Surds and indices, quadratic functions and the discriminant, simultaneous equations, inequalities, polynomial division and the factor theorem, and graph transformations.

What this dot point is asking

WJEC wants fluent algebra: simplifying surds and applying the laws of indices, solving and analysing quadratics (including completing the square and the discriminant), solving simultaneous equations and inequalities, dividing polynomials and using the factor theorem, and applying transformations to graphs. This is the toolkit that every later topic relies on, so accuracy here pays off across the whole paper.

The answer

Surds and indices

A surd is an irrational root left in exact form. Simplify using $\sqrt{ab} = \sqrt{a}\,\sqrt{b}$, so $\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$. To rationalise a denominator, multiply numerator and denominator by the conjugate: $\dfrac{1}{\sqrt{a}+\sqrt{b}} \times \dfrac{\sqrt{a}-\sqrt{b}}{\sqrt{a}-\sqrt{b}}$ removes the surd from the bottom via the difference of two squares.

Quadratics and the discriminant

A quadratic $ax^2 + bx + c$ can be solved by factorising, by the formula, or by completing the square, which also reveals the vertex.

Completing the square writes $ax^2+bx+c$ as $a\left(x+\dfrac{b}{2a}\right)^2 + \left(c - \dfrac{b^2}{4a}\right)$, so the minimum (or maximum) point is read off directly.

Simultaneous equations and inequalities

Solve a linear and a quadratic simultaneously by substitution, then solve the resulting quadratic. For inequalities, sketch the quadratic and read off where it is above or below the axis; remember to reverse the inequality when multiplying or dividing by a negative number, and use a number line for the solution set of a quadratic inequality.

Polynomials and the factor theorem

Polynomial long division (or comparing coefficients) divides one polynomial by another. The factor theorem is the quick test for factors.

Graph transformations

Transformations move or stretch a known curve $y = f(x)$:

• $y = f(x) + a$: translation up by $a$.
• $y = f(x + a)$: translation left by $a$ (note the sign).
• $y = a\,f(x)$: vertical stretch, scale factor $a$.
• $y = f(ax)$: horizontal stretch, scale factor $\dfrac{1}{a}$.
• $y = -f(x)$: reflection in the $x$-axis; $y = f(-x)$: reflection in the $y$-axis.

Examples in context

Example 1. Discriminant for a tangent. The line $y = mx + 1$ is a tangent to $y = x^2 + 3$. Setting them equal gives $x^2 - mx + 2 = 0$, and tangency means the discriminant is zero: $m^2 - 8 = 0$, so $m = \pm 2\sqrt{2}$. This links the discriminant directly to coordinate geometry.

Example 2. Factorising a cubic. Factorise $f(x) = x^3 - 4x^2 + x + 6$. Test $x = -1$: $f(-1) = -1 - 4 - 1 + 6 = 0$, so $(x+1)$ is a factor. Dividing gives $x^2 - 5x + 6 = (x-2)(x-3)$, so $f(x) = (x+1)(x-2)(x-3)$. The factor theorem found the first root and division finished the job.

Try this

Q1. Simplify $\sqrt{75} - \sqrt{12}$. [2 marks]

• Cue. $\sqrt{75} = 5\sqrt{3}$ and $\sqrt{12} = 2\sqrt{3}$, so the answer is $3\sqrt{3}$.

Q2. Find the set of values of $x$ for which $x^2 - x - 6 < 0$. [3 marks]

• Cue. Factorise to $(x-3)(x+2) < 0$, so the curve is below the axis between the roots: $-2 < x < 3$.

Q3. Show that $(x - 2)$ is a factor of $x^3 - 3x^2 + 4$. [2 marks]

• Cue. $f(2) = 8 - 12 + 4 = 0$, so by the factor theorem $(x-2)$ is a factor.