CIE AS/A Level Maths-1.1 Quadratics- Study Notes- New Syllabus - 2026-2027
CIE AS/A Level Maths-1.1 Quadratics- Study Notes- New Syllabus
Ace AS/A Level Maths Exam with CIE AS/A Level Maths-1.1 Quadratics- Study Notes
Key Concepts:
- Completing the Square for a Quadratic Polynomial
- The Discriminant of a Quadratic Polynomial
- Solving Quadratic Equations and Inequalities in One Unknown
- Solving a Linear–Quadratic Pair by Substitution
- Recognising & Solving Equations Quadratic in a Function of \( x \)
Completing the Square for a Quadratic Polynomial
Completing the Square for a Quadratic Polynomial
A quadratic expression is of the form \( ax^2 + bx + c \), where \( a \neq 0 \).
Completing the square is a technique used to rewrite this quadratic into a form that makes it easier to analyze and solve equations, find the vertex of a parabola, or determine minimum/maximum values.
Step 1: Factor out the coefficient of \( x^2 \) (if \( a \neq 1 \))
Start with the quadratic polynomial:
\( ax^2 + bx + c \)
Factor out \( a \) from the first two terms:
\( a \big( x^2 + \dfrac{b}{a}x \big) + c \)
Step 2: Complete the square inside the brackets
Take the coefficient of \( x \), which is \( \dfrac{b}{a} \), divide it by 2, and square it:
\( \left( \dfrac{b}{2a} \right)^2 = \dfrac{b^2}{4a^2} \)
Add and subtract this value inside the bracket to keep the expression balanced:
\( a \left( x^2 + \dfrac{b}{a}x + \dfrac{b^2}{4a^2} – \dfrac{b^2}{4a^2} \right) + c \)
Step 3: Rewrite as a perfect square
The first three terms form a perfect square trinomial:
\( a \left( \bigg(x + \dfrac{b}{2a}\bigg)^2 – \dfrac{b^2}{4a^2} \right) + c \)
Step 4: Simplify
Expand the brackets:
\( a \bigg(x + \dfrac{b}{2a}\bigg)^2 – a \cdot \dfrac{b^2}{4a^2} + c \)
\( = a \bigg(x + \dfrac{b}{2a}\bigg)^2 – \dfrac{b^2}{4a} + c \)
\( = a \bigg(x + \dfrac{b}{2a}\bigg)^2 + \bigg(c – \dfrac{b^2}{4a}\bigg) \)
Final Completed Square Form:
\( ax^2 + bx + c = a \bigg(x + \dfrac{b}{2a}\bigg)^2 + \Big(c – \dfrac{b^2}{4a}\Big) \)
Uses of the Completed Square Form
1. Finding the Vertex of a Parabola
The quadratic in completed square form is:
\( a(x + \dfrac{b}{2a})^2 + \bigg(c – \dfrac{b^2}{4a}\bigg) \)
The vertex (turning point) of the parabola is:
\( \Big( -\dfrac{b}{2a}, \; c – \dfrac{b^2}{4a} \Big) \)
If \( a > 0 \), this point is a minimum. If \( a < 0 \), this point is a maximum.
2. Solving Quadratic Equations
Setting the quadratic equal to zero and solving from completed square form makes it easier to apply square roots.
3. Analyzing the Range of a Quadratic Function
The completed square form immediately shows the minimum (or maximum) value of the quadratic function.
Example:
Rewrite \( 2x^2 + 8x + 5 \) in completed square form.
▶️ Answer/Explanation
\( 2x^2 + 8x + 5 = 2(x^2 + 4x) + 5 \)
Half of 4 is 2, square it: \( 4 \).
\( 2(x^2 + 4x + 4 – 4) + 5 \)
\( = 2((x+2)^2 – 4) + 5 \)
\( = 2(x+2)^2 – 8 + 5 \)
\( = 2(x+2)^2 – 3 \)
Answer: \( 2(x+2)^2 – 3 \), with vertex at \( (-2, -3) \).
Example:
Use the completed square form of \( x^2 + 6x + 5 \) to:
(a) Solve \( x^2 + 6x + 5 = 0 \)
(b) Find the minimum value of the quadratic function.
▶️ Answer/Explanation
Step 1: Rewrite in completed square form
\( x^2 + 6x + 5 = (x^2 + 6x + 9) – 9 + 5 \)
\( = (x+3)^2 – 4 \)
So: \( x^2 + 6x + 5 = (x+3)^2 – 4 \).
