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Quadratic functions

Graphs, the discriminant, completing the square, solving quadratic equations, roots and turning points.

Spec MM1.3, M4.5, M4.11-M4.14Section 1 Part 1: AS-level pure mathematics26 questions18 real past questions
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1Quadratic functions · spec MM1.3 · Paper 1 style · ★☆☆

The equation \(x^2 + (k-2)x + (k+1) = 0\) has two distinct real roots. Find the complete set of possible values of \(k\).

  1. A\(0 < k < 8\)
  2. B\(-2 < k < 4\)
  3. C\(k < -2\) or \(k > 4\)
  4. D\(k < 0\) or \(k > 4\)
  5. E\(k < 0\) or \(k > 8\)
  6. Fall real \(k\)
Worked solution

Discriminant \((k-2)^2 - 4(k+1) = k^2 - 8k = k(k-8) > 0\), so \(k < 0\) or \(k > 8\).

Full method — how and why
'Two distinct real roots' is always a discriminant condition. The unknown is in the coefficients, so the discriminant becomes a quadratic in \(k\) and you finish with an inequality.
  1. Identify \(a = 1\), \(b = k - 2\), \(c = k + 1\) and write the discriminant: \((k-2)^2 - 4(k+1)\).
    'Two distinct real roots' \(\Leftrightarrow\) \(b^2 - 4ac > 0\). Everything else in the question is decoration.
  2. Expand and simplify: \(k^2 - 4k + 4 - 4k - 4 = k^2 - 8k\).
    The constants cancel, which is the sign you have expanded correctly.
  3. Factorise and solve \(k(k - 8) > 0\).
    A quadratic inequality needs its roots and its shape: roots \(0\) and \(8\), opening upwards.
  4. The upward parabola is positive outside its roots: \(k < 0\) or \(k > 8\).
    Sketch it if in doubt - 'outside the roots' for \(> 0\), 'between the roots' for \(< 0\).
Where it goes wrong
  • Expanding \((k-2)^2\) as \(k^2 - 4\), which gives \(k^2 - 4k - 8\) and a different answer set.
  • Reading the sign chart backwards and choosing \(0 < k < 8\) - that is where the discriminant is negative and there are no real roots.
Takeaway Conditions on the number of roots are conditions on \(b^2 - 4ac\); if the coefficients contain a parameter, you end up solving a quadratic inequality in that parameter.
Answer E · Quadratic functions (MM1.3) · question QUA-01
2Quadratic functions · spec MM1.3 · Paper 1 style · ★☆☆

By completing the square, find the minimum value of \(2x^2 - 12x + 23\).

  1. A\(-13\)
  2. B\(5\)
  3. C\(23\)
  4. D\(7\)
  5. E\(-5\)
  6. F\(3\)
Worked solution

\(2x^2 - 12x + 23 = 2(x-3)^2 - 18 + 23 = 2(x-3)^2 + 5\), so the minimum is \(5\) (at \(x = 3\)).

Full method — how and why
Minimum value of a quadratic: complete the square. The leading coefficient \(2\) must be factored out before completing the square.
  1. Factor the \(2\) out of the \(x\) terms: \(2(x^2 - 6x) + 23\).
    Completing the square only works on \(x^2 + (\text{something})x\); the coefficient of \(x^2\) must be \(1\) inside the bracket.
  2. Complete the square inside: \(x^2 - 6x = (x-3)^2 - 9\), so the expression is \(2\left[(x-3)^2 - 9\right] + 23\).
    Halve the coefficient of \(x\) (\(-6 \to -3\)) and subtract the square of that half.
  3. Multiply out the \(2\): \(2(x-3)^2 - 18 + 23 = 2(x-3)^2 + 5\).
    The \(-9\) is inside the bracket, so it is multiplied by \(2\) as well - this is where marks are lost.
  4. \(2(x-3)^2 \geq 0\), so the least value is \(5\), at \(x = 3\).
    A square is never negative, and it is zero exactly at the vertex.
Where it goes wrong
  • Forgetting to multiply the \(-9\) by \(2\), giving \(2(x-3)^2 + 14\) and the answer \(14\).
  • Quoting the \(x\)-coordinate \(3\) instead of the minimum value \(5\) - both are offered.
Takeaway Take the leading coefficient outside first, complete the square, then multiply back in; the constant left outside is the minimum.
Answer B · Quadratic functions (MM1.3) · question QUA-02
3Quadratic functions · spec MM1.3 · Paper 1 style · ★☆☆

The quadratic \(y = ax^2 + bx + c\) has its vertex at \((2, -3)\) and passes through \((0, 5)\). Find \(a + b + c\).

  1. A\(-1\)
  2. B\(7\)
  3. C\(1\)
  4. D\(5\)
  5. E\(-3\)
  6. F\(3\)
Worked solution

\(y = a(x-2)^2 - 3\); at \(x = 0\): \(4a - 3 = 5\), so \(a = 2\). Then \(y = 2(x-2)^2 - 3 = 2x^2 - 8x + 5\), and \(a + b + c = 2 - 8 + 5 = -1\) (which is just \(y(1)\)).

Full method — how and why
Given a vertex, the completed-square form is the natural one. The clever final step is noticing that \(a + b + c\) is just \(y(1)\).
  1. Write the completed-square form using the vertex: \(y = a(x - 2)^2 - 3\).
    The vertex \((h, k)\) appears directly in \(y = a(x-h)^2 + k\), so two of the three unknowns are already known.
  2. Use the point \((0, 5)\): \(a(0-2)^2 - 3 = 5 \Rightarrow 4a = 8 \Rightarrow a = 2\).
    One extra point is exactly what is needed to pin down the remaining constant.
  3. Note that \(a + b + c\) is the value of \(ax^2 + bx + c\) at \(x = 1\).
    Substituting \(x = 1\) makes every power of \(x\) equal to \(1\) - a standard shortcut worth remembering.
  4. Evaluate: \(y(1) = 2(1-2)^2 - 3 = 2 - 3 = -1\).
    No need to expand into \(2x^2 - 8x + 5\) at all.
Where it goes wrong
  • Expanding and then mis-collecting: \(2(x-2)^2 - 3 = 2x^2 - 8x + 5\), and slips in the \(-8\) give \(3\) or \(7\).
  • Using the vertex as \((-2, -3)\) - in \(a(x - h)^2 + k\) the sign of \(h\) is already flipped.
Takeaway \(f(1)\) is the sum of the coefficients; and a given vertex means you should start from \(a(x-h)^2 + k\).
Answer A · Quadratic functions (MM1.3) · question QUA-03
4Quadratic functions · spec MM1.3 · Paper 1 style · ★☆☆

For what value of \(c\) does the line \(y = 2x + c\) touch the parabola \(y = x^2 - 4x + 7\)?

