Quadratics at A-Level
You already know how to solve quadratics from GCSE — by factorising, completing the square, and using the quadratic formula. A-Level Maths uses these as a starting point and pushes further: completing the square for any value of \(a\) (not just \(a = 1\)), using the discriminant to find an unknown coefficient, proving an expression is always positive, and sketching a parabola directly from its algebraic form.
Five skills make up this topic:
- Solving — factorising and the quadratic formula, including \(a \neq 1\)
- Completing the square — in general form, to find the turning point
- The discriminant — determining the number of real roots, and finding unknown coefficients
- Always-positive proofs — using a completed square to prove an expression is never negative
- Sketching — showing the roots, y-intercept and turning point on one diagram
Solving quadratic equations
For \(ax^2 + bx + c = 0\), the quadratic formula is \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). It always works, even when the expression won't factorise.
Worked Example 1
Solve \(2x^2 + 5x - 3 = 0\).
1
Identify \(a = 2\), \(b = 5\), \(c = -3\)
2
Discriminant: \(b^2 - 4ac = 25 - 4(2)(-3) = 25 + 24 = 49\)
3
\(x = \dfrac{-5 \pm \sqrt{49}}{4} = \dfrac{-5 \pm 7}{4}\)
4
\(x = \dfrac{2}{4} = 0.5\), or \(x = \dfrac{-12}{4} = -3\)
Answer\(x = 0.5\) or \(x = -3\)
Completing the square
For \(a \neq 1\), factor \(a\) out of the \(x^2\) and \(x\) terms first, complete the square inside the bracket, then multiply back out. The result, \(a(x + p)^2 + q\), gives the turning point \((-p, q)\) directly.
Worked Example 2
Write \(2x^2 - 8x + 3\) in the form \(a(x + p)^2 + q\), and state the minimum value of the expression.
1
Factor 2 out of the \(x^2\) and \(x\) terms: \(2(x^2 - 4x) + 3\)
2
Complete the square inside the bracket: \(x^2 - 4x = (x - 2)^2 - 4\)
3
Substitute back: \(2[(x - 2)^2 - 4] + 3 = 2(x - 2)^2 - 8 + 3\)
4
Simplify: \(2(x - 2)^2 - 5\)
Answer\(2(x - 2)^2 - 5\); minimum value \(-5\) at \(x = 2\)
The discriminant
The discriminant, \(b^2 - 4ac\), tells you how many real roots a quadratic has without solving it:
| Discriminant | Number of real roots |
|---|---|
| \(b^2 - 4ac > 0\) | Two distinct real roots |
| \(b^2 - 4ac = 0\) | One repeated root (the curve touches the x-axis) |
| \(b^2 - 4ac < 0\) | No real roots |
At A-Level, the discriminant is also used the other way round — to find the values of an unknown coefficient that give a particular type of root.
Worked Example 3
Find the values of \(k\) for which \(x^2 + kx + 9 = 0\) has equal roots.
1
Equal roots means the discriminant is zero: \(k^2 - 4(1)(9) = 0\)
2
\(k^2 - 36 = 0\)
3
\(k^2 = 36\)
Answer\(k = 6\) or \(k = -6\)
Proving a quadratic is always positive
Completing the square proves an expression is always positive (or always negative) for every real value of x, because a squared term can never be negative.
Worked Example 4
Prove that \(x^2 - 6x + 11\) is positive for all real values of \(x\).
1
Complete the square: \(x^2 - 6x + 11 = (x - 3)^2 - 9 + 11 = (x - 3)^2 + 2\)
2
\((x - 3)^2 \geq 0\) for every real \(x\), since a square can never be negative
3
So \((x - 3)^2 + 2 \geq 2\), which is always positive
Answer\(x^2 - 6x + 11 = (x - 3)^2 + 2 \geq 2 > 0\) for all real \(x\)
Sketching quadratic graphs
A good sketch shows three things clearly: where the curve crosses the x-axis (the roots), where it crosses the y-axis, and its turning point (from the completed-square form).
Worked Example 5
Sketch the graph of \(y = x^2 - 2x - 8\), showing the roots, the y-intercept and the turning point.
