Mathematics · AQA 8300 §A5
Quadratic Equations
- Solve quadratic equations by factorising, including when $a \neq 1$
- Solve quadratic equations using the quadratic formula, giving exact surd answers
- Complete the square to solve equations and find turning points
- Use the discriminant to determine the number of solutions and find unknown constants
- Solve simultaneous linear and quadratic equations and interpret geometrically
🔑 Core Concepts
The Standard Form of a Quadratic
A quadratic equation always involves an $x^2$ term and can always be rearranged into a single standard form. Understanding this structure is essential before choosing a solution method.
DEFINITION — Quadratic Equation
A quadratic equation has the standard form $ax^2 + bx + c = 0$, where $a \neq 0$ and $a$, $b$, $c$ are real constants. The term $ax^2$ is the quadratic term, $bx$ is the linear term, and $c$ is the constant. Note that $b$ or $c$ (but not $a$) may equal zero.
EXAM TIP
Always rearrange to $ax^2 + bx + c = 0$ before solving. If you see $3x^2 = 7x - 2$, rewrite as $3x^2 - 7x + 2 = 0$ first. If $a$ is negative, multiply the entire equation by $-1$ to make $a$ positive — this reduces sign errors when using the quadratic formula.
Method 1: Solving by Factorising
Factorising relies on the zero product property: if two factors multiply to zero, then at least one of them must be zero. This transforms a quadratic equation into two simple linear equations.
FACTORISING WHEN $a = 1$
To factorise $x^2 + bx + c$, find two integers $p$ and $q$ satisfying:
$$p + q = b \quad \text{and} \quad pq = c$$
Then $x^2 + bx + c = (x + p)(x + q)$, giving solutions $x = -p$ or $x = -q$.
FACTORISING WHEN $a \neq 1$ — AC Method
To factorise $ax^2 + bx + c$:
- Calculate $ac$.
- Find two numbers $p$ and $q$ such that $p + q = b$ and $pq = ac$.
- Rewrite the middle term: $ax^2 + px + qx + c$.
- Factorise by grouping.
COMMON MISTAKE — Wrong Sign in Solutions
After factorising to $(x + 3)(x - 2) = 0$, many students write $x = 3$ or $x = -2$. This is wrong. Set each factor equal to zero: $x + 3 = 0 \Rightarrow x = -3$, and $x - 2 = 0 \Rightarrow x = 2$. The signs always flip.
Rearrange to
$ax^2+bx+c=0$
$ax^2+bx+c=0$
→
Find $p$, $q$:
$p+q=b$, $pq=ac$
$p+q=b$, $pq=ac$
→
Factorise into
two brackets
two brackets
→
Set each factor
$= 0$ and solve
$= 0$ and solve
Method 2: The Quadratic Formula
The quadratic formula is derived by completing the square on the general form $ax^2 + bx + c = 0$. It always works and gives exact surd answers. Examiners specifically ask for exact answers in Grade 9 questions — never round unless told to.
The Quadratic Formula
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
$a$ = coefficient of $x^2$ (must be $\neq 0$)
$b$ = coefficient of $x$ (may be negative)
$c$ = constant term (may be zero)
$\pm$ gives two separate solutions
EXAM TIP
Always write $a = \ldots$, $b = \ldots$, $c = \ldots$ before substituting — this earns a method mark even if you make an arithmetic error. If $b = -5$, then $-b = +5$ and $b^2 = 25$ (squaring always gives a positive result, regardless of the sign of $b$).
IMPORTANT — Structure of the Formula
The denominator $2a$ divides the entire numerator $-b \pm \sqrt{b^2 - 4ac}$. A critical error is writing $x = -b \pm \frac{\sqrt{b^2-4ac}}{2a}$, which is incorrect. Always write the formula as a single fraction with a long horizontal bar.
