Quadratic Equations in IGCSE Maths
Three methods for solving quadratic equations: factorising, the quadratic formula, and completing the square. Know when to use each one and how to set out your working for full marks.
0580 · E2.5What is a quadratic equation?
A quadratic equation has the form:
\[ax^2 + bx + c = 0\]where \(a\), \(b\) and \(c\) are numbers and \(a \neq 0\). The highest power of \(x\) is 2.
The solutions (also called roots) are the values of \(x\) that make the equation true. A quadratic equation can have two solutions, one repeated solution, or no real solutions.
Always rearrange the equation into the form \(ax^2 + bx + c = 0\) before solving. Everything must be on one side, with zero on the other. This is the single most important step — skip it and every method fails.
Solving by factorising
Factorising means writing the quadratic as a product of two brackets. It relies on one key fact: if two things multiply to give zero, then one of them must be zero. So if \((x + 2)(x + 3) = 0\), then either \(x + 2 = 0\) or \(x + 3 = 0\).
Factorising only works when the solutions are whole numbers or simple fractions. If you can't find the factors, switch to the formula.
For equations of the form \(x^2 +\) \(b\)\(x +\) \(c\) \(= 0\):
m \(\times\) n \(=\) \(c\) (they multiply to give the constant term)
m \(+\) n \(=\) \(b\) (they add to give the coefficient of \(x\))
\((x +\) \(m\)\()(x +\) \(n\)\() = 0\)
Set each bracket equal to zero to find the two solutions.
Solve \(x^2 +\) \(5\)\(x +\) \(6\) \(= 0\).
We need two numbers where m \(\times\) n \(=\) \(6\) and m \(+\) n \(=\) \(5\).
Try \(2\) and \(3\). Check: \(2\) \(\times\) \(3\) \(=\) \(6\) ✓ and \(2\) \(+\) \(3\) \(=\) \(5\) ✓
When the coefficient of \(x^2\) isn't 1, we need to think about what goes in both brackets. The idea is straightforward: try combinations and check by expanding.
Write two brackets with spaces to fill: \((\text{?}x + \text{?})(\text{?}x + \text{?}) = 0\). The first terms in each bracket must multiply to give \(a\)\(x^2\), and the last terms must multiply to give \(c\).
List the factor pairs of \(a\) and \(c\). Try different combinations in the brackets.
For each attempt, mentally expand to see if the middle term gives \(b\)\(x\). When it does, you've found it.
Solve \(2\)\(x^2 +\) \(7\)\(x +\) \(3\) \(= 0\).
The \(x^2\) coefficient is \(2\), so the brackets must start with \(2x\) and \(x\):
\[(2x + \text{?})(x + \text{?}) = 0\]The constant term is \(3\), so the two numbers in the gaps must multiply to give \(3\). The only factor pair of 3 is \(1\) and \(3\).
Try 1: \((2x + 1)(x + 3)\) → expand the middle terms: \(6x + x = \) \(7x\) ✓ That's it!
Solve \(3\)\(x^2 -\) \(11\)\(x +\) \(6\) \(= 0\).
The \(x^2\) coefficient is \(3\), so the brackets start with \(3x\) and \(x\):
\[(3x + \text{?})(x + \text{?}) = 0\]The constant term is \(+6\) and the middle term is negative, so both numbers in the gaps will be negative. Factor pairs of 6: \(1 \times 6\) or \(2 \times 3\).
Try 1: \((3x - 1)(x - 6)\) → expand the middle terms: \(-18x - x = -19x\) ✗
Try 2: \((3x - 6)(x - 1)\) → expand the middle terms: \(-3x - 6x = -9x\) ✗
Try 3: \((3x - 2)(x - 3)\) → expand the middle terms: \(-9x - 2x = \) \(-11x\) ✓
This method works well when \(a\) and \(c\) only have a few factor pairs. But with something like \(6x^2 + x - 2 = 0\), there are many possible combinations to try. For those, the grid method below is more systematic.
The grid method uses a 2×2 box to organise the factorisation. It's the same maths as the "ac method" but laid out visually so you can see the common factors appear.
This is the key step. Multiply the coefficient of \(x^2\) by the constant, then find two numbers that multiply to give that product and add to give the coefficient of \(x\).
Place \(a\)\(x^2\) in the top-left and \(c\) in the bottom-right. The other two cells are \(m\)\(x\) and \(n\)\(x\) (your two numbers from step 1, each multiplied by \(x\)).
Find the common factor of each row and each column. These give you the brackets.
Solve \(6\)\(x^2 +\) \(1\)\(x\) \(- 2\) \(= 0\).