Part (a): Solve equation
\( (x+3)^2 – 4 = 0 \)
\( (x+3)^2 = 4 \)
\( x+3 = \pm 2 \)
\( x = -3 + 2 = -1 \) or \( x = -3 – 2 = -5 \).
Solutions: \( x = -1, -5 \).
Part (b): Minimum value
The quadratic is \( (x+3)^2 – 4 \).
The term \( (x+3)^2 \geq 0 \) for all real \( x \).
So the minimum value occurs when \( (x+3)^2 = 0 \), giving minimum \( -4 \).
It occurs at \( x = -3 \).
Final Answer:
(a) \( x = -1, -5 \)
(b) Minimum value = \( -4 \) at \( x = -3 \).
Example:
Find the vertex (turning point) of the parabola \( y = 2x^2 – 8x + 3 \) using the completed square form.
▶️ Answer/Explanation
\( y = 2x^2 – 8x + 3 \)
\( = 2(x^2 – 4x) + 3 \)
\( = 2(x^2 – 4x + 4 – 4) + 3 \)
\( = 2((x-2)^2 – 4) + 3 \)
\( = 2(x-2)^2 – 8 + 3 \)
\( = 2(x-2)^2 – 5 \)
Completed square form: \( y = 2(x-2)^2 – 5 \)
The vertex is at \( (2, -5) \).
Since \( a = 2 > 0 \), this is a minimum point.
Example:
Locate the vertex of the graph of \( y = x^2 – 4x + 3 \) and sketch the graph.
▶️ Answer/Explanation
Step 1: Rewrite in completed square form
\( y = x^2 – 4x + 3 \)
\( = (x^2 – 4x + 4) – 4 + 3 \)
\( = (x-2)^2 – 1 \)
So: \( y = (x-2)^2 – 1 \).
Step 2: Identify the vertex
The vertex is at \( (2, -1) \).
Since \( a = 1 > 0 \), the parabola opens upwards and the vertex is the minimum point.
Step 3: Sketching the graph
- Axis of symmetry: vertical line \( x = 2 \).
- Y-intercept: when \( x=0 \), \( y = 0^2 – 4(0) + 3 = 3 \). So the graph crosses at \( (0, 3) \).
- X-intercepts: solve \( y = 0 \):
\( (x-2)^2 – 1 = 0 \)
\( (x-2)^2 = 1 \)
\( x-2 = \pm 1 \)
\( x = 1 \) or \( x = 3 \)
- So the parabola crosses the x-axis at \( (1, 0) \) and \( (3, 0) \).
Final Answer: The vertex is at \( (2, -1) \). The parabola opens upwards, has axis of symmetry \( x=2 \), crosses the x-axis at \( (1,0) \) and \( (3,0) \), and the y-axis at \( (0,3) \).
The Discriminant of a Quadratic Polynomial
The Discriminant of a Quadratic Polynomial
A quadratic polynomial is of the form:
\( ax^2 + bx + c \), where \( a \neq 0 \).
Its roots are given by the quadratic formula:
\( x = \dfrac{-b \pm \sqrt{b^2 – 4ac}}{2a} \).
The expression under the square root is called the discriminant:
\( \Delta = b^2 – 4ac \).
Uses of the Discriminant
| Value of \( \Delta \) | Nature of Roots | Graph of Parabola |
|---|---|---|
| \( \Delta > 0 \) | Two distinct real roots | Cuts the x-axis at two points |
| \( \Delta = 0 \) | One repeated real root | Touches the x-axis at one point |
| \( \Delta < 0 \) | No real roots (complex roots) | Does not intersect the x-axis |
Example:
Determine the nature of the roots of \( x^2 – 5x + 6 \), and solve the quadratic equation.
▶️ Answer/Explanation
\( a = 1, \; b = -5, \; c = 6 \).
\( \Delta = b^2 – 4ac \)
\( = (-5)^2 – 4(1)(6) \)
\( = 25 – 24 = 1 \).
\( \Delta > 0 \), so there are two distinct real roots.
\( x = \dfrac{-b \pm \sqrt{\Delta}}{2a} \)
\( = \dfrac{5 \pm \sqrt{1}}{2} \)
\( = \dfrac{5 \pm 1}{2} \)
\( x = 2 \) or \( x = 3 \).
