The laws of indices and surds, completing the square and the quadratic formula, the discriminant, simultaneous equations, inequalities, and polynomial manipulation including the factor and remainder theorems.

What this dot point is asking

CCEA wants you to manipulate algebraic expressions confidently: apply the laws of indices and surds, solve quadratic equations by every method, use the discriminant to count and classify roots, solve simultaneous equations and inequalities, and divide and factorise polynomials using the factor and remainder theorems. This is the algebraic toolkit every later topic relies on, and it appears throughout AS 1.

The answer

Indices and surds

For example $\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$, and to rationalise $\frac{2}{\sqrt{5} + 1}$ multiply by $\frac{\sqrt{5} - 1}{\sqrt{5} - 1}$ to get $\frac{2(\sqrt{5} - 1)}{4} = \frac{\sqrt{5} - 1}{2}$.

Quadratic equations

A quadratic $ax^2 + bx + c = 0$ can be solved three ways. Factorising finds two numbers multiplying to $ac$ and adding to $b$. Completing the square writes it as $a(x + p)^2 + q$, which also gives the turning point. The quadratic formula

Simultaneous equations and inequalities

To solve a linear and a quadratic simultaneously, substitute the linear equation into the quadratic to get a single quadratic in one variable. A linear inequality is solved like an equation, but the sign reverses when you multiply or divide by a negative. A quadratic inequality such as $x^2 - x - 6 > 0$ is solved by finding the roots ($x = 3$, $x = -2$), sketching the parabola, and reading off where it lies above or below the axis: here $x < -2$ or $x > 3$.

Polynomials, the factor and remainder theorems

These let you factorise a cubic: find one root by trying small values, then divide out the linear factor (by algebraic long division or comparing coefficients) to leave a quadratic.

Worked example: factorising a cubic

Examples in context

Example 1. A bridging algebra error. A common slip is writing $(x + 3)^2 = x^2 + 9$. Completing the square forces you to expand carefully: $(x + 3)^2 = x^2 + 6x + 9$, so $x^2 + 6x + 1 = (x + 3)^2 - 8$. This appears whenever you find a minimum point without calculus.

Example 2. Counting intersections. Asking "for what values of $k$ does $y = x^2 + 3$ meet $y = kx$?" reduces to $x^2 - kx + 3 = 0$, and the discriminant $k^2 - 12$ decides: two points if $k^2 > 12$, a tangent if $k^2 = 12$, none if $k^2 < 12$. The discriminant is the link between algebra and geometry.

Try this

Q1. Simplify $\sqrt{72} + \sqrt{8}$. [2 marks]

• Cue. $6\sqrt{2} + 2\sqrt{2} = 8\sqrt{2}$.

Q2. Solve $2x^2 - 7x + 3 = 0$ by factorising. [2 marks]

• Cue. $(2x - 1)(x - 3) = 0$, so $x = \tfrac{1}{2}$ or $x = 3$.

Q3. Find the set of values of $x$ for which $x^2 - 4x - 5 \le 0$. [3 marks]

• Cue. Roots at $x = 5$ and $x = -1$; the parabola is below the axis between them, so $-1 \le x \le 5$.