Method 3: Completing the Square
Completing the square is the most versatile algebraic technique for quadratics. It rewrites $ax^2 + bx + c$ in the form $a(x + p)^2 + q$, from which you can read off the turning point directly, solve the equation exactly, and prove properties about roots.
COMPLETING THE SQUARE when $a = 1$
$$x^2 + bx + c = \left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c$$
Process: Halve the coefficient of $x$ → write as $(x + \frac{b}{2})^2$ → subtract the square of the half-coefficient → add back $c$.
Example: $x^2 + 6x + 2 = (x + 3)^2 - 9 + 2 = (x + 3)^2 - 7$
COMPLETING THE SQUARE when $a \neq 1$
Factor out $a$ from the $x^2$ and $x$ terms only, then complete the square inside the brackets:
$$ax^2 + bx + c = a\!\left(x^2 + \frac{b}{a}x\right) + c = a\!\left[\left(x + \frac{b}{2a}\right)^{\!2} - \frac{b^2}{4a^2}\right] + c = a\!\left(x + \frac{b}{2a}\right)^{\!2} - \frac{b^2}{4a} + c$$
Example: $2x^2 - 8x + 3 = 2(x^2 - 4x) + 3 = 2[(x-2)^2 - 4] + 3 = 2(x-2)^2 - 5$
Turning Point from Completed Square Form
$$y = a(x + p)^2 + q \;\implies\; \text{Turning point: } (-p,\; q)$$
$x$-coordinate of turning point: $-p$ (opposite sign to $p$)
$y$-coordinate of turning point (min/max value): $q$
$a > 0$ → minimum turning point (U-shaped parabola)
$a < 0$ → maximum turning point ($\cap$-shaped parabola)
EXAM TIP — Turning Point Sign
From $y = (x + 3)^2 - 7$, the turning point is $(-3, -7)$, not $(3, -7)$. Ask yourself: “What value of $x$ makes the bracket equal to zero?” For $(x + 3)^2$: $x + 3 = 0 \Rightarrow x = -3$. The $x$-coordinate always has the opposite sign to the number inside the bracket.
The Discriminant
The discriminant is the expression $b^2 - 4ac$ found under the square root in the quadratic formula. Its value tells us how many real solutions exist without solving the equation. This is an essential Grade 9 tool for proving properties of equations and finding unknown constants.
The Discriminant
$$\Delta = b^2 - 4ac$$
$\Delta > 0$ → two distinct real roots
$\Delta = 0$ → one repeated (equal) root: $x = -\frac{b}{2a}$
$\Delta < 0$ → no real roots (parabola does not cross $x$-axis)
| Discriminant | Number of Real Roots | Graph of $y = ax^2 + bx + c$ |
|---|---|---|
| $b^2 - 4ac > 0$ | Two distinct real roots | Parabola crosses the $x$-axis at two distinct points |
| $b^2 - 4ac = 0$ | One repeated (equal) root | Parabola is tangent to the $x$-axis (touches at one point) |
| $b^2 - 4ac < 0$ | No real roots | Parabola does not intersect the $x$-axis |
EXAM TIP — Grade 9 Discriminant Language
Learn to recognise these code phrases in exam questions:
- "two distinct real solutions" → set $\Delta > 0$
- "equal roots" or "the line is a tangent to the curve" → set $\Delta = 0$
- "prove the equation has no real solutions" → show $\Delta < 0$
- "how many intersection points?" → find $\Delta$ and interpret
MEMORY TRICK — The Discriminant Triangle
Think of the Greek letter $\Delta$ (delta) as a triangle with three corners: one corner for each case.
Corner 1 (> 0): Two roots — the line passes through
Corner 2 (= 0): One root — the line just touches
Corner 3 (< 0): None — the line misses completely
Simultaneous Linear and Quadratic Equations
A straight line and a parabola can intersect at 0, 1, or 2 points. To find intersection points algebraically, substitute the linear equation into the quadratic. The number of solutions is determined by the discriminant of the resulting quadratic.