Final Answer: Two distinct real roots at \( x = 2 \) and \( x = 3 \).
Example :
Determine the nature of the roots of \( x^2 – 4x + 4 \), and solve the quadratic equation.
▶️ Answer/Explanation
\( a = 1, \; b = -4, \; c = 4 \).
\( \Delta = b^2 – 4ac \)
\( = (-4)^2 – 4(1)(4) \)
\( = 16 – 16 = 0 \).
\( \Delta = 0 \), so there is one repeated real root.
\( x = \dfrac{-b}{2a} = \dfrac{4}{2} = 2 \).
Final Answer: One repeated root at \( x = 2 \).
Example:
Determine the nature of the roots of \( 2x^2 – 4x + 3 \), and state whether the parabola intersects the x-axis.
▶️ Answer/Explanation
\( a = 2, \; b = -4, \; c = 3 \).
\( \Delta = b^2 – 4ac \)
\( = (-4)^2 – 4(2)(3) \)
\( = 16 – 24 = -8 \).
\( \Delta < 0 \), so there are no real roots (the roots are complex).
The parabola does not intersect the x-axis.
Final Answer: No real roots. The graph lies entirely above the x-axis (since \( a = 2 > 0 \)).
Example:
Find the values of \( k \) for which the quadratic equation \( x^2 + (k+1)x + k = 0 \) has real roots.
▶️ Answer/Explanation
\( a = 1, \; b = (k+1), \; c = k \).
For real roots: \( \Delta \geq 0 \).
\( \Delta = b^2 – 4ac \)
\( = (k+1)^2 – 4(1)(k) \)
\( = (k^2 + 2k + 1) – 4k \)
\( = k^2 – 2k + 1 \).
\( \Delta = (k – 1)^2 \geq 0 \).
This inequality is always true for all real \( k \), since a square is never negative.
Final Answer: The quadratic has real roots for all real values of \( k \). If \( k = 1 \), the discriminant is 0, giving a repeated root.
Solving Quadratic Equations and Inequalities in One Unknown
Solving Quadratic Equations and Inequalities in One Unknown
A quadratic has the form \( ax^2 + bx + c = 0 \) with \( a \neq 0 \). Three core methods are used in A-Level: factorising, completing the square, and the quadratic formula. For inequalities, interpret the sign of the quadratic using its roots and the parabola shape.
Method 1: Factorising
Write \( ax^2 + bx + c \) as a product of two linear factors if possible. Then use the zero product rule.
Method 2: Completing the Square
Rewrite \( ax^2 + bx + c \) as \( a\Big(x + \dfrac{b}{2a}\Big)^2 + \Big(c – \dfrac{b^2}{4a}\Big) \).
Solve by isolating the square and taking square roots. This also reveals the vertex for sketching and sign analysis.
Method 3: Quadratic Formula
If factorising is difficult, use \( \displaystyle x = \dfrac{-b \pm \sqrt{b^2 – 4ac}}{2a} \). The discriminant \( \Delta = b^2 – 4ac \) indicates the nature of the roots.
Inequalities Strategy
- 1. Solve the related equation \( ax^2 + bx + c = 0 \) to find critical points.
- 2. Use a sign chart or the parabola shape. If \( a > 0 \) the parabola opens up. If \( a < 0 \) it opens down.
- 3. For \( \lt 0 \) choose intervals where the graph is below the x-axis. For \( \leq 0 \) include the roots. For \( \gt 0 \) choose intervals where it is above the x-axis. For \( \geq 0 \) include the roots.
Example (Equation by factorising):
Solve \( x^2 – 5x + 6 = 0 \) by factorising.
▶️ Answer/Explanation
Find numbers with sum \( -5 \) and product \( 6 \). They are \( -2 \) and \( -3 \).
\( x^2 – 5x + 6 = (x – 2)(x – 3) \).
Set factors to zero: \( x – 2 = 0 \) or \( x – 3 = 0 \).
Answer: \( x = 2 \) or \( x = 3 \).
Example (Equation by completing the square):
Solve \( x^2 – 4x – 5 = 0 \) by completing the square.
▶️ Answer/Explanation
\( x^2 – 4x – 5 = (x^2 – 4x + 4) – 4 – 5 = (x – 2)^2 – 9 \).
