derive an equation relating \(x\) and \(k\).

Show that there are no real values of \(x\) if \(k < 5\).

Recall that for a quadratic equation \(ax^2 + bx + c = 0\), the discriminant is the expression \(b^2 - 4ac\), and that the quadratic has no real solutions if and only if \(\Delta < 0\).

The previous part of the question led to the quadratic equation \(15x^2-30x+25=2k\), which we can rewrite as \(15x^2-30x+(25-2k)=0\).

The associated discriminant is then \[\begin{align*} (-30)^2 - 4 \times 15(25-2k) &= 900-60(25-2k)\\ &= -600+120k. \end{align*}\]Therefore there are no real values of \(x\) if and only if \(-600+120k<0\), or \(120k<600\). Dividing by \(120\) gives us \(k<5\) as required.

Geometrically, the two equations represent a fixed straight line and an ellipse of size determined by \(k\). When the ellipse is small enough, the line will not intersect it, as can be seen in this interactivity.