METHOD — Substitution
- From the linear equation, express $y$ in terms of $x$ (or vice versa).
- Substitute into the quadratic equation to eliminate $y$.
- Rearrange the resulting equation to standard form $ax^2 + bx + c = 0$.
- Solve the quadratic to find all $x$-values.
- Substitute each $x$-value back into the linear equation to find the corresponding $y$-values.
- State solutions as coordinate pairs $(x, y)$.
COMMON MISTAKE — Incomplete Solutions
After finding $x = 1$ and $x = 3$, many students stop. Always find the $y$-values too and write full coordinate pairs. Substitute back into the linear equation (it is simpler than the quadratic). Writing only $x$-values loses marks.
Geometric Problems Leading to Quadratics
Area and perimeter problems frequently produce quadratic equations. The key skill is translating a geometric scenario into algebra, then solving and interpreting the solutions in context.
EXAM TIP — Rejecting Invalid Solutions
Quadratic equations often produce two solutions, but only one may be valid in a geometric context. If $x$ represents a length, width, or distance, it must be positive. Always state explicitly: "Reject $x = -3$ since lengths must be positive." This line earns a mark.
MEMORY TRICK — SAFE for Geometry Problems
S — Set up the equation (write the area/perimeter formula with the unknown $x$)A — Arrange into standard form $ax^2 + bx + c = 0$
F — Factorise or use the quadratic formula to solve
E — Examine solutions: reject any that are physically impossible
🗺 Visual Notes
Quadratic
Equations
Equations
Solving Methods
- Factorising (quickest when it works)
- Quadratic formula (always works)
- Completing the square (most powerful)
- Graphically (estimate only)
The Discriminant $\Delta$
- $\Delta = b^2 - 4ac$
- $\Delta > 0$: two distinct real roots
- $\Delta = 0$: one repeated root
- $\Delta < 0$: no real roots
Completed Square Form
- Form: $a(x + p)^2 + q$
- Turning point: $(-p,\, q)$
- Min/max value: $q$
- Axis of symmetry: $x = -p$
Applications
- Area and perimeter problems
- Simultaneous equations
- Line tangent to a curve
- Finding unknown constants
Grade 9 Skills
- Exact surd answers
- Complete square when $a \neq 1$
- Discriminant inequalities
- Proving no real solutions
Comparison of Solving Methods
| Method | Best Used When | Exact Surd Answers? | Always Works? |
|---|---|---|---|
| Factorising | Coefficients are integers; you can spot the factors quickly | Yes (rational) | No — not all quadratics factorise over integers |
| Quadratic Formula | Factorising is not obvious; surd answers expected | Yes | Yes |
| Completing the Square | Turning point also needed; proof questions | Yes | Yes |
| Graphical / Calculator | Checking or estimating only | No (decimal approx.) | Approximately |
Decision Tree: Which Method to Choose?
Can you spot
integer factors?
integer factors?
YES →
Use Factorising
Quickest method
Quickest method
Do you also need
the turning point?
the turning point?
YES →
Complete
the Square
the Square
NO →
Use the
Quadratic Formula
Quadratic Formula
Discriminant and Graph Relationship
| Condition | $\Delta$ Value | Graph Behaviour | Number of Solutions |
|---|---|---|---|
| Two distinct real roots | $\Delta > 0$ | Parabola crosses $x$-axis at 2 points | $x = \dfrac{-b + \sqrt{\Delta}}{2a}$ and $x = \dfrac{-b - \sqrt{\Delta}}{2a}$ |
| One repeated root | $\Delta = 0$ | Parabola is tangent to $x$-axis | $x = \dfrac{-b}{2a}$ (one value only) |
| No real roots | $\Delta < 0$ | Parabola does not reach $x$-axis | No real solutions exist |
Completing the Square — Step-by-Step Process ($a \neq 1$)
Factor out $a$ from
$x^2$ and $x$ terms
$x^2$ and $x$ terms
→
Halve the new
$x$ coefficient
$x$ coefficient
→
Write as $(x + \frac{b}{2a})^2$
minus its square
minus its square
→
Multiply out the
bracket; add $c$
bracket; add $c$
→
Read off turning
point $(-p,\, q)$
point $(-p,\, q)$
✏ Worked Examples
Grade 4–5 · Solving by Factorising
Solve $x^2 + 7x + 10 = 0$.