  1. A\(7\)
  2. B\(2\)
  3. C\(-2\)
  4. D\(-1\)
  5. E\(0\)
  6. F\(5\)
Worked solution

Equate: \(x^2 - 6x + (7 - c) = 0\). Tangency needs discriminant zero: \(36 - 4(7-c) = 0\), so \(8 + 4c = 0\) and \(c = -2\).

Full method — how and why
'Touches' means exactly one intersection, which after substitution means a repeated root: discriminant zero.
  1. Set the two expressions equal: \(2x + c = x^2 - 4x + 7\).
    Intersections of \(y = f(x)\) and \(y = g(x)\) are the solutions of \(f(x) = g(x)\).
  2. Collect into a standard quadratic: \(x^2 - 6x + (7 - c) = 0\).
    Everything to one side, so the discriminant test can be applied.
  3. Set the discriminant to zero: \((-6)^2 - 4(1)(7 - c) = 0\).
    Touching means a repeated root, i.e. the line meets the curve once - exactly the \(b^2 = 4ac\) case.
  4. Solve: \(36 - 28 + 4c = 0 \Rightarrow 4c = -8 \Rightarrow c = -2\).
    Watch the double negative: \(-4(7 - c) = -28 + 4c\).
Where it goes wrong
  • Sign slip in \(-4(7-c)\) giving \(c = 2\), which is offered.
  • Using \(b^2 - 4ac > 0\) (crossing) rather than \(= 0\) (touching).
Takeaway Tangency between a line and a curve: substitute, then set the discriminant of the resulting quadratic to zero.
Answer C · Quadratic functions (MM1.3) · question QUA-05
5Quadratic functions · spec MM1.3 · Paper 1 style · ★☆☆

How many real solutions does the equation \(x^4 - 5x^2 + 4 = 0\) have?

  1. A\(2\)
  2. B\(4\)
  3. C\(5\)
  4. D\(0\)
  5. E\(1\)
  6. F\(3\)
Worked solution

Let \(u = x^2\): \(u^2 - 5u + 4 = (u-1)(u-4) = 0\), so \(x^2 = 1\) or \(x^2 = 4\), giving \(x = \pm1, \pm2\): four solutions.

Full method — how and why
A quartic that is quadratic in \(x^2\). The trap is counting the solutions of the substituted equation rather than of the original.
  1. Substitute \(u = x^2\): \(u^2 - 5u + 4 = 0\).
    Only even powers of \(x\) appear, which is what makes the substitution work.
  2. Factorise: \((u - 1)(u - 4) = 0\), so \(u = 1\) or \(u = 4\).
    Two values of \(u\) - but \(u\) is not the answer.
  3. Undo the substitution: \(x^2 = 1\) gives \(x = \pm1\); \(x^2 = 4\) gives \(x = \pm2\).
    Each positive value of \(u\) produces two values of \(x\); a negative value would produce none.
  4. Count: four real solutions.
    The question asks how many, not what they are.
Where it goes wrong
  • Answering \(2\) because the quadratic in \(u\) has two roots.
  • Forgetting the negative square roots and answering \(2\) again.
Takeaway After a substitution, always translate back: each positive value of \(x^2\) gives two values of \(x\), zero gives one, negative gives none.
Answer B · Quadratic functions (MM1.3) · question QUA-07
6Quadratic functions · spec MM1.3 · Paper 1 style · ★☆☆

The equation \(kx^2 + 4x + k = 0\) has no real roots. Find the complete set of values of \(k\).

  1. A\(0 < k < 2\)
  2. B\(k > 2\)
  3. C\(k < -2\) or \(k > 2\)
  4. D\(-2 < k < 2\)
  5. E\(k < -2\)
  6. F\(-2 < k < 0\) or \(0 < k < 2\)
Worked solution

Need \(16 - 4k^2 < 0\), i.e. \(k^2 > 4\), so \(k < -2\) or \(k > 2\). (If \(k = 0\) the equation is linear with one root, so \(k = 0\) is excluded anyway.)

Full method — how and why
'No real roots' is the discriminant \(< 0\). The extra subtlety is the coefficient \(k\) appearing in the \(x^2\) term, so \(k = 0\) needs a thought.
  1. Write the discriminant: \(4^2 - 4(k)(k) = 16 - 4k^2\).
    Here \(a = k\) and \(c = k\) - both the first and last coefficients are the parameter.
  2. Set it negative: \(16 - 4k^2 < 0 \Rightarrow k^2 > 4\).
    No real roots \(\Leftrightarrow b^2 - 4ac < 0\).
  3. Solve \(k^2 > 4\): \(k < -2\) or \(k > 2\).
    \(k^2 > 4\) means \(|k| > 2\), which is two separate intervals, not \(-2 < k < 2\).
  4. Check \(k = 0\) separately: the equation becomes \(4x = 0\), which has the root \(x = 0\).
    If \(k = 0\) it is not a quadratic at all, so the discriminant test does not apply - but \(k = 0\) is excluded by the answer anyway.
Where it goes wrong
  • Solving \(k^2 > 4\) as \(-2 < k < 2\) - that is the solution of \(k^2 < 4\).
  • Ignoring the possibility \(k = 0\) entirely (harmless here, fatal in questions where \(0\) lies inside the answer set).
Takeaway \(k^2 > c\) gives two outer intervals; and whenever the leading coefficient contains the parameter, check the value that makes it zero.
Answer C · Quadratic functions (MM1.3) · question QUA-08
7Quadratic functions · spec MM1.3 · Paper 1 style · ★☆☆

The quadratic \(x^2 + px + q\) has minimum value \(-4\) and one of its roots is \(0\). Which of the following could be the quadratic?

  1. A\(x^2 - 6x + 8\)
  2. B\(x^2 - 2x - 3\)
  3. C\(x^2 + 2x - 8\)
  4. D\(x^2 - 4x\)
  5. E\(x^2 + 6x + 5\)
Worked solution

Minimum \(-4\) means \(x^2 + px + q = (x-h)^2 - 4\), with roots \(h \pm 2\). One root is \(0\), so \(h = \pm 2\): the quadratic is \(x^2 - 4x\) or \(x^2 + 4x\). Only \(x^2 - 4x\) is offered.