\(a\) \(\times\) \(c\) \(=\) \(6\) \(\times\) \((-2)\) \(= -12\).
Two numbers that multiply to \(-12\) and add to \(1\): that's \(4\) and \(-3\).
Check: \(4\) \(\times\) \((-3)\) \(= -12\) ✓ and \(4\) \(+\) \((-3)\) \(=\) \(1\) ✓
Place \(6x^2\) top-left and \(-2\) bottom-right. The other two cells are \(4x\) and \(-3x\).
Row 1: the common factor of \(6x^2\) and \(4x\) is \(2x\).
Row 2: the common factor of \(-3x\) and \(-2\) is \(-1\).
Column 1: the common factor of \(6x^2\) and \(-3x\) is \(3x\).
Column 2: the common factor of \(4x\) and \(-2\) is \(2\).
The row factors give one bracket: \((2x - 1)\). The column factors give the other: \((3x + 2)\).
Solve \(x^2 - 16 = 0\).
There is no \(x\) term (\(b = 0\)). This is the pattern \(a^2 - b^2 = (a+b)(a-b)\):
\[\begin{aligned}x^2 - 16 &= (x + 4)(x - 4) = 0 \\[6pt] \color{#dc2626}{x = -4} \quad &\text{or} \quad \color{#dc2626}{x = 4}\end{aligned}\]If every term shares a common factor, take it out before factorising. For example, \(3x^2 + 12x + 9 = 0\) becomes \(3(x^2 + 4x + 3) = 0\). Now you only need to factorise \(x^2 + 4x + 3\), which is much simpler. The \(3\) out the front doesn't affect the solutions.
The quadratic formula
For any quadratic equation \(ax^2 + bx + c = 0\):
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula always works, whether the equation factorises or not. The \(\pm\) gives two solutions: one using \(+\) and one using \(-\).
The formula is given on the Cambridge 0580 formula sheet, but you still need to know how to use it accurately.
Solve \(3x^2 - 5x - 1 = 0\). Give your answers correct to 2 decimal places.
\(a = 3\), \(b = -5\), \(c = -1\)
Work out the discriminant (\(b^2 - 4ac\)) first and write it down. Then calculate the two solutions separately. This reduces errors and makes your working easy for the examiner to follow. Don't try to do it all in one go on the calculator.
Completing the square
Completing the square rewrites \(ax^2 + bx + c\) in the form:
\[a(x + p)^2 + q\]This is useful for solving the equation (rearrange for \(x\)) and finding the turning point of the graph (it's at \((-p, q)\)).
To write \(x^2 + bx + c\) in the form \((x + p)^2 + q\):
\(\dfrac{b}{2} = p\). Write \((x + p)^2\).
\((x + p)^2\) produces an extra \(p^2\) when expanded. Subtract it, then add the original constant:
\[x^2 + bx + c = (x + p)^2 - p^2 + c\]Write \(x^2 + 6x + 2\) in the form \((x + p)^2 + q\).
\(\dfrac{6}{2} = 3\). So \(p = 3\). Write \((x + 3)^2\).
So \(p = 3\) and \(q = -7\). The turning point of the graph is \((-3, -7)\).
Solve \(x^2 + 6x + 2 = 0\) by completing the square. Give exact answers.
Extra first step before completing the square:
Take it out of the \(x^2\) and \(x\) terms only. Leave the constant outside.
Write \(2x^2 + 12x + 5\) in the form \(a(x + p)^2 + q\).
\(\dfrac{6}{2} = 3\). Write \((x + 3)^2\).
The discriminant
The expression under the square root in the quadratic formula is called the discriminant:
\[\Delta = b^2 - 4ac\]It tells you how many solutions the equation has, without solving it:
\(\Delta > 0\): two distinct real solutions (the parabola crosses the x-axis twice).
\(\Delta = 0\): one repeated solution (the parabola just touches the x-axis).
\(\Delta < 0\): no real solutions (the parabola doesn't reach the x-axis).
Show that \(2x^2 + 3x + 5 = 0\) has no real solutions.
Since \(\Delta < 0\), the equation has no real solutions.
Quadratic graphs
The graph of \(y = ax^2 + bx + c\) is a parabola. If \(a > 0\) it's U-shaped; if \(a < 0\) it's ∩-shaped.
The roots (solutions) are where the parabola crosses the x-axis. The turning point (vertex) is the minimum or maximum. Completing the square gives the turning point directly: if \(y = a(x + p)^2 + q\), the vertex is at \((-p, q)\).