1
Find the factor pair
We need two integers $p$ and $q$ such that $p + q = 7$ (coefficient of $x$) and $pq = 10$ (constant).
Factor pairs of 10: $(1, 10)$ gives sum $11$; $(2, 5)$ gives sum $7$. ✓
So $p = 2$ and $q = 5$.
Factor pairs of 10: $(1, 10)$ gives sum $11$; $(2, 5)$ gives sum $7$. ✓
So $p = 2$ and $q = 5$.
2
Factorise
$$x^2 + 7x + 10 = (x + 2)(x + 5) = 0$$
Check by expanding: $(x+2)(x+5) = x^2 + 5x + 2x + 10 = x^2 + 7x + 10$ ✓
3
Apply the zero product property
If $(x + 2)(x + 5) = 0$, then either $x + 2 = 0$ or $x + 5 = 0$.
4
Solve each linear equation
$x + 2 = 0 \Rightarrow x = -2$ and $x + 5 = 0 \Rightarrow x = -5$
$x = -2$ or $x = -5$
Grade 6–7 · Quadratic Formula (Exact Surd Answer)
Solve $2x^2 - 4x - 3 = 0$, giving your answers in the form $a \pm b\sqrt{c}$.
1
Identify $a$, $b$, $c$ and check it does not factorise
$a = 2$, $b = -4$, $c = -3$.
For integer factorisation, need $p + q = -4$ and $pq = 2 \times (-3) = -6$. No integer pair exists — use the formula.
For integer factorisation, need $p + q = -4$ and $pq = 2 \times (-3) = -6$. No integer pair exists — use the formula.
2
Calculate the discriminant
$$\Delta = b^2 - 4ac = (-4)^2 - 4(2)(-3) = 16 + 24 = 40$$
Since $\Delta = 40 > 0$, there are two distinct real roots. Simplify: $\sqrt{40} = \sqrt{4 \times 10} = 2\sqrt{10}$.
3
Apply the quadratic formula
$$x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-(-4) \pm 2\sqrt{10}}{2(2)} = \frac{4 \pm 2\sqrt{10}}{4}$$
4
Simplify by dividing numerator and denominator by 2
$$x = \frac{4 \pm 2\sqrt{10}}{4} = \frac{2 \pm \sqrt{10}}{2} = 1 \pm \frac{\sqrt{10}}{2}$$
$x = 1 + \dfrac{\sqrt{10}}{2}$ or $x = 1 - \dfrac{\sqrt{10}}{2}$
Grade 9 · Completing the Square ($a \neq 1$), Turning Point & Discriminant Proof
(a) Write $3x^2 - 12x + 7$ in the form $a(x + p)^2 + q$, where $a$, $p$, and $q$ are integers.
(b) Hence state the coordinates of the turning point of $y = 3x^2 - 12x + 7$ and whether it is a minimum or maximum.
(c) Show that the equation $3x^2 - 12x + 7 = k$ has no real solutions when $k < -5$.
(b) Hence state the coordinates of the turning point of $y = 3x^2 - 12x + 7$ and whether it is a minimum or maximum.
(c) Show that the equation $3x^2 - 12x + 7 = k$ has no real solutions when $k < -5$.
1
Part (a): Factor out 3 from the first two terms
$$3x^2 - 12x + 7 = 3(x^2 - 4x) + 7$$
Key: Only factor out 3 from the terms containing $x$. The $+7$ is left outside.
