Solve quadratic equations by factorising, by the quadratic formula and by completing the square, and form and solve quadratics from problems (Higher tier for formula and completing the square).

What this dot point is asking

Quadratic equations contain an $x^2$ term and are a major Higher-tier CCEA Algebra topic. You must solve them by factorising, by the quadratic formula, and by completing the square, and form and solve quadratics from worded or geometric problems. A quadratic usually has two solutions (roots), and choosing the most efficient method, then giving both roots, is the key exam skill. These connect directly to quadratic graphs and to the linear-and-quadratic simultaneous case.

Solving by factorising

If a quadratic factorises, this is the quickest method. First rearrange so one side is zero, then factorise into two brackets. Because a product equals zero only when one of its factors is zero, set each bracket to zero and solve.

For $x^2 + 2x - 15 = 0$, find two numbers multiplying to $-15$ and adding to $+2$: these are $+5$ and $-3$, so $(x + 5)(x - 3) = 0$, giving $x = -5$ or $x = 3$. When the coefficient of $x^2$ is not 1, you may need to split the middle term or test bracket combinations.

The quadratic formula

When a quadratic does not factorise neatly, the formula always works.

Substitute $a$, $b$ and $c$ carefully, keeping signs. The $\pm$ gives the two roots from the same calculation. Round only at the end, to the accuracy asked for.

Completing the square

Completing the square rewrites $x^2 + bx + c$ as $(x + \tfrac{b}{2})^2 - (\tfrac{b}{2})^2 + c$. It both solves the equation and reveals the minimum point of the matching graph.

Forming quadratics from problems

Many marks come from setting up a quadratic from a context, such as an area or a product of consecutive numbers, then solving it and rejecting any answer that does not fit (for example a negative length). Define the unknown, write the equation, rearrange to equal zero, solve, and interpret.

For example, a rectangle is 3 cm longer than it is wide and has an area of 40 cm squared. Let the width be $x$, so the length is $x + 3$ and the area is $x(x + 3) = 40$. Expanding gives $x^2 + 3x - 40 = 0$, which factorises to $(x + 8)(x - 5) = 0$, so $x = -8$ or $x = 5$. A width cannot be negative, so the width is 5 cm and the length is 8 cm. The step of rejecting the impossible root is exactly the AO3 interpretation CCEA rewards, so always state which solution you keep and why. The same logic applies to consecutive-number and right-angled-triangle problems, where the negative or non-integer root is discarded once the context is considered.

Why this matters

Quadratics model situations where a quantity rises and falls, such as projectile height, area and optimisation, and they are central to the Higher-tier Algebra strand. The three methods connect to graphs (roots are where the curve meets the $x$-axis, and completing the square gives the turning point) and to simultaneous equations. CCEA rewards choosing the efficient method and presenting both roots clearly.