\( (x – 2)^2 = 9 \Rightarrow x – 2 = \pm 3 \).
Answer: \( x = 5 \) or \( x = -1 \).
Example (Equation by quadratic formula):
Solve \( 2x^2 + 3x – 7 = 0 \) using the formula.
▶️ Answer/Explanation
\( a = 2, \; b = 3, \; c = -7 \).
\( \Delta = b^2 – 4ac = 9 – 4(2)(-7) = 9 + 56 = 65 \).
\( \displaystyle x = \dfrac{-3 \pm \sqrt{65}}{4} \).
Answer: \( \displaystyle x = \dfrac{-3 + \sqrt{65}}{4}, \; \dfrac{-3 – \sqrt{65}}{4} \).
Example (Inequality, factorisable, \( a > 0 \)):
Solve \( x^2 – 5x + 6 \leq 0 \).
▶️ Answer/Explanation
Factorise: \( x^2 – 5x + 6 = (x – 2)(x – 3) \).
Roots are \( x = 2 \) and \( x = 3 \). Parabola opens up.
For \( \leq 0 \) take the portion on or below the x-axis, which is between the roots.
Answer: \( 2 \leq x \leq 3 \).
Example (Inequality, non-factorisable, use formula or vertex):
Solve \( 2x^2 + 3x – 7 > 0 \).
▶️ Answer/Explanation
Solve \( 2x^2 + 3x – 7 = 0 \) first. From the formula example, roots are \( \displaystyle \alpha = \dfrac{-3 – \sqrt{65}}{4} \) and \( \displaystyle \beta = \dfrac{-3 + \sqrt{65}}{4} \) with \( \alpha < \beta \).
Since \( a = 2 > 0 \), the parabola opens up. It is above the x-axis outside the roots.
Answer: \( x < \dfrac{-3 – \sqrt{65}}{4} \) or \( x > \dfrac{-3 + \sqrt{65}}{4} \).
Example (Inequality using completing the square quickly):
Solve \( x^2 – 4x – 5 \geq 0 \).
▶️ Answer/Explanation
\( x^2 – 4x – 5 = (x – 2)^2 – 9 \).
\( (x – 2)^2 – 9 \geq 0 \Rightarrow (x – 2 – 3)(x – 2 + 3) \geq 0 \Rightarrow (x – 5)(x + 1) \geq 0 \).
Critical points \( -1 \) and \( 5 \). Parabola opens up, so \( \geq 0 \) outside the roots.
Answer: \( x \leq -1 \) or \( x \geq 5 \).
Solving a Linear–Quadratic Pair by Substitution
Solving a Linear–Quadratic Pair by Substitution
Strategy: Rearrange the linear equation to get one variable in terms of the other, substitute into the quadratic, solve the resulting quadratic, then back-substitute to get the paired values.
Example:
Solve the system \( x + y + 1 = 0 \) and \( x^2 + y^2 = 25 \) by substitution.
▶️ Answer/Explanation
From the linear equation: \( y = -x – 1 \).
Substitute into the quadratic: \( x^2 + (-x – 1)^2 = 25 \).
\( x^2 + (x^2 + 2x + 1) = 25 \Rightarrow 2x^2 + 2x + 1 = 25 \).
\( 2x^2 + 2x – 24 = 0 \Rightarrow x^2 + x – 12 = 0 \).
Factor: \( (x + 4)(x – 3) = 0 \Rightarrow x = 3 \) or \( x = -4 \).
Back-substitute: if \( x = 3 \), \( y = -4 \); if \( x = -4 \), \( y = 3 \).
Answer: \( (x, y) = (3, -4) \) or \( (-4, 3) \).
Example:
Solve the system \( 2x + 3y = 7 \) and \( 3x^2 = 4 + 4xy \) by substitution.
▶️ Answer/Explanation
From the linear equation: \( y = \dfrac{7 – 2x}{3} \).
Substitute into \( 3x^2 = 4 + 4xy \):
\( 3x^2 = 4 + 4x \cdot \dfrac{7 – 2x}{3} = 4 + \dfrac{28x – 8x^2}{3} \).
Multiply by 3: \( 9x^2 = 12 + 28x – 8x^2 \Rightarrow 17x^2 – 28x – 12 = 0 \).
Discriminant \( \Delta = (-28)^2 – 4 \cdot 17 \cdot (-12) = 1600 \Rightarrow \sqrt{\Delta} = 40 \).