2
Complete the square inside the brackets
For $x^2 - 4x$: halve $-4$ to get $-2$, so $(x - 2)^2 = x^2 - 4x + 4$.
$$x^2 - 4x = (x - 2)^2 - 4$$
Substitute: $3\left[(x-2)^2 - 4\right] + 7$
3
Expand the outer bracket and simplify
$$3(x-2)^2 - 12 + 7 = 3(x-2)^2 - 5$$
So $a = 3$, $p = -2$, $q = -5$. (The form $a(x+p)^2+q$ with $p=-2$ gives $a(x+(-2))^2+q = 3(x-2)^2-5$.)
4
Part (b): Read off the turning point
From $y = 3(x - 2)^2 - 5$, the turning point occurs when $(x-2)^2 = 0$, i.e.\ $x = 2$.
At $x = 2$: $y = 3(0) - 5 = -5$.
Since $a = 3 > 0$, the parabola is U-shaped, so this is a minimum.
Since $a = 3 > 0$, the parabola is U-shaped, so this is a minimum.
5
Part (c): Prove no real solutions when $k < -5$
Setting $3(x-2)^2 - 5 = k$:
$$3(x - 2)^2 = k + 5$$
For real solutions, we need $(x-2)^2 \geq 0$, which requires the right-hand side to also be $\geq 0$:
$$k + 5 \geq 0 \implies k \geq -5$$
Therefore, when $k < -5$, we would need $(x-2)^2$ to be negative, which is impossible for real $x$.
Hence the equation has no real solutions when $k < -5$. $\blacksquare$
(a) $3(x-2)^2 - 5$ | (b) Minimum turning point at $(2,\, -5)$ | (c) Requires $k \geq -5$ for real solutions — proved above
❓ Exam Questions
Q1
1 mark
Calculate the discriminant of the equation $x^2 + 4x + 7 = 0$ and state the number of real solutions it has.
Mark scheme:
$\Delta = b^2 - 4ac = 4^2 - 4(1)(7) = 16 - 28 = -12$ [1 mark]Since $\Delta = -12 < 0$, the equation has no real solutions.
Q2
2 marks
Solve $x^2 - 9x + 18 = 0$ by factorising.
Mark scheme:
Find two numbers multiplying to 18 and adding to $-9$: these are $-3$ and $-6$.$(x - 3)(x - 6) = 0$ [1 mark for correct factorisation]
$x = 3$ or $x = 6$ [1 mark for both solutions]
Q3
3 marks
Solve $3x^2 - 10x + 3 = 0$, giving your answers as fractions where necessary.
Mark scheme:
$a = 3$, $b = -10$, $c = 3$$\Delta = (-10)^2 - 4(3)(3) = 100 - 36 = 64$ [1 mark]
$x = \dfrac{10 \pm \sqrt{64}}{6} = \dfrac{10 \pm 8}{6}$ [1 mark for correct substitution]
$x = \dfrac{18}{6} = 3$ or $x = \dfrac{2}{6} = \dfrac{1}{3}$ [1 mark for both values]
Alternatively: $(3x - 1)(x - 3) = 0$ gives the same solutions.
Q4
4 marks
(a) Complete the square on $x^2 - 6x + 2$, writing it in the form $(x + p)^2 + q$. [2 marks]
(b) Hence solve $x^2 - 6x + 2 = 0$, giving your answers in exact form. [2 marks]
Mark scheme:
Part (a):$x^2 - 6x + 2 = (x - 3)^2 - 9 + 2$ [1 mark for $(x-3)^2$]
$= (x - 3)^2 - 7$ [1 mark for $-7$]
Part (b):
$(x - 3)^2 - 7 = 0$
$(x - 3)^2 = 7$ [1 mark]
$x - 3 = \pm\sqrt{7}$
$x = 3 + \sqrt{7}$ or $x = 3 - \sqrt{7}$ [1 mark for both exact answers]
Q5
4 marks
Solve simultaneously $y = x^2 - 4x + 3$ and $y = x - 1$.