1
Roots \((y = 0)\): \(x^2 - 2x - 8 = (x - 4)(x + 2) = 0\), so \(x = 4\) or \(x = -2\)
2
y-intercept \((x = 0)\): \(y = -8\)
3
Turning point: \(x^2 - 2x - 8 = (x - 1)^2 - 1 - 8 = (x - 1)^2 - 9\), so the vertex is \((1, -9)\)
AnswerRoots \((-2, 0)\) and \((4, 0)\); y-intercept \((0, -8)\); minimum \((1, -9)\)
The parabola \(y = x^2 - 2x - 8\), showing both roots, the y-intercept and the turning point found in Worked Example 5.
Worked Example 6
Find the range of values of \(k\) for which \(3x^2 + kx + 3 = 0\) has no real roots.
1
No real roots means the discriminant is negative: \(k^2 - 4(3)(3) < 0\)
2
\(k^2 - 36 < 0\)
3
\(k^2 < 36\)
4
Taking the square root of an inequality like this gives a range either side of zero
Answer\(-6 < k < 6\)
Exam tips
💡 Tip 1
"Equal roots", "repeated root" and "tangent to the x-axis" all mean discriminant = 0
Exam questions phrase this condition in several different ways — learn to recognise all of them as the same instruction: set \(b^2 - 4ac = 0\).
💡 Tip 2
Read off the turning point directly from \(a(x + p)^2 + q\)
The vertex is \((-p, q)\) — note the sign flip on \(p\). For \(2(x - 2)^2 - 5\), the turning point is \((2, -5)\), not \((-2, -5)\).
💡 Tip 3
State the reason in "prove always positive" questions
Don't just write the completed square — explicitly state that a squared term is never negative, then conclude. This reasoning line carries marks on its own.
💡 Tip 4
Sketches don't need a table of values
Once you have the roots, y-intercept and turning point, you have enough to sketch an accurate parabola — you don't need to plot extra points.
💡 Tip 5
Squaring an inequality flips it if both sides could be negative
When solving \(k^2 < 36\), don't just take the square root of both sides — the correct result is \(-6 < k < 6\), not \(k < 6\) alone. Sketching \(y = k^2 - 36\) as a quick mental check avoids this error.
Common mistakes
Common Mistake 1
Forgetting to multiply back out after completing the square with \(a \neq 1\)
For \(2x^2 - 8x + 3\), writing \(2(x - 2)^2 - 4 + 3\) forgets that the \(-4\) inside the bracket was itself multiplied by 2. Correctly: \(2[(x - 2)^2 - 4] + 3 = 2(x - 2)^2 - 8 + 3 =\) \(2(x - 2)^2 - 5\).
Common Mistake 2
Reading the turning point sign incorrectly
For \(y = (x + 3)^2 - 7\), the turning point is \((-3, -7)\), not \((3, -7)\) — the x-coordinate is the value that makes the bracket zero, which is the opposite sign of the number inside it.
Common Mistake 3
Using the wrong sign convention in the discriminant
For \(ax^2 + bx + c\), the discriminant is \(b^2 - 4ac\), using the coefficients exactly as they appear including their signs. For \(x^2 - 5x + 6\), this is \((-5)^2 - 4(1)(6) = 25 - 24 = 1\), not \(-5^2 - 24\).
Common Mistake 4
Only finding one solution to \(k^2 = 36\)
\(k^2 = 36\) has two solutions, \(k = 6\) and \(k = -6\) — square rooting both sides always gives a \(\pm\) result. Dropping the negative solution loses marks in "find the values of \(k\)" questions.
Common Mistake 5
Sketching a parabola the wrong way up
If the coefficient of \(x^2\) is negative, the parabola opens downward (a maximum, not a minimum). Always check the sign of \(a\) before sketching — \(y = 6 + x - x^2\) has a maximum, not a minimum, because \(a = -1\).
Practice questions
Work through each question before checking the worked solution.