Full method — how and why
Two pieces of information about a quadratic - its minimum and one root - pin the whole thing down. Working in completed-square form makes the roots visible.
  1. Minimum \(-4\) and leading coefficient \(1\) means \(x^2 + px + q = (x - h)^2 - 4\) for some \(h\).
    A monic quadratic with minimum \(m\) must be \((x - h)^2 + m\); only the position \(h\) is unknown.
  2. Its roots satisfy \((x - h)^2 = 4\), i.e. \(x = h \pm 2\).
    Solve the completed-square form directly rather than expanding.
  3. One root is \(0\), so \(h + 2 = 0\) or \(h - 2 = 0\): \(h = -2\) or \(h = 2\).
    Both cases are legitimate; the options decide which one is wanted.
  4. That gives \(x^2 + 4x\) or \(x^2 - 4x\); the list offers \(x^2 - 4x\).
    Check it: \(x^2 - 4x = (x-2)^2 - 4\), minimum \(-4\), roots \(0\) and \(4\).
Where it goes wrong
  • Picking \(x^2 - 2x - 3\): its minimum is \(-4\) but its roots are \(-1\) and \(3\), so \(0\) is not a root.
  • Checking only the minimum (several options have minimum \(-4\)) or only the roots.
Takeaway Test every stated condition - an option that satisfies one of them is designed to catch you.
Answer D · Quadratic functions (MM1.3) · question QUA-10
8Quadratic functions · spec MM1.3 · Paper 1 style · ★☆☆

The curve \(y = x^2 - 2ax + a^2 + a - 6\) has its vertex on the \(x\)-axis. Find the value of \(a\).

  1. A\(0\)
  2. B\(6\)
  3. C\(-6\)
  4. D\(3\)
  5. E\(2\)
  6. F\(-2\)
Worked solution

Complete the square: \(y = (x-a)^2 + a - 6\). The vertex is \((a, a-6)\), which lies on the \(x\)-axis when \(a = 6\).

Full method — how and why
'Vertex on the \(x\)-axis' means the minimum value is zero, i.e. the quadratic has a repeated root. Completing the square is quicker than the discriminant here.
  1. Complete the square in \(x\), treating \(a\) as a constant: \(x^2 - 2ax + a^2 = (x - a)^2\), so \(y = (x-a)^2 + (a - 6)\).
    The first three terms are already a perfect square - that is why the question is built with \(a^2\) in it.
  2. Read off the vertex: \((a,\; a - 6)\).
    In \((x - h)^2 + k\) the vertex is \((h, k)\).
  3. On the \(x\)-axis means the \(y\)-coordinate is zero: \(a - 6 = 0\), so \(a = 6\).
    'On the \(x\)-axis' is a statement about \(y\), not \(x\) - the commonest misread in this question.
Where it goes wrong
  • Setting the \(x\)-coordinate to zero and answering \(a = 0\).
  • Using the discriminant but mis-expanding \(4a^2 - 4(a^2 + a - 6)\), which should collapse to \(-4(a - 6)\).
Takeaway Vertex on the \(x\)-axis \(\Leftrightarrow\) repeated root \(\Leftrightarrow\) discriminant zero - three descriptions of the same situation.
Answer B · Quadratic functions (MM1.3) · question QUA-11
9Quadratic functions · spec MM1.3 · Paper 1 style · ★☆☆

The roots of \(2x^2 + bx + 18 = 0\) are equal. Find the sum of the possible values of \(b\).

  1. A\(36\)
  2. B\(6\)
  3. C\(0\)
  4. D\(12\)
  5. E\(24\)
  6. F\(-12\)
Worked solution

\(b^2 - 144 = 0\), so \(b = \pm 12\); their sum is \(0\).

Full method — how and why
'Equal roots' is the discriminant-zero condition; the twist is that it gives two values of \(b\) and the question wants their sum.
  1. Write the discriminant with \(a = 2\), \(c = 18\): \(b^2 - 4(2)(18) = b^2 - 144\).
    Substitute carefully - \(4ac\) is \(144\), not \(36\).
  2. Set it to zero: \(b^2 = 144\), so \(b = 12\) or \(b = -12\).
    Equal (repeated) roots means the discriminant vanishes; a square equal to a positive number always has two solutions.
  3. Add them: \(12 + (-12) = 0\).
    The two values are symmetric about zero, so the sum is \(0\) without any further work.
Where it goes wrong
  • Giving only the positive root and answering \(12\).
  • Reading 'sum of the possible values of \(b\)' as 'the value of \(b\)'.
Takeaway \(b^2 = k\) always has two solutions \(\pm\sqrt k\); read whether the question wants one of them, both, or a combination.
Answer C · Quadratic functions (MM1.3) · question QUA-12
10Quadratic functions · spec MM1.3 · Paper 1 style · ★☆☆

Which of the following quadratics has its vertex in the second quadrant (where \(x < 0\) and \(y > 0\))?

  1. A\(y = -x^2 - 2x - 3\)
  2. B\(y = -x^2 + 2x + 1\)
  3. C\(y = x^2 - 4x + 5\)
  4. D\(y = x^2 + 6x + 10\)
  5. E\(y = x^2 + 4x + 1\)
Worked solution

Complete the square: A vertex \((-2, -3)\); B \((2, 1)\); C \((-1, -2)\); D \((-3, 1)\), which has \(x < 0\) and \(y > 0\); E \((1, 2)\). So D.

Full method — how and why
A 'which one' question about vertices. Complete the square on each option - but use the sign of the \(x\)-coordinate to eliminate most of them first.
  1. Recall that for \(y = x^2 + bx + c\) the vertex is at \(x = -\frac{b}{2}\).
    You only need the vertex position, not the whole completed square, to test the first condition \(x < 0\).
  2. Second quadrant needs \(x < 0\), so \(-\frac b2 < 0\), i.e. \(b > 0\): only \(y = x^2 + 4x + 1\) and \(y = x^2 + 6x + 10\) survive among the upward parabolas.
    Eliminating on one condition first halves the work.
  3. Compute the \(y\)-coordinates: \(x^2 + 4x + 1 = (x+2)^2 - 3\), vertex \((-2, -3)\): fails \(y > 0\). \(x^2 + 6x + 10 = (x+3)^2 + 1\), vertex \((-3, 1)\): passes.
    Now apply the second condition to the short list.
  4. Check the downward ones quickly: \(-x^2 - 2x - 3\) has vertex \((-1, -2)\) and \(-x^2 + 2x + 1\) has vertex \((1, 2)\) - neither is in the second quadrant.
    Worth 10 seconds of confirmation on a 'which one' question.
Where it goes wrong
  • Reading the vertex of \((x+3)^2 + 1\) as \((3, 1)\) - the sign flips.
  • Mixing up the quadrants: the second quadrant is up and to the left.
Takeaway On 'which of these' questions, test the cheapest condition first and eliminate before doing any heavy algebra.
Answer D · Quadratic functions (MM1.3) · question QUA-13
11Quadratic functions · spec MM1.3 · Paper 1 style · ★☆☆

The graph of \(y = x^2 + bx + c\) crosses the \(x\)-axis at \((1, 0)\) and \((5, 0)\). What is the \(y\)-coordinate of the vertex?

