Mathematics 9709/12 — February/March 2023

Walkthrough

At an intersection of the line and the curve, the $y$-values must be equal. Therefore set the two equations equal to one another:

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

Bring every term to the left-hand side:

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

Combining the $x$-terms gives a quadratic in $x$:

$$x^2 - (k+3)x + (2+2k) = 0$$

The line and curve meet if this equation has at least one real solution for $x$. For a quadratic $ax^2 + bx + c = 0$, real solutions exist exactly when the discriminant $b^2 - 4ac$ is non-negative. Here $a = 1$, $b = -(k+3)$ and $c = 2+2k$.

Compute the discriminant:

$$b^2 - 4ac = (k+3)^2 - 4(1)(2+2k)$$ $$= (k+3)^2 - 8(1+k)$$ $$= k^2 + 6k + 9 - 8 - 8k$$ $$= k^2 - 2k + 1$$ $$= (k-1)^2$$

Because every square is non-negative, $(k-1)^2 \geq 0$ for all real $k$. Hence the quadratic always has real roots, so the line and the curve always meet. In the special case $k = 1$, the discriminant is $0$, so the line is tangent to the curve, but they still meet.

Key Takeaways

• The intersection of a line and a quadratic curve is equivalent to solving simultaneous equations by substituting the two expressions for $y$.
• The discriminant is the most direct way to prove whether a quadratic equation has real roots.
• A perfect square such as $(k-1)^2$ is automatically non-negative, which is exactly what is needed to show that the discriminant is non-negative for every parameter value.

Common Mistakes

• After rearrangement, a common sign error is writing $b$ as $k+3$ instead of $-(k+3)$. The correct coefficient of $x$ is $-(k+3)$.
• Be careful with the constant term: $2+2k$, so $4ac = 8(1+k)$, not $4+4k$ or $8+4k$.
• Do not write the discriminant as being equal to $0$ prematurely while simplifying it. The mark scheme says to ignore an equals $0$ sign at the discriminant stage.
• A conclusion is required: you must state that $(k-1)^2 \geq 0$, so the discriminant is non-negative and the graphs meet for all $k$.

Things to Be Careful About

• The discriminant is calculated for the quadratic equation in $x$, but it is expressed in terms of the parameter $k$.
• The condition for the line and curve to meet is discriminant $\geq 0$, not necessarily $> 0$. A zero discriminant means the line touches the curve, and they still meet.
• The discriminant must be evaluated explicitly; it is not enough to quote the quadratic formula and leave the discriminant hidden inside it.
• The question asks only to show that they meet, so there is no need to solve for the actual $x$-coordinates of intersection.