\( x = \dfrac{28 \pm 40}{34} \Rightarrow x = 2 \) or \( x = \dfrac{-12}{34} = \dfrac{-6}{17} \).
Back-substitute to find \( y \):
If \( x = 2 \), \( 2(2) + 3y = 7 \Rightarrow y = 1 \).
If \( x = \dfrac{-6}{17} \), \( 2x = \dfrac{-12}{17} \) so \( 3y = 7 – 2x = 7 + \dfrac{12}{17} = \dfrac{131}{17} \Rightarrow y = \dfrac{131}{51} \).
Answer: \( (x, y) = (2, 1) \) or \( \Big(\dfrac{-6}{17}, \dfrac{131}{51}\Big) \).
Notes:
- Always isolate from the linear equation first to keep algebra manageable.
- Expect up to two real solution pairs, one pair, or none, depending on the discriminant that arises after substitution.
- If fractions appear, keep them exact using \( \dfrac{\cdot}{\cdot} \) and simplify at the end.
Recognising & Solving Equations Quadratic in a Function of \( x \)
Recognising & Solving Equations Quadratic in a Function of \( x \)
Idea: Spot when the equation is quadratic in a simpler expression (like \( x^2 \), \( \sqrt{x} \), \( \tan x \), etc.). Make a substitution \( u = \text{that expression} \), solve the quadratic in \( u \), then reverse the substitution and apply any domain restrictions.
Quick recognition tips
- 1) Forms like \( x^4, x^2 \Rightarrow \) try \( u = x^2 \).
- 2) Mixed \( x \) with \( \sqrt{x} \Rightarrow \) try \( u = \sqrt{x} \) so \( x = u^2 \).
- 3) Trig powers like \( \tan^2 x \Rightarrow \) try \( u = \tan x \).
Always check domains such as \( \sqrt{x} \ge 0 \) and periodic trig solutions.
Example:
Solve \( x^4 – 5x^2 + 4 = 0 \) by recognising it is quadratic in \( x^2 \).
▶️ Answer/Explanation
Let \( u = x^2 \). Then \( u^2 – 5u + 4 = 0 \).
Factor: \( (u – 1)(u – 4) = 0 \Rightarrow u = 1 \) or \( u = 4 \).
Reverse: \( x^2 = 1 \Rightarrow x = \pm 1 \); \( x^2 = 4 \Rightarrow x = \pm 2 \).
Answer: \( x \in \{ -2, -1, 1, 2 \} \).
Example:
Solve \( 6x + \sqrt{x} – 1 = 0 \) by substituting \( \sqrt{x} \).
▶️ Answer/Explanation
Domain: \( x \ge 0 \). Let \( u = \sqrt{x} \Rightarrow x = u^2 \) with \( u \ge 0 \).
Substitute: \( 6u^2 + u – 1 = 0 \).
\( \displaystyle u = \dfrac{-1 \pm \sqrt{1 + 24}}{12} = \dfrac{-1 \pm 5}{12} \Rightarrow u = \dfrac{1}{3} \) or \( u = \dfrac{-1}{2} \).
But \( u \ge 0 \), so take \( u = \dfrac{1}{3} \Rightarrow x = u^2 = \dfrac{1}{9} \).
Answer: \( \displaystyle x = \dfrac{1}{9} \).
Example :
Solve \( \tan^2 x = 1 + \tan x \) by substituting \( \tan x \).
▶️ Answer/Explanation
Let \( u = \tan x \). Then \( u^2 – u – 1 = 0 \).
\( \displaystyle u = \dfrac{1 \pm \sqrt{1 + 4}}{2} = \dfrac{1 \pm \sqrt{5}}{2} \).
Reverse substitution: \( \tan x = \dfrac{1 \pm \sqrt{5}}{2} \).
General solutions: \( \displaystyle x = \arctan\!\Big(\dfrac{1 + \sqrt{5}}{2}\Big) + n\pi \) or \( \displaystyle x = \arctan\!\Big(\dfrac{1 – \sqrt{5}}{2}\Big) + n\pi \), where \( n \in \mathbb{Z} \).
Answer: \( \displaystyle x = \arctan\!\Big(\dfrac{1 \pm \sqrt{5}}{2}\Big) + n\pi \), \( n \in \mathbb{Z} \).