  1. A\(-9\)
  2. B\(5\)
  3. C\(-4\)
  4. D\(-3\)
  5. E\(4\)
  6. F\(-5\)
Worked solution

\(y = (x-1)(x-5)\), vertex at \(x = 3\): \(y = (2)(-2) = -4\).

Full method — how and why
Given the two roots, use the factorised form and the symmetry of the parabola - no expansion needed.
  1. Write the factorised form: \(y = (x - 1)(x - 5)\).
    Roots at \(1\) and \(5\) with leading coefficient \(1\) (given, since the equation is \(x^2 + bx + c\)).
  2. The vertex lies midway between the roots: \(x = \frac{1 + 5}{2} = 3\).
    A parabola is symmetric, so its turning point sits on the axis of symmetry, exactly halfway between the roots.
  3. Evaluate there: \(y = (3-1)(3-5) = 2 \times (-2) = -4\).
    Substituting into the factorised form is far quicker than expanding to \(x^2 - 6x + 5\).
Where it goes wrong
  • Averaging the roots but then substituting into the wrong expression, e.g. getting \(-9\) from \((3)^2 - 6(3) + 5\) miscomputed.
  • Giving the \(x\)-coordinate \(3\) instead of the \(y\)-coordinate.
Takeaway Midpoint of the roots gives the axis of symmetry; substitute into the factorised form for the vertex height.
Answer C · Quadratic functions (MM1.3) · question QUA-15
12Quadratic functions · spec MM1.3 · Paper 1 style · ★☆☆

Given that \(x^2 + y^2 = 10\) and \(x + y = 4\), find the value of \(xy\).

  1. A\(4\)
  2. B\(2\)
  3. C\(3\)
  4. D\(6\)
  5. E\(-3\)
  6. F\(5\)
Worked solution

\((x+y)^2 = x^2 + y^2 + 2xy\), so \(16 = 10 + 2xy\) and \(xy = 3\).

Full method — how and why
A symmetric pair of equations. Expanding \((x+y)^2\) links the two given quantities to \(xy\) directly.
  1. Write the identity \((x + y)^2 = x^2 + y^2 + 2xy\).
    This is the only identity that connects a sum, a sum of squares and a product.
  2. Substitute the given values: \(4^2 = 10 + 2xy\).
    \(x + y = 4\) and \(x^2 + y^2 = 10\) are both given, so only \(xy\) is unknown.
  3. Solve: \(16 - 10 = 2xy \Rightarrow xy = 3\).
    Halve at the end - forgetting to divide by \(2\) gives \(6\), which is offered.
Where it goes wrong
  • Answering \(6\) by omitting the factor \(2\) in \(2xy\).
  • Trying to solve for \(x\) and \(y\) individually (they are \(1\) and \(3\)) - slower and unnecessary.
Takeaway \((x+y)^2 = x^2 + y^2 + 2xy\) converts between the three symmetric quantities; recognise which two you are given.
Answer C · Quadratic functions (MM1.3) · question QUA-16
13Quadratic functions · spec MM1.3 · Paper 1 style · ★☆☆

Find the range of values of \(x\) for which \(x^2 - 7x + 10 < 0\) and \(x^2 - 4 > 0\).

  1. A\(2 < x < 5\)
  2. B\(-2 < x < 2\)
  3. C\(x > 5\)
  4. D\(x < -2\)
  5. E\(2 < x < 5\) or \(x < -2\)
  6. Fno values
Worked solution

\((x-2)(x-5) < 0\) gives \(2 < x < 5\); \(x^2 > 4\) gives \(x < -2\) or \(x > 2\). The intersection is \(2 < x < 5\).

Full method — how and why
Two simultaneous quadratic inequalities. Solve each separately, then intersect on a number line.
  1. First: \(x^2 - 7x + 10 < 0\) factorises as \((x-2)(x-5) < 0\), giving \(2 < x < 5\).
    An upward parabola is below the axis between its roots.
  2. Second: \(x^2 - 4 > 0\) gives \(x < -2\) or \(x > 2\).
    Above the axis means outside the roots - two intervals.
  3. Intersect: \((2, 5)\) overlaps \((2, \infty)\) entirely and misses \((-\infty, -2)\) completely, so the answer is \(2 < x < 5\).
    'And' means intersection: a value must satisfy both, so draw both solution sets on one line and take the common part.
Where it goes wrong
  • Taking the union and including \(x < -2\) - the word is 'and', not 'or'.
  • Getting the first inequality backwards (\(x < 2\) or \(x > 5\)) by forgetting that \(< 0\) means between the roots.
Takeaway \(<0\) between the roots, \(>0\) outside them (for an upward parabola); then combine solution sets on a number line.
Answer A · Quadratic functions (MM1.3) · question QUA-18
14Quadratic functions · spec MM1.3 · Paper 1 style · ★☆☆

The expression \(x^2 + 6x + c\) can be written as \((x + p)^2 + q\). Given that \(q = -c\), find the value of \(c\).

  1. A\(9\)
  2. B\(3\)
  3. C\(-\dfrac{9}{2}\)
  4. D\(\dfrac{9}{2}\)
  5. E\(0\)
  6. F\(-9\)
Worked solution

\((x+3)^2 + (c - 9)\), so \(q = c - 9 = -c\), giving \(2c = 9\) and \(c = \frac92\).