Core Skills
Q1Solve by factorising: \(x^2 + 7x + 12 = 0\).Core Skills
Q2Solve using the quadratic formula: \(x^2 - 3x - 10 = 0\).Core Skills
Q3Complete the square for \(x^2 + 6x + 2\).Core Skills
Q4Calculate the discriminant of \(2x^2 - 3x + 5\) and state the number of real roots.Core Skills
Q5Write down the coordinates of the turning point of \(y = (x - 4)^2 + 1\).Core Skills
Exam-Style
Q6Complete the square for \(3x^2 + 12x - 2\), and state the minimum value of the expression.Exam-Style
Q7Find the values of \(k\) for which \(x^2 + kx + 16 = 0\) has equal roots.Exam-Style
Q8Show that \(x^2 + 4x + 9\) is positive for all real values of \(x\).Exam-Style
Q9Sketch \(y = x^2 + 2x - 3\), stating the roots, y-intercept and turning point.Exam-Style
Q10Find the range of values of \(k\) for which \(x^2 + 6x + k = 0\) has two distinct real roots.Exam-Style
A* Challenge
Q11Find the values of \(k\) for which the line \(y = kx + 4\) is a tangent to the curve \(y = x^2 + 3x + 7\).A* Challenge
Q12Show that \(x^2 + (k + 3)x + k = 0\) has two distinct real roots for all values of \(k\).A* Challenge
Q13Sketch \(y = 6 + x - x^2\), stating the roots, y-intercept and turning point.A* Challenge
Q14The curve \(y = x^2 + px + q\) has a minimum point at \((3, -5)\). Find the values of \(p\) and \(q\).A* Challenge
Q15Given \(k \neq 0\), find the values of \(k\) for which the curve \(y = kx^2 - 4x + k\) touches the x-axis.A* Challenge
Answers — full worked solutions
Core Skills (Q1–Q5)
Q1 — \(x = -3\) or \(x = -4\)
1
Find two numbers that multiply to 12 and add to 7: 3 and 4
2
\((x + 3)(x + 4) = 0\)
\(x = -3\) or \(x = -4\)
Q2 — \(x = 5\) or \(x = -2\)
1
\(a = 1\), \(b = -3\), \(c = -10\). Discriminant \(= 9 + 40 = 49\)
2
\(x = \dfrac{3 \pm 7}{2}\)
\(x = 5\) or \(x = -2\)
Q3 — \((x + 3)^2 - 7\)
1
Half of 6 is 3: \((x + 3)^2 = x^2 + 6x + 9\)
2
Subtract the extra 9, keeping the original +2: \((x + 3)^2 - 9 + 2\)
\((x + 3)^2 - 7\)
Q4 — Discriminant \(= -31\), no real roots
1
\(a = 2\), \(b = -3\), \(c = 5\): \((-3)^2 - 4(2)(5) = 9 - 40\)
Discriminant \(= -31 < 0\), so no real roots
Q5 — \((4, 1)\)
1
For \(a(x + p)^2 + q\), the vertex is \((-p, q)\). Here \(p = -4\), \(q = 1\)
\((4, 1)\)
Exam-Style (Q6–Q10)
Q6 — \(3(x + 2)^2 - 14\); minimum \(-14\)
1
Factor 3 out: \(3(x^2 + 4x) - 2\)
2
Complete the square: \(x^2 + 4x = (x + 2)^2 - 4\)
3
\(3[(x + 2)^2 - 4] - 2 = 3(x + 2)^2 - 12 - 2\)
\(3(x + 2)^2 - 14\); minimum value \(-14\) at \(x = -2\)
Q7 — \(k = 8\) or \(k = -8\)
1
Equal roots: \(k^2 - 4(1)(16) = 0\)
2
\(k^2 = 64\)
\(k = 8\) or \(k = -8\)
Q8 — \((x + 2)^2 + 5 \geq 5 > 0\)
1
Complete the square: \(x^2 + 4x + 9 = (x + 2)^2 - 4 + 9\)
2
\(= (x + 2)^2 + 5\)
3
\((x + 2)^2 \geq 0\) for all real \(x\), so the expression \(\geq 5\)
\((x + 2)^2 + 5 \geq 5 > 0\) for all real \(x\)
Q9 — Roots \((-3, 0)\), \((1, 0)\); y-int \((0, -3)\); min \((-1, -4)\)
1
