answer in the original condition. If $k = 7$ makes the line cut the curve, the answer containing $k = 7$ is wrong.
Use strict inequalities: at $k = 2$ and $k = 6$ the line is a tangent, so it does meet the curve.