**Translate the condition into algebra.** The line and the parabola intersect where their $y$-values are equal. "Does not intersect" means that equation has **no real solutions**, which requires $$\Delta < 0$$ **Set the two expressions equal:** $$x^2 + 3 = mx - 2$$ **Rearrange to standard form:** $$x^2 - mx + 5 = 0$$ **Identify the coefficients:** $$a = 1, \qquad b = -m, \qquad c = 5$$ **Write the discriminant:** $$\Delta = (-m)^2 - 4(1)(5) = m^2 - 20$$ Note that $(-m)^2 = m^2$ — the sign disappears on squaring, which is why the final answer will be symmetric about zero. **Apply the condition:** $$m^2 - 20 < 0$$ $$m^2 < 20$$ **Solve the inequality.** A square is less than 20 exactly when the quantity lies **between** $-\sqrt{20}$ and $\sqrt{20}$: $$-\sqrt{20} < m < \sqrt{20}$$ **Simplify the surd.** $20 = 4 \times 5$, so $\sqrt{20} = 2\sqrt{5}$: $$-2\sqrt{5} < m < 2\sqrt{5}$$ **Verify at three points.** With $2\sqrt{5} \approx 4.472$: - **$m = 0$** (inside the range): the line $y = -2$ and the parabola $y = x^2 + 3$, whose minimum is 3. Since $-2 < 3$, the horizontal line passes below the entire parabola and never meets it ✓ - **$m = 5$** (outside): $\Delta = 25 - 20 = 5 > 0$, so there are two intersection points ✓ correctly excluded - **$m = 2\sqrt{5}$** (the boundary): $\Delta = 20 - 20 = 0$, so the line is **tangent** — it touches at exactly one point. Since touching counts as intersecting, the boundary must be **excluded**, which is why the inequality is strict ✓ **The answer is $-2\sqrt{5} < m < 2\sqrt{5}$.** **Interpreting the result geometrically.** All the lines $y = mx - 2$ pass through the fixed point $(0, -2)$, which sits **below** the parabola's vertex at $(0, 3)$. Rotating a line about that point, there is a range of shallow gradients for which it slips underneath the parabola entirely, bounded by the two **tangent** lines at $m = \pm 2\sqrt{5}$. Beyond those gradients the line is steep enough to cut through. The symmetry of the answer about $m = 0$ reflects the symmetry of the parabola about the $y$-axis: a line of gradient $m$ and one of gradient $-m$ are mirror images, so they must behave identically.

Algebraic Methods — practice questions

Manipulating Expressions

Expanding and factorising

Rational expressions

Exponents, roots and surds

Equations & Their Roots

Linear equations and rearranging formulae

Solving quadratic equations

The discriminant and the nature of the roots

Exponential Equations & Logs

Logarithms and the log laws

Solving exponential equations

Forming & Using Models

Forming equations from a context

Growth, decay and compound interest