Roots: \(x^2 + 2x - 3 = (x + 3)(x - 1) = 0\), so \(x = -3\) or \(x = 1\)
2
y-intercept: \(x = 0\) gives \(y = -3\)
3
Turning point: \(x^2 + 2x - 3 = (x + 1)^2 - 1 - 3 = (x + 1)^2 - 4\), vertex \((-1, -4)\)
Roots \((-3, 0)\) and \((1, 0)\); y-intercept \((0, -3)\); minimum \((-1, -4)\)
Q10 — \(k < 9\)
1
Two distinct real roots: discriminant \(> 0\): \(36 - 4k > 0\)
2
\(36 > 4k\)
\(k < 9\)
A* Challenge (Q11–Q15)
Q11 — \(k = 3 + 2\sqrt{3}\) or \(k = 3 - 2\sqrt{3}\)
1
A tangent touches at exactly one point, so set the curves equal: \(x^2 + 3x + 7 = kx + 4\)
2
Rearrange: \(x^2 + (3 - k)x + 3 = 0\)
3
Tangency means discriminant = 0: \((3 - k)^2 - 4(1)(3) = 0\)
4
\((3 - k)^2 = 12\), so \(3 - k = \pm 2\sqrt{3}\)
\(k = 3 - 2\sqrt{3}\) or \(k = 3 + 2\sqrt{3}\)
Q12 — Discriminant \(= (k + 1)^2 + 8\), always positive
1
Discriminant: \((k + 3)^2 - 4(1)(k) = k^2 + 6k + 9 - 4k\)
2
\(= k^2 + 2k + 9\)
3
Complete the square in \(k\): \((k + 1)^2 - 1 + 9 = (k + 1)^2 + 8\)
4
\((k + 1)^2 \geq 0\) for every real \(k\), so the discriminant \(\geq 8 > 0\) always
Discriminant \(= (k + 1)^2 + 8 \geq 8 > 0\), so there are always two distinct real roots
Q13 — Roots \((-2, 0)\), \((3, 0)\); y-int \((0, 6)\); max \((0.5, 6.25)\)
1
Rewrite: \(y = -x^2 + x + 6 = -(x^2 - x - 6) = -(x - 3)(x + 2)\)
2
Roots: \(x = 3\) or \(x = -2\). y-intercept: \(x = 0\) gives \(y = 6\)
3
Turning point: \(-x^2 + x + 6 = -(x - 0.5)^2 + 0.25 + 6 = -(x - 0.5)^2 + 6.25\)
4
Since \(a = -1 < 0\), this is a maximum
Roots \((-2, 0)\) and \((3, 0)\); y-intercept \((0, 6)\); maximum \((0.5, 6.25)\)
Q14 — \(p = -6\), \(q = 4\)
1
A minimum at \((3, -5)\) means the vertex form is \(y = (x - 3)^2 - 5\)
2
Expand: \((x - 3)^2 - 5 = x^2 - 6x + 9 - 5\)
3
\(= x^2 - 6x + 4\)
\(p = -6\), \(q = 4\)
Q15 — \(k = 2\) or \(k = -2\)
1
"Touches the x-axis" means the discriminant is zero, with \(a = k\), \(b = -4\), \(c = k\)
2
\((-4)^2 - 4(k)(k) = 0\)
3
\(16 - 4k^2 = 0\), so \(k^2 = 4\)
\(k = 2\) or \(k = -2\)
Summary
- Solve any quadratic with the formula \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) — it works even when factorising doesn't.
- Complete the square for \(a \neq 1\) by factoring \(a\) out first; the vertex of \(a(x + p)^2 + q\) is \((-p, q)\).
- The discriminant \(b^2 - 4ac\) tells you the number of real roots: positive (two), zero (one repeated), negative (none).
- A completed square proves an expression is always positive, since a squared term can never be negative.
- Sketch a parabola from its roots, y-intercept and turning point — no table of values needed.
- Check the sign of a before sketching: positive opens upward (minimum), negative opens downward (maximum).
These techniques underpin the next topic in Pure Year 1, Equations & Inequalities, and reuse the algebraic manipulation skills from earlier in the course.
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