Full method — how and why
Completing the square symbolically, then imposing a relationship between the two constants. It is an equation in \(c\), not in \(x\).
  1. Complete the square: \(x^2 + 6x + c = (x + 3)^2 - 9 + c\).
    Half of \(6\) is \(3\); subtracting \(9\) compensates for the \(+9\) the square introduces.
  2. Identify \(p = 3\) and \(q = c - 9\).
    Compare with the required form \((x+p)^2 + q\) term by term.
  3. Impose \(q = -c\): \(c - 9 = -c\).
    The condition given in the question is a link between \(q\) and \(c\), so substitute your expression for \(q\).
  4. Solve: \(2c = 9\), so \(c = \frac92\).
    A simple linear equation - just collect the \(c\) terms.
Where it goes wrong
  • Writing \(q = c + 9\) (sign error) and getting \(c = -\frac92\).
  • Solving \(c - 9 = c\) and concluding there is no solution.
Takeaway Complete the square with the constant left as a letter; the extra condition then becomes an ordinary equation in that letter.
Answer D · Quadratic functions (MM1.3) · question QUA-19
15Quadratic functions · spec MM1.3 · Paper 1 style · ★☆☆

The parabola \(y = x^2 - 4x + 3\) is reflected in the \(x\)-axis and then translated \(2\) units in the positive \(y\)-direction. What is the maximum value of the resulting function?

  1. A\(3\)
  2. B\(1\)
  3. C\(-1\)
  4. D\(-3\)
  5. E\(5\)
  6. F\(2\)
Worked solution

\(x^2 - 4x + 3 = (x-2)^2 - 1\), minimum \(-1\). Reflecting gives maximum \(1\); translating up by \(2\) gives maximum \(3\).

Full method — how and why
Transformations applied to a quadratic. Track what happens to the vertex rather than to the whole equation.
  1. Complete the square: \(x^2 - 4x + 3 = (x - 2)^2 - 1\), so the minimum is \(-1\) at \(x = 2\).
    The turning point is the only feature the transformations act on in a way that matters.
  2. Reflect in the \(x\)-axis: \(y \to -y\), so the minimum \(-1\) becomes a maximum of \(1\).
    Reflection in the \(x\)-axis flips the sign of every \(y\)-value, turning a minimum into a maximum.
  3. Translate \(2\) up: the maximum becomes \(1 + 2 = 3\).
    A vertical translation adds to every \(y\)-value, including the extreme one.
Where it goes wrong
  • Forgetting that the reflection converts the minimum into a maximum and reporting \(-1 + 2 = 1\).
  • Reflecting the \(x\)-coordinate as well - the turning point stays at \(x = 2\) throughout.
Takeaway For transformation questions, follow one key point (usually the vertex) through each step in order.
Answer A · Quadratic functions (MM1.3) · question QUA-21
16Quadratic functions · spec MM1.3 · Paper 1 style · ★☆☆

The line \(y = mx\) meets the curve \(y = x^2 + 9\) at two distinct points. Find the complete set of values of \(m\).

  1. A\(m < -3\) or \(m > 3\)
  2. B\(m < -6\) or \(m > 6\)
  3. C\(-6 < m < 6\)
  4. D\(-3 < m < 3\)
  5. E\(m > 6\)
  6. Fall real \(m\)
Worked solution

\(x^2 - mx + 9 = 0\) needs \(m^2 - 36 > 0\), so \(m < -6\) or \(m > 6\).

Full method — how and why
A line meeting a curve twice: substitute and demand a positive discriminant.
  1. Substitute \(y = mx\) into \(y = x^2 + 9\): \(mx = x^2 + 9\).
    Intersections are the common solutions, found by equating the two expressions for \(y\).
  2. Rearrange: \(x^2 - mx + 9 = 0\).
    Standard form is needed before the discriminant can be read off.
  3. Two distinct intersections \(\Rightarrow\) discriminant \(> 0\): \(m^2 - 36 > 0\).
    Here \(b = -m\), and \(b^2 = m^2\) regardless of sign.
  4. Solve \(m^2 > 36\): \(m < -6\) or \(m > 6\).
    Two intervals, since \(m^2 > 36\) means \(|m| > 6\); geometrically, steep lines in either direction cut the parabola twice.
Where it goes wrong
  • Answering \(-6 < m < 6\) (the region where the line misses the parabola entirely).
  • Writing the discriminant as \(-m^2 - 36\) by mishandling the sign of \(b\).
Takeaway \(b^2\) kills the sign of \(b\): conditions like this are almost always symmetric in \(\pm m\).
Answer B · Quadratic functions (MM1.3) · question QUA-23
17Quadratic functions · spec MM1.3 · Paper 1 style · ★☆☆

Let \(f(x) = x^2 - 2x - 8\). For how many integers \(n\) is \(f(n) < 0\)?

  1. A\(8\)
  2. B\(6\)
  3. C\(5\)
  4. D\(7\)
  5. E\(3\)
  6. F\(4\)
Worked solution

\((x-4)(x+2) < 0\) gives \(-2 < x < 4\); the integers are \(-1, 0, 1, 2, 3\): five of them.

Full method — how and why
Counting integers in the region where a quadratic is negative. Solve the inequality, then count carefully at the ends.
  1. Factorise: \(n^2 - 2n - 8 = (n - 4)(n + 2)\).
    Two numbers multiplying to \(-8\) and adding to \(-2\): \(-4\) and \(+2\).
  2. Negative between the roots: \(-2 < n < 4\).
    Upward parabola, so \(f < 0\) strictly between the roots.
  3. List the integers strictly inside: \(-1, 0, 1, 2, 3\) - five of them.
    The endpoints \(-2\) and \(4\) give \(f = 0\), which is not less than zero, so they are excluded; and \(0\) is easy to forget.
Where it goes wrong
  • Including \(-2\) and \(4\) and answering \(7\).
  • Forgetting \(0\) is an integer and answering \(4\).
Takeaway After solving an inequality, list the integers explicitly - endpoints and zero are the two things people drop.
Answer C · Quadratic functions (MM1.3) · question QUA-24
18Quadratic functions · spec MM1.3 · Paper 2 style · ★☆☆

The equation \(x^2 + bx + c = 0\) has real coefficients. Which of the following is a sufficient condition for it to have two distinct real roots?

  1. A\(b \neq 0\)
  2. B\(b > 0\)
  3. C\(c > 0\)
  4. D\(b^2 \geq 4c\)
  5. E\(c < 0\)
Worked solution

\(c < 0\) makes the discriminant \(b^2 - 4c\) positive. \(b^2 \geq 4c\) allows equality (a repeated root); the others do not control the discriminant.

