Solving quadratics by factorising, using the quadratic formula, completing the square at Higher tier, and interpreting the roots and turning point.

What this dot point is asking

AQA wants you to solve quadratic equations by three methods (factorising, the quadratic formula, and at Higher tier completing the square) and to read the roots as the points where the curve $y = ax^2 + bx + c$ crosses the $x$-axis. Quadratics are one of the highest-value Higher topics: they reappear in graph sketching, simultaneous equations, area problems and rates of change, so fluency here pays off across the whole paper.

Solving by factorising

Factorising is the fastest method when it works, and it is the expected method on non-calculator papers. The key fact is the zero-product principle: if two things multiply to give zero, at least one of them must be zero.

For a monic quadratic $x^2 + bx + c$ (where $a = 1$), find two numbers that multiply to $c$ and add to $b$. To solve $x^2 + 5x + 6 = 0$, you need two numbers multiplying to $6$ and adding to $5$: those are $2$ and $3$, so $(x + 2)(x + 3) = 0$ and $x = -2$ or $x = -3$.

When $a \ne 1$ you split the middle term. For $6x^2 + 11x - 10 = 0$, multiply $a$ by $c$ to get $-60$, then find two numbers multiplying to $-60$ and adding to $11$: those are $15$ and $-4$. Rewrite as $6x^2 + 15x - 4x - 10 = 0$, factor by grouping into $3x(2x + 5) - 2(2x + 5) = 0$, so $(3x - 2)(2x + 5) = 0$ and $x = \tfrac{2}{3}$ or $x = -\tfrac{5}{2}$.

The quadratic formula

When a quadratic does not factorise over the integers, the formula always works and is the standard tool on calculator papers.

The quantity $b^2 - 4ac$ is the discriminant. If it is positive there are two distinct real roots; if it is zero there is one repeated root (the curve just touches the $x$-axis); if it is negative there are no real roots (the curve never crosses the $x$-axis). Examiners often hide a discriminant test inside a "how many solutions" question.

Completing the square at Higher tier

Completing the square is the Higher-tier method that both solves the equation and exposes the turning point of the parabola.

For example $x^2 + 6x + 5 = (x + 3)^2 - 9 + 5 = (x + 3)^2 - 4$, which has a minimum at $(-3, -4)$. To solve $x^2 + 6x + 5 = 0$ you then write $(x + 3)^2 = 4$, take square roots to get $x + 3 = \pm 2$, and finish with $x = -1$ or $x = -5$.

When $a \ne 1$, factor out $a$ first: $2x^2 + 8x + 3 = 2\left[(x + 2)^2 - 4\right] + 3 = 2(x + 2)^2 - 5$, minimum at $(-2, -5)$.

Interpreting roots and the turning point

The roots are the $x$-intercepts of $y = ax^2 + bx + c$. A positive $a$ gives a U-shaped curve with a minimum; a negative $a$ gives an n-shaped curve with a maximum. The line of symmetry passes through the turning point at $x = -\tfrac{b}{2a}$, which sits exactly halfway between the two roots. Completing the square is the cleanest route to the turning point because the bracket gives the $x$-coordinate and the constant gives the $y$-coordinate directly.