Solve quadratic equations for real roots using factorisation, the quadratic formula or completing the square

1. Functions – Core Concepts (Syllabus 1.1‑1.8)

1.1 Definition and Basic Terminology

• Function \$f\$: a rule that assigns to each element \$x\$ in a set called the domain a unique element \$f(x)\$ in a set called the range.
• Domain – all admissible input values.
• Range – all possible output values.
• One‑to‑one (injective) – different inputs give different outputs.
• Many‑to‑one – two or more inputs can give the same output.
• Inverse function \$f^{-1}\$ – exists only when \$f\$ is one‑to‑one; it swaps the roles of domain and range.
• Composite function \$f\circ g\$ – defined by \$(f\circ g)(x)=f\bigl(g(x)\bigr)\$; the order matters.

1.2 When a Function Does Not Have an Inverse

A function fails to be one‑to‑one when at least two different \$x\$‑values give the same \$y\$‑value. Example:

\[

f(x)=x^{2},\qquad \text{Domain } \mathbb R.

\]

Both \$x=2\$ and \$x=-2\$ give \$f(x)=4\$, so \$f\$ is not one‑to‑one and \$f^{-1}\$ does not exist on \$\mathbb R\$. Restricting the domain to \$x\ge0\$ (or \$x\le0\$) makes \$f\$ one‑to‑one, and then \$f^{-1}(x)=\sqrt{x}\$ (or \$-\sqrt{x}\$).

1.3 Composite Functions – Order Matters

In general \$f\circ g\neq g\circ f\$. Example:

\[

f(x)=2x-1,\qquad g(x)=\sqrt{x+3}.

\]

\[

(f\circ g)(x)=2\sqrt{x+3}-1,\qquad (g\circ f)(x)=\sqrt{2x-1+3}=\sqrt{2x+2}.

\]

The two composites have different domains and different expressions, illustrating that the order of composition is essential.

1.4 Notation Box

1.5 Simple Worked Examples

1. Linear function: \$f(x)=2x+3\$

    Domain \$=\mathbb R\$, Range \$=\mathbb R\$, one‑to‑one.

    Inverse: \$f^{-1}(x)=\dfrac{x-3}{2}\$.

2. Square‑root function: \$g(x)=\sqrt{x-2}\$

    Domain \$x\ge2\$, Range \$y\ge0\$.

    Inverse: \$g^{-1}(x)=x^{2}+2\$ (valid for \$x\ge0\$).

3. Composition: \$h(x)=f\bigl(g(x)\bigr)=2\sqrt{x-2}+3\$.

1.6 Graphical Insight

The graph of a function and its inverse are mirror images in the line \$y=x\$. A sketch of \$y=2x+3\$ together with its inverse \$y=\frac{x-3}{2}\$ illustrates this symmetry.

2. Quadratic Functions – Form, Discriminant & Inequalities (Syllabus 2.1‑2.5)

2.1 Standard Form

A quadratic equation is written as

\[

ax^{2}+bx+c=0\qquad(a\neq0),

\]

where \$a,b,c\in\mathbb R\$.

2.2 Discriminant

The discriminant \$D=b^{2}-4ac\$ determines the nature of the roots:

2.3 Vertex Form & Extrema

Completing the square rewrites the quadratic as

\[

a\bigl(x-h\bigr)^{2}+k=0\quad\text{with}\quad

h=-\frac{b}{2a},\;k=\frac{4ac-b^{2}}{4a}.

\]

The point \$(h,k)\$ is the vertex. If \$a>0\$, \$k\$ is a minimum; if \$a<0\$, \$k\$ is a maximum. The sign of \$D\$ tells whether the vertex lies above, on, or below the \$x\$‑axis.

2.4 Solving Quadratics – Three Standard Methods

2.4.1 Factorisation

Best used when the quadratic can be expressed as a product of two linear factors with integer or simple rational coefficients.

1. Write the equation in standard form.
2. Find two numbers that multiply to \$ac\$ and add to \$b\$.
3. Split the middle term, factor by grouping, and obtain \$(px+q)(rx+s)=0\$.
4. Set each factor to zero and solve for \$x\$.

2.4.2 Quadratic Formula

Applicable to any quadratic.

\[

x=\frac{-b\pm\sqrt{D}}{2a},\qquad D=b^{2}-4ac.

\]

2.4.3 Completing the Square

Useful for deriving the vertex form, solving when \$a\neq1\$, or when factorisation is inconvenient.

1. If \$a\neq1\$, divide the whole equation by \$a\$.
2. Move the constant term to the right‑hand side.
3. Add \$\bigl(\frac{b}{2a}\bigr)^{2}\$ to both sides.
4. Factor the left side as a perfect square.
5. Take square roots and solve for \$x\$.

2.5 Quadratic Inequalities

Two common approaches:

• Graphical method: Sketch the parabola \$y=ax^{2}+bx+c\$ and read off the \$x\$‑intervals where the curve lies above ( \$>\$ ) or below ( \$<\$ ) the \$x\$‑axis.
• Algebraic method: Solve the corresponding equation \$ax^{2}+bx+c=0\$ to obtain the critical points, then test a point in each interval to determine the sign.

Example: Solve \$x^{2}-5x+6\ge0\$.

1. Factorise: \$(x-2)(x-3)=0\;\Rightarrow\;x=2,\;3\$.
2. Number line test gives \$x\le2\$ or \$x\ge3\$.

3. Factors of Polynomials – Remainder & Factor Theorems (Syllabus 3.1‑3.3)

3.1 Remainder Theorem

If a polynomial \$P(x)\$ is divided by \$(x-a)\$, the remainder is \$P(a)\$.

Example: \$P(x)=x^{3}-4x^{2}+x+6\$, find the remainder on division by \$(x-2)\$.

\[

P(2)=2^{3}-4\cdot2^{2}+2+6=8-16+2+6=0,

\]

so the remainder is \$0\$; consequently \$(x-2)\$ is a factor.

3.2 Factor Theorem

If \$P(a)=0\$, then \$(x-a)\$ is a factor of \$P(x)\$. Repeated use allows us to factor higher‑degree polynomials.

3.3 Synthetic Division (Quick Division by a Linear Factor)

For \$P(x)\$ divided by \$(x-a)\$:

1. Write the coefficients of \$P(x)\$.
2. Bring down the leading coefficient.
3. Multiply by \$a\$, add to the next coefficient, and repeat.
4. The final number is the remainder; the row of numbers (except the remainder) gives the coefficients of the quotient.

Worked example: Factor \$x^{3}-6x^{2}+11x-6\$.

1. Test \$x=1\$: \$P(1)=0\;\Rightarrow\;(x-1)\$ is a factor.
2. Synthetic division by \$a=1\$ yields coefficients \$1,\; -5,\; 6\$ → quotient \$x^{2}-5x+6\$.
3. Factor the quadratic: \$(x-2)(x-3)\$. Hence

\[

x^{3}-6x^{2}+11x-6=(x-1)(x-2)(x-3).

\]

4. Equations, Inequalities & Graphs (Syllabus 4.1‑4.5)

4.1 Absolute‑Value Equations & Inequalities

Four standard forms (with \$a>0\$):

• \$|x|=k\$