Full method — how and why
'Sufficient' means the condition alone guarantees the conclusion. Translate into the discriminant and check each option guarantees positivity.
  1. Two distinct real roots requires discriminant \(b^2-4c>0\) (strictly).
    Zero discriminant gives a repeated root, not two distinct ones.
  2. If \(c<0\) then \(-4c>0\), so \(b^2-4c \geq -4c > 0\). Guaranteed - sufficient.
    \(b^2 \geq 0\) whatever \(b\) is, so the whole expression is positive.
  3. \(b^2 \geq 4c\) allows equality, which gives a repeated root - not sufficient for distinct roots.
    Read the word 'distinct'; the boundary case ruins it.
  4. \(b \neq 0\), \(c>0\), \(b>0\) give no control: e.g. \(b=1, c=1\) has discriminant \(-3<0\), no real roots.
    One failing example shows a condition is not sufficient.
Where it goes wrong
  • Choosing \(b^2 \geq 4c\) because it looks like the discriminant condition - it is the condition for at least one real root.
  • Confusing 'sufficient' with 'necessary': \(c<0\) is not necessary (\(x^2-5x+6\) has two distinct roots with \(c=6\)).
Takeaway Distinct real roots \(\iff b^2-4c>0\). A sufficient condition must force that strictly, with no exceptions.
Answer E · Quadratic functions (MM1.3) · question X2-30
19Quadratic functions · spec MM1.3 · Paper 1 style · ★★☆

The roots of \(x^2 - 5x + 2 = 0\) are \(\alpha\) and \(\beta\). Find the value of \(\alpha^2 + \beta^2\).

  1. A\(25\)
  2. B\(29\)
  3. C\(23\)
  4. D\(27\)
  5. E\(21\)
  6. F\(17\)
Worked solution

\(\alpha + \beta = 5\) and \(\alpha\beta = 2\). So \(\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = 25 - 4 = 21\).

Full method — how and why
A symmetric function of the roots. The roots themselves are irrational, so use the sum and product rather than the quadratic formula.
  1. Read off \(\alpha + \beta = -\frac{b}{a} = 5\) and \(\alpha\beta = \frac{c}{a} = 2\).
    These come straight from the coefficients; you never need the actual roots.
  2. Use the identity \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\).
    Expanding \((\alpha+\beta)^2\) gives \(\alpha^2 + 2\alpha\beta + \beta^2\), so subtracting \(2\alpha\beta\) leaves the sum of squares.
  3. Substitute: \(25 - 2(2) = 21\).
    Arithmetic only.
Where it goes wrong
  • Using \((\alpha+\beta)^2 = \alpha^2 + \beta^2\) and answering \(25\).
  • Subtracting \(\alpha\beta\) once instead of twice, giving \(23\) - both wrong values are offered.
Takeaway Any symmetric expression in the roots can be built from \(\alpha + \beta = -b/a\) and \(\alpha\beta = c/a\); the quadratic formula is rarely needed.
Answer E · Quadratic functions (MM1.3) · question QUA-04
20Quadratic functions · spec MM1.3 · Paper 2 style · ★★☆

The function \(f(x) = x^2 - 6x + 13\) is defined for all real \(x\). Which of the following statements is/are true?

  1. I\(f(x) > 0\) for all real \(x\)
  2. IIThe graph of \(y = f(x)\) is symmetric about the line \(x = 3\)
  3. IIIThe equation \(f(x) = 4\) has exactly one real solution
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

\(f(x) = (x-3)^2 + 4 \geq 4 > 0\), so I is true. The axis of symmetry is \(x = 3\), so II is true. \(f(x) = 4\) only when \((x-3)^2 = 0\), i.e. \(x = 3\): exactly one solution, so III is true.

Full method — how and why
Three statements about one quadratic. Completing the square once answers all three at a stroke.
  1. Complete the square: \(x^2 - 6x + 13 = (x - 3)^2 + 4\).
    This single form contains the range, the line of symmetry and the vertex - so do it first, before looking at the statements.
  2. I: \((x-3)^2 \geq 0\) so \(f(x) \geq 4 > 0\) for every \(x\). True.
    The minimum value is \(4\), comfortably positive; equivalently the discriminant \(36 - 52 < 0\), so the curve never reaches the axis.
  3. II: the vertex is at \(x = 3\), and a parabola is symmetric about the vertical line through its vertex. True.
    Symmetry about \(x = h\) is immediate from the \((x - h)^2\) form.
  4. III: \(f(x) = 4\) means \((x-3)^2 = 0\), so \(x = 3\) only - exactly one solution. True.
    A horizontal line through the vertex meets the parabola once; any higher line meets it twice.
  5. All three hold, so the answer is 'I, II and III'.
    Decide each statement independently, then match to the option list.
Where it goes wrong
  • Assuming \(f(x) = 4\) has two solutions because quadratics 'usually' do - \(y = 4\) is the tangent line at the vertex.
  • Reading the symmetry line off as \(x = -3\).
Takeaway Completing the square answers questions about sign, range, symmetry and repeated roots all at once.
Answer H · Quadratic functions (MM1.3) · question QUA-06
21Quadratic functions · spec MM1.3 · Paper 1 style · ★★☆

Find the sum of all real values of \(x\) satisfying \((x^2 - 3x)^2 - 2(x^2 - 3x) - 8 = 0\).

  1. A\(0\)
  2. B\(9\)
  3. C\(3\)
  4. D\(-3\)
  5. E\(6\)
  6. F\(2\)
Worked solution

Let \(u = x^2 - 3x\): \((u-4)(u+2) = 0\). \(x^2 - 3x - 4 = 0\) gives \(x = 4, -1\) (sum 3). \(x^2 - 3x + 2 = 0\) gives \(x = 1, 2\) (sum 3). Total \(6\).