Mark scheme:
Substitute $y = x - 1$ into $y = x^2 - 4x + 3$:$x - 1 = x^2 - 4x + 3$ [1 mark for substitution]
$0 = x^2 - 5x + 4$ [1 mark for correct rearrangement]
$0 = (x - 1)(x - 4)$
$x = 1$ or $x = 4$ [1 mark for both $x$-values]
When $x = 1$: $y = 1 - 1 = 0$ → $(1, 0)$
When $x = 4$: $y = 4 - 1 = 3$ → $(4, 3)$ [1 mark for both coordinate pairs]
Q6
6 marks
The equation $kx^2 + 6x + k = 0$ has two distinct real solutions.
Find the range of values of $k$. [6 marks]
Mark scheme:
For two distinct real solutions, the discriminant must be strictly positive: $\Delta > 0$ [1 mark — correct condition stated]With $a = k$, $b = 6$, $c = k$:
$b^2 - 4ac > 0 \Rightarrow 36 - 4k^2 > 0$ [1 mark for correct expression]
$4k^2 < 36$ [1 mark]
$k^2 < 9$ [1 mark]
$|k| < 3$, giving $-3 < k < 3$ [1 mark]
For the equation to be quadratic, we also need $k \neq 0$ (otherwise the equation becomes $6x = 0$, which is linear).
Therefore: $k \in (-3, 0) \cup (0, 3)$ [1 mark for $k \neq 0$ condition]
⭐ Grade 9 Model Answers
Full Annotated Solution: Q6 — Discriminant with Unknown Coefficient
THE QUESTION
The equation $kx^2 + 6x + k = 0$ has two distinct real solutions. Find the range of values of $k$.
Grade 9 · Full 6-Mark Annotated Solution
Annotated model answer with examiner commentary at each step
1
Identify the condition — [1 mark]
"Two distinct real solutions" is the key phrase. This means the discriminant must be strictly greater than zero:
$$\Delta > 0 \implies b^2 - 4ac > 0$$
EXAMINER'S NOTE
Many students confuse "two distinct solutions" ($\Delta > 0$) with "at least one real solution" ($\Delta \geq 0$). The word distinct means strictly more than one, so $\Delta$ must be strictly positive. This is the most common mark lost in this type of question.2
Correctly identify $a$, $b$, $c$ — [1 mark]
From $kx^2 + 6x + k = 0$: $a = k$, $b = 6$, $c = k$.
Note: Both $a$ and $c$ are equal to $k$. This symmetry is deliberate and leads to a clean quadratic inequality in $k$.
Note: Both $a$ and $c$ are equal to $k$. This symmetry is deliberate and leads to a clean quadratic inequality in $k$.
3
Set up the inequality — [1 mark]
$$b^2 - 4ac > 0 \implies 6^2 - 4(k)(k) > 0 \implies 36 - 4k^2 > 0$$
4
Solve the quadratic inequality — [2 marks]
$$36 - 4k^2 > 0 \implies 4k^2 < 36 \implies k^2 < 9$$
Taking square roots: $|k| < 3$, which gives $\mathbf{-3 < k < 3}$.
Alternatively: $k^2 - 9 < 0$ has roots $k = \pm 3$; since the coefficient of $k^2$ is positive, the parabola $y = k^2 - 9$ is below zero for $-3 < k < 3$.
Alternatively: $k^2 - 9 < 0$ has roots $k = \pm 3$; since the coefficient of $k^2$ is positive, the parabola $y = k^2 - 9$ is below zero for $-3 < k < 3$.
5
Apply the hidden constraint — [1 mark: the Grade 9 mark]
For the equation to be quadratic, we need $k \neq 0$. If $k = 0$, the equation reduces to $6x = 0$ (linear, not quadratic).