The discriminant tells you how the parabola sits relative to the x-axis. Two roots means it crosses twice (\(\Delta > 0\)). One repeated root means it just touches (\(\Delta = 0\)). No real roots means it floats above or below the x-axis entirely (\(\Delta < 0\)).
Which method should I use?
The question wording tells you which method to use:
| The question says... | Use this method | Why |
|---|---|---|
| "Solve by factorising" | Factorising | The question is telling you it factorises. If you can't spot the factors, the question wouldn't phrase it this way. |
| "Give answers to 2 d.p." or "to 3 s.f." | Quadratic formula | Decimal answers mean the equation doesn't factorise neatly. The formula handles everything. |
| "Give exact answers" or "in the form \(p \pm \sqrt{q}\)" | Completing the square or formula | Leave surds in your answer. Either method works — completing the square gives the form directly. |
| "Write in the form \((x+p)^2+q\)" or "find the turning point" | Completing the square | Only this method gives you the completed square form and the turning point \((-p, q)\). |
| No method specified | Try factorising first | Spend 30 seconds looking for factors. If you can't find them, switch to the formula. |
Exam-style questions
Solve \(3x^2 - 10x + 8 = 0\).
Show solution
The brackets must start with \(3x\) and \(x\) (since \(3x \times x = 3x^2\)). The constant is \(8\), and we need the middle term to be \(-10x\). Trying factors of 8:
\((3x - 4)(x - 2)\) → expand middle: \(-6x - 4x = -10x\) ✓
\[\begin{aligned}(3x - 4)(x - 2) &= 0 \\[6pt] 3x - 4 = 0 \quad &\Rightarrow \quad x = \tfrac{4}{3} \\[6pt] x - 2 = 0 \quad &\Rightarrow \quad x = 2\end{aligned}\]Solve \(x^2 - 7x + 2 = 0\). Give your answers correct to 2 decimal places.
Show solution
\(a = 1\), \(b = -7\), \(c = 2\)
\[\begin{aligned}x &= \frac{7 \pm \sqrt{49 - 8}}{2} \\[8pt] &= \frac{7 \pm \sqrt{41}}{2}\end{aligned}\] \[\begin{aligned}x &= \frac{7 + 6.403...}{2} = 6.70 \text{ (2 d.p.)} \\[8pt] x &= \frac{7 - 6.403...}{2} = 0.30 \text{ (2 d.p.)}\end{aligned}\](a) Write \(x^2 - 8x + 10\) in the form \((x - p)^2 + q\). [2]
(b) Hence find the minimum value of \(x^2 - 8x + 10\) and the value of \(x\) at which it occurs. [2]
Show solution (a)
Half of \(-8\) is \(-4\). Write \((x - 4)^2\), which expands to \(x^2 - 8x + 16\). Adjust:
\[\begin{aligned}x^2 - 8x + 10 &= (x - 4)^2 - 16 + 10 \\[6pt] &= (x - 4)^2 - 6\end{aligned}\]Show solution (b)
Since \((x - 4)^2 \geq 0\) for all \(x\), the minimum value of the expression is \(-6\), occurring when \(x = 4\).
The turning point is \((4, -6)\).
A rectangle has a length that is 3 cm more than its width. Its area is 54 cm². Find the dimensions of the rectangle.
Show solution
Let the width be \(x\) cm. Then the length is \((x + 3)\) cm.
\[\begin{aligned}x(x + 3) &= 54 \\[6pt] x^2 + 3x - 54 &= 0 \\[6pt] (x + 9)(x - 6) &= 0 \\[6pt] x = -9 \quad &\text{or} \quad x = 6\end{aligned}\]Since \(x\) is a length, \(x = -9\) is rejected. The width is \(6\) cm and the length is \(9\) cm.
Common mistakes
If you try to factorise \(x^2 + 5x = 6\) without rearranging, you'll get stuck or get the wrong answer. Always move everything to one side first: \(x^2 + 5x - 6 = 0\).
In the formula, the first term is \(-b\), not \(b\). If \(b = -5\), then \(-b = 5\). Write \(-(-5) = 5\) explicitly in your working to avoid sign errors.
Given \(x^2 = 5x\), never divide both sides by \(x\) — you lose the solution \(x = 0\). Instead, rearrange: \(x^2 - 5x = 0\), then factorise: \(x(x - 5) = 0\).
When you take the square root of both sides, you must write \(\pm\). If \((x + 3)^2 = 7\), then \(x + 3 = +\sqrt{7}\) or \(x + 3 = -\sqrt{7}\). Missing the negative root loses half the answer.
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