Full method — how and why
A quadratic in the expression \(x^2 - 3x\). After substituting back you get two quadratics, and the question only wants the total of all four roots.
  1. Let \(u = x^2 - 3x\): the equation becomes \(u^2 - 2u - 8 = 0\).
    The same expression appears squared and linearly, which is the signal for a substitution.
  2. Factorise: \((u - 4)(u + 2) = 0\), so \(u = 4\) or \(u = -2\).
    Both values are usable - neither is ruled out, because \(x^2 - 3x\) can take any value \(\geq -\frac94\).
  3. Case \(u = 4\): \(x^2 - 3x - 4 = 0\). Its roots sum to \(3\) (sum \(= -b/a\)). Discriminant \(9 + 16 > 0\), so both are real.
    You never need the roots themselves - the sum of roots comes straight from the coefficients.
  4. Case \(u = -2\): \(x^2 - 3x + 2 = 0\). Roots sum to \(3\); discriminant \(9 - 8 > 0\), so both are real.
    Check the discriminant in each case: a case with no real roots would contribute nothing to the total.
  5. Total: \(3 + 3 = 6\).
    Four real values of \(x\), summing to \(6\).
Where it goes wrong
  • Answering \(3\) by solving only one of the two cases.
  • Adding the \(u\) values (\(4 + (-2) = 2\)) instead of the \(x\) values.
Takeaway With a substitution, each value of \(u\) gives its own equation in \(x\); check each for real roots and use 'sum of roots \(= -b/a\)' to avoid solving.
Answer E · Quadratic functions (MM1.3) · question QUA-09
22Quadratic functions · spec MM1.3 · Paper 1 style · ★★☆

The equation \(x^2 - 2mx + m + 6 = 0\) has two distinct positive roots. Find the complete set of values of \(m\).

  1. A\(-2 < m < 3\)
  2. B\(m > 3\)
  3. C\(-6 < m < -2\)
  4. D\(m < -2\) or \(m > 3\)
  5. E\(m > 0\)
  6. F\(m > -6\)
Worked solution

Distinct real roots: \(4m^2 - 4(m + 6) > 0 \Rightarrow m^2 - m - 6 > 0 \Rightarrow (m-3)(m+2) > 0\), so \(m < -2\) or \(m > 3\). Both roots positive needs sum \(2m > 0\) and product \(m + 6 > 0\), so \(m > 0\). Combined: \(m > 3\).

Full method — how and why
'Two distinct positive roots' is three conditions at once: discriminant positive, sum of roots positive, product of roots positive. All three must hold.
  1. Discriminant: \(4m^2 - 4(m + 6) > 0 \Rightarrow m^2 - m - 6 > 0 \Rightarrow (m-3)(m+2) > 0\), so \(m < -2\) or \(m > 3\).
    Distinctness and reality come from \(b^2 - 4ac > 0\); this alone allows negative \(m\).
  2. Sum of roots \(= 2m > 0\), so \(m > 0\).
    If both roots are positive their sum is positive. This is the condition that kills the \(m < -2\) branch.
  3. Product of roots \(= m + 6 > 0\), so \(m > -6\).
    If both roots are positive their product is positive; this also rules out one positive and one negative root.
  4. Intersect all three: \((m < -2 \text{ or } m > 3)\) and \(m > 0\) and \(m > -6\) gives \(m > 3\).
    Take the overlap on a number line - the answer must satisfy every condition simultaneously.
Where it goes wrong
  • Stopping after the discriminant and choosing \(m < -2\) or \(m > 3\) - which is offered.
  • Forgetting that sum and product must both be positive: either one alone is not enough.
Takeaway Both roots positive \(\Leftrightarrow\) discriminant \(\geq 0\), sum \(> 0\) and product \(> 0\); sign conditions on roots always come in this package.
Answer B · Quadratic functions (MM1.3) · question QUA-14
23Quadratic functions · spec MM1.3 · Paper 2 style · ★★☆

The quadratic equation \(ax^2 + bx + c = 0\) (with \(a \neq 0\)) has real coefficients. Which of the following statements is/are always true?

  1. IIf \(ac < 0\), the equation has two distinct real roots.
  2. IIIf \(b = 0\), the equation has no real roots.
  3. IIIIf the equation has a repeated root, then \(b^2 = 4ac\).
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

I: \(b^2 - 4ac > 0\) whenever \(ac < 0\), so true. II: \(x^2 - 4 = 0\) has \(b = 0\) and real roots, so false. III: a repeated root means the discriminant is zero, so true.

Full method — how and why
Three general claims about \(ax^2 + bx + c = 0\). Each is a discriminant statement in disguise; the false one needs a counterexample.
  1. I: if \(ac < 0\) then \(-4ac > 0\), so \(b^2 - 4ac \geq -4ac > 0\): two distinct real roots. True.
    \(b^2 \geq 0\) always, so a positive \(-4ac\) forces the whole discriminant positive. Geometrically, \(a\) and \(c\) of opposite signs means the curve crosses the axis.
  2. II: test \(b = 0\) with \(x^2 - 4 = 0\), which has roots \(\pm 2\). False.
    One counterexample settles a 'for all' claim. With \(b = 0\) the discriminant is \(-4ac\), whose sign depends on \(a\) and \(c\).
  3. III: a repeated root means the discriminant is zero, i.e. \(b^2 - 4ac = 0\), i.e. \(b^2 = 4ac\). True.
    This is the definition of the repeated-root case, just rearranged.
  4. I and III only.
    Match the verdicts to the options.
Where it goes wrong
  • Believing II because \(b = 0\) 'removes' the linear term and the graph 'sits on the axis' - it does not; \(x^2 - 4\) is a counterexample.
  • Mis-stating III as \(b^2 = -4ac\).
Takeaway Statements about roots translate into statements about \(b^2 - 4ac\); to refute one, look for the simplest numbers that fit the hypothesis.
Answer F · Quadratic functions (MM1.3) · question QUA-17
24Quadratic functions · spec MM1.3 · Paper 1 style · ★★☆

For how many integer values of \(k\) does \(x^2 + kx + 12 = 0\) have integer roots?

  1. A\(8\)
  2. B\(4\)
  3. C\(2\)
  4. D\(12\)
  5. E\(3\)
  6. F\(6\)
Worked solution

Integer roots \(r, s\) with \(rs = 12\): pairs \((1,12), (2,6), (3,4)\) and their negatives, giving \(k = -(r+s) = -13, -8, -7, 13, 8, 7\): six values.