Therefore the complete answer is: $k \in (-3, 0) \cup (0, 3)$.
Therefore the complete answer is: $k \in (-3, 0) \cup (0, 3)$.
WHY THIS IS THE GRADE 9 MARK
This final mark separates Grade 8 from Grade 9. The examiner specifically includes $k$ as the leading coefficient to test whether students remember the constraint $a \neq 0$. Grade 9 students check: "Does my answer maintain the conditions stated in the question?"
$-3 < k < 3$, $k \neq 0$ (equivalently, $k \in (-3, 0) \cup (0, 3)$)
Why This Question Requires Grade 9 Thinking
SYNOPTIC SKILLS TESTED
This question tests three algebraic skills simultaneously: (1) knowing when to use the discriminant and the correct inequality direction, (2) solving a quadratic inequality in an unknown parameter, and (3) recognising that the leading coefficient must be non-zero. Grade 9 students plan before they write: they see "two distinct real solutions" and immediately write $\Delta > 0$ as their opening line, earning the first mark before any calculation.
📋 Revision Sheet
Key Definitions
| Term | Meaning |
|---|---|
| Quadratic equation | $ax^2+bx+c=0$, $a \neq 0$ |
| Discriminant $\Delta$ | $b^2-4ac$; determines root count |
| Completing the square | Rewriting as $a(x+p)^2+q$ |
| Turning point | Vertex of parabola: $(-p, q)$ |
| Repeated root | $\Delta=0$; parabola tangent to $x$-axis |
| Surd | Exact irrational answer, e.g. $3+\sqrt{7}$ |
| Zero product property | $AB=0 \Rightarrow A=0$ or $B=0$ |
Essential Formulae
Quadratic Formula:
$$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$Discriminant: $\Delta = b^2 - 4ac$
Completing the square ($a=1$):
$$x^2+bx+c = \!\left(x+\tfrac{b}{2}\right)^{\!2} - \tfrac{b^2}{4} + c$$Turning point from $a(x+p)^2+q$:
$$\text{Turning point: } (-p,\; q)$$Repeated root: $x = \dfrac{-b}{2a}$ when $\Delta = 0$
Memory Hooks
- "Minus $b$, plus-or-minus..." — say the formula aloud as you write it
- $\Delta$ has three corners — one for each case: $>0$, $=0$, $<0$
- SAFE for geometry: Set up, Arrange, Factorise, Examine
- "Halve it, square it, subtract it" — the completing the square mantra
- "The sign flips" — $(x+3)^2$ gives turning point $x=-3$
- "Equal roots = tangent" — $\Delta=0$ means the line just touches
- "Distinct = strictly greater" — two distinct roots means $\Delta > 0$, not $\geq 0$
Exam Tips
- Write $a = \ldots$, $b = \ldots$, $c = \ldots$ before using the formula
- Calculate $b^2$ and $4ac$ separately before subtracting
- $b^2$ is always positive; $-4ac$ can be positive or negative
- Give exact surd answers unless the question says "decimal"
- In simultaneous equations: use the linear equation to find $y$
- In geometric problems: always reject negative lengths
- "Prove no real solutions" → calculate $\Delta$ and show $\Delta < 0$
- Check $k \neq 0$ when $k$ appears as the leading coefficient
🔄 Flashcards
Click each card to reveal the answer. Work through all 15 cards until you can answer instantly.
✗ Common Mistakes
MISTAKE 1 — Wrong Sign Before $b$ in the Quadratic Formula
What students do: Write $x = \frac{b \pm \sqrt{b^2-4ac}}{2a}$ instead of $\frac{-b \pm \sqrt{b^2-4ac}}{2a}$.
Why marks are lost: Both solutions are completely wrong; typically 0 out of 3 marks for the question.
How to avoid it: Memorise the formula as starting with "negative $b$". Before substituting, explicitly write $-b = -(b\text{-value})$. If $b = -5$, write $-b = -(-5) = +5$ in your working.
MISTAKE 2 — Not Dividing the Entire Numerator by $2a$
What students do: Write $x = -b \pm \frac{\sqrt{b^2-4ac}}{2a}$, dividing only the square root part by $2a$.
Why marks are lost: The long fraction bar extends under the whole of $-b \pm \sqrt{b^2-4ac}$. This error gives wrong final values even when the discriminant is correct.
How to avoid it: Always draw a long horizontal fraction bar. Bracket the numerator: write $\frac{(-b) \pm \sqrt{b^2-4ac}}{2a}$ explicitly.
MISTAKE 3 — Forgetting $\pm$ When Taking a Square Root
What students do: From $(x - 3)^2 = 7$, write only $x = 3 + \sqrt{7}$, missing the second solution $x = 3 - \sqrt{7}$.
Why marks are lost: An answer that is not written cannot earn marks. This loses 1 mark for the missing solution — the most preventable error in the chapter.
How to avoid it: Whenever you take a square root in algebra, always write $\pm$. Make it a reflex: the square root step is the $\pm$ step.
MISTAKE 4 — Not Factoring Out $a$ Before Completing the Square
What students do: For $2x^2 - 8x + 3$, write $(2x - 4)^2 - \ldots$ without first factoring out 2.
Why marks are lost: The resulting completed square form is wrong; the turning point and solutions derived from it will also be wrong. Multiple marks are lost.
How to avoid it: Before completing the square, always check the value of $a$. If $a \neq 1$, your first line must be $a(x^2 + \frac{b}{a}x) + c$ with $a$ factored out.
MISTAKE 5 — Wrong Sign for the $x$-Coordinate of the Turning Point
What students do: From $y = (x + 4)^2 - 3$, state the turning point as $(4, -3)$ instead of $(-4, -3)$.
Why marks are lost: The $x$-coordinate of the turning point is the value that makes the bracket zero. For $(x + 4) = 0$, we get $x = -4$. The sign always flips.
How to avoid it: Ask: "What value of $x$ makes this bracket equal to zero?" Write that value. Do not just copy the number from inside the bracket.
MISTAKE 6 — Accepting Negative Solutions in Geometric Contexts
What students do: For a problem where $x$ represents a length, they write both $x = -2$ and $x = 5$ as valid answers.
Why marks are lost: A length cannot be negative. Accepting $x = -2$ shows a failure to interpret the solution in context, losing 1 mark for "selecting the appropriate solution".
How to avoid it: After solving, always re-read what $x$ represents. Write: "Reject $x = -2$ since length cannot be negative." This statement itself can earn a mark.
✅ Final Checklist
Click each item when you can do it confidently without notes. Aim for 15/15 before your exam.
0 / 15- I can rearrange any equation into standard form $ax^2 + bx + c = 0$
- I can solve a quadratic by factorising when $a = 1$ (e.g. $x^2 + 5x + 6 = 0$)
- I can solve a quadratic by factorising when $a \neq 1$ using the AC method
- I can correctly identify $a$, $b$, $c$ before using the quadratic formula
- I can apply the quadratic formula and simplify surds in the answer (e.g. $\sqrt{40} = 2\sqrt{10}$)
- I can complete the square for $x^2 + bx + c$ when $a = 1$
- I can complete the square when $a \neq 1$ by factoring out $a$ first
- I can read off the turning point $(-p, q)$ from $a(x+p)^2 + q$
- I can calculate the discriminant $\Delta = b^2 - 4ac$ and interpret the result
- I can use $\Delta = 0$ to find unknown constants (tangency / equal roots problems)
- I can prove a quadratic has no real solutions by showing $\Delta < 0$
- I can solve simultaneous linear and quadratic equations by substitution
- I can form and solve a quadratic equation from a geometric area problem
- I can reject negative solutions that represent lengths and justify doing so
- I can find the range of values of an unknown constant using a discriminant inequality