Full method — how and why
Integer roots means an integer factorisation of the constant term. Count systematically and do not forget negative pairs.
  1. If the roots are integers \(r\) and \(s\) then \(rs = 12\) and \(r + s = -k\).
    For a monic quadratic, product of roots \(= c\) and sum \(= -b\).
  2. List the positive factor pairs of \(12\): \((1, 12), (2, 6), (3, 4)\).
    Work through divisors in order so none is missed.
  3. Each pair also has a negative version \((-1, -12)\), \((-2, -6)\), \((-3, -4)\), since two negatives also multiply to \(+12\).
    The product is positive, so the roots share a sign - both positive or both negative.
  4. Each of the six pairs gives a different \(k = -(r+s)\): \(-13, -8, -7, 13, 8, 7\). So six values.
    Check they are distinct before counting.
Where it goes wrong
  • Listing only the positive pairs and answering \(3\).
  • Counting \((1,12)\) and \((12,1)\) as different, which double-counts.
Takeaway Integer-root problems become divisor-counting problems; the sign of the product tells you whether the roots share a sign.
Answer F · Quadratic functions (MM1.3) · question QUA-20
25Quadratic functions · spec MM1.3 · Paper 1 style · ★★☆

The equation \(x^2 + ax + b = 0\) has roots \(3\) and \(-5\). The equation \(x^2 + bx + a = 0\) has roots \(p\) and \(q\). Find \(p^2 + q^2\).

  1. A\(229\)
  2. B\(195\)
  3. C\(4\)
  4. D\(221\)
  5. E\(225\)
  6. F\(255\)
Worked solution

Sum of roots \(3 + (-5) = -a\), so \(a = 2\); product \(-15 = b\). Then \(x^2 - 15x + 2 = 0\) has \(p + q = 15\) and \(pq = 2\), so \(p^2 + q^2 = 15^2 - 2(2) = 221\).

Full method — how and why
Two linked quadratics: the roots of the first give the coefficients of the second. Sum-and-product relations do all the work.
  1. For \(x^2 + ax + b = 0\) with roots \(3\) and \(-5\): sum \(= 3 + (-5) = -2 = -a\), so \(a = 2\).
    Sum of roots \(= -b/a\); with a monic quadratic that is just \(-a\) - mind the double negative.
  2. Product \(= 3 \times (-5) = -15 = b\).
    Product of roots \(= c/a = b\) here.
  3. The second equation is \(x^2 - 15x + 2 = 0\), so \(p + q = 15\) and \(pq = 2\).
    Substitute the values found, being careful that the roles of \(a\) and \(b\) swap between the two equations.
  4. Use \(p^2 + q^2 = (p+q)^2 - 2pq = 225 - 4 = 221\).
    The same identity as in QUA-04.
Where it goes wrong
  • Getting \(a = -2\) by forgetting the minus in 'sum \(= -a\)', which turns the second equation into \(x^2 - 15x - 2\) and gives \(229\).
  • Computing \((p+q)^2\) and forgetting to subtract \(2pq\), giving \(225\).
Takeaway Read coefficients from roots with sum \(= -b/a\) and product \(= c/a\), and keep track of which equation you are in.
Answer D · Quadratic functions (MM1.3) · question QUA-22
26Quadratic functions · spec MM1.3 · Paper 2 style · ★★☆

\(p\) and \(q\) are real numbers. Which of the following statements is/are true?

  1. IIf \(p^2 < 4q\) then \(x^2 + px + q > 0\) for all real \(x\).
  2. IIIf \(x^2 + px + q > 0\) for all real \(x\) then \(q > 0\).
  3. IIIIf \(q < 0\) then \(x^2 + px + q = 0\) has two real roots of opposite sign.
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

I: negative discriminant and positive leading coefficient. II: put \(x = 0\). III: the product of the roots is \(q < 0\) (and the discriminant \(p^2 - 4q > 0\)), so the roots are real with opposite signs.

Full method — how and why
Three claims about a quadratic. Each is settled by the discriminant, by a well-chosen substitution, or by the root-product relationship.
  1. I: \(p^2 < 4q\) means the discriminant \(p^2 - 4q < 0\), so \(x^2+px+q\) has no real roots; with a positive \(x^2\) coefficient the graph lies entirely above the axis. True.
    No roots plus an upward parabola means always positive - the graph never reaches the axis.
  2. II: if \(x^2+px+q > 0\) for all \(x\), then in particular at \(x = 0\): \(q > 0\). True.
    Substituting a convenient value is the quickest way to extract information from a universal statement.
  3. III: the product of the roots is \(q\). If \(q<0\) then \(p^2 - 4q > p^2 \geq 0\), so the roots are real and distinct, and their product being negative makes them opposite in sign. True.
    For \(x^2+px+q\) the sum of roots is \(-p\) and the product is \(q\).
  4. So all three.
    Match to the list.
Where it goes wrong
  • Reading I's condition as \(p^2 > 4q\) - that would give two real roots and the opposite conclusion.
  • Forgetting to check in III that the roots are real; a negative product alone would be meaningless for complex roots.
  • Trying to prove II with a general argument when \(x=0\) settles it in one line.
Takeaway Discriminant controls the number of roots; \(x=0\) extracts the constant term; sum and product of roots are \(-b/a\) and \(c/a\).
Answer H · Quadratic functions (MM1.3) · question X2-01

How this topic has been examined

Real TMUA questions on Quadratic functions, newest first. Links open the official paper at the question and the online marking page.

2023 · Paper 1Question 2line and parabola do not meet: range of mquestion papersit online
2023 · Paper 2Question 18x^4 + bx^2 + c = 0 has four real roots iffquestion papersit online
2022 · Paper 1Question 12family of quadratics y_kquestion papersit online
2021 · Paper 1Question 8two tangent lines to a parabolaquestion papersit online
2021 · Paper 1Question 12minimum of x^4 - p^2 x^2 is -9question papersit online
2020 · Paper 1Question 5curve y = px^2 + 6x - qquestion papersit online
2020 · Paper 1Question 8maximum of (p - x)(x + 2)question papersit online
2020 · Paper 1Question 9roots alpha, beta of x^2 - 14x + 9question papersit online
2020 · Paper 1Question 11quadratic through (2, 0) and (q, 0)question papersit online
2020 · Paper 2Question 1line crosses or touches a parabolaquestion papersit online
2019 · Paper 1Question 1quadratic through (1, -1) with vertex (-1, 3)question papersit online
2019 · Paper 1Question 2positive for all x: range of kquestion papersit online
2019 · Paper 2Question 16y = px^2 + qx + p: graph of p against rootsquestion papersit online
October 2018 · Paper 1Question 16stationary point closest to the originquestion papersit online
November 2017 · Paper 2Question 14graph of x^2 - 2Bx + c with B > bquestion papersit online
Specimen papers · Paper 1Question 9roots differ by 2question papersit online
Specimen papers · Paper 1Question 17real distinct roots for all non-zero kquestion papersit online
Practice papers · Paper 2Question 15roots differ by between 2 and 4: iff conditionquestion papersit online