Year 12

AQA A level Mathematics [Pure 1] Quadratics Notes

Quadratics

— Pure Mathematics 1 · AS & A Level (9709) · Chapter 1 —
★ Exam Tip: Quadratics sit under almost everything in P1. The end-of-chapter past-paper questions (9709 Paper 11) are all about line meets curve, tangents, discriminant, completing the square — master sections 8 & 9 especially.

Quadratic function: y = ax² + bx + c (a ≠ 0). Graph = parabola — has a max or min and a line of symmetry. e.g. path (trajectory) of a ball — Galileo showed it is a quadratic.

Chapter goals: completing the square · discriminant · quadratic equations and inequalities · linear + quadratic simultaneous · equations quadratic in a function of x · graph ↔ algebra link.

1. Solving by Factorisation

Method

  1. Rearrange so that one side is 0: ax² + bx + c = 0.
  2. Take out any common factor first (÷ it from both sides).
  3. Factorise into two brackets.
  4. Use the zero rule, then solve each bracket.
Zero ruleif pq = 0 then p = 0 or q = 0
Tip: Divide by a common factor first, if possible.
Common mistake (Rosa, Explore 1.1): solve 2x² + 3x − 5 = (x−1)(x−2). Factorising LHS: (x−1)(2x+5) = (x−1)(x−2). She divides both sides by (x−1) ⇒ x = −7 — but this loses the solution x = 1 (dividing by 0!). Never divide by something containing x — move everything to one side and factorise.
(x−1)(2x+5) − (x−1)(x−2) = 0(x−1)[(2x+5) − (x−2)] = 0(x−1)(x+7) = 0x = 1 or x = −7
Worked Ex 1.1factorise

Solve a) 6x² + 5 = 17x b) 9x² − 39x − 30 = 0

a)6x² − 17x + 5 = 0(2x − 5)(3x − 1) = 02x−5 = 0 or 3x−1 = 0x = 52 or 13
b)9x² − 39x − 30 = 0÷3: 3x² − 13x − 10 = 0(3x + 2)(x − 5) = 0x = −23 or 5
Worked Ex 1.2fractions

Solve 212x − 2x+3 = 1

Multiply both sides by the common denominator 2x(x+3), then expand & rearrange to = 0.

21(x+3) − 4x = 2x(x+3)21x + 63 − 4x = 2x² + 6x2x² − 11x − 63 = 0(2x + 7)(x − 9) = 0x = −72 or 9
Worked Ex 1.3fraction = 0

Solve 3x² + 26x + 35x² + 8 = 0

A fraction is 0 only when its numerator is 0. Denominator x²+8 is never 0, so safe to multiply through.

3x² + 26x + 35 = 0(3x + 5)(x + 7) = 0x = −53 or −7
(6x − 7) cmx cmarea = 90 cm²x = 4½ ✓ (reject −10/3)
Ex 1.4 — rectangle
Worked Ex 1.4word problem

Rectangle sides x cm and (6x − 7) cm, area 90 cm². Find the sides.

x(6x − 7) = 906x² − 7x − 90 = 0(2x − 9)(3x + 10) = 0x = 92 or x = −103length must be positive ⇒ reject −10/3x = 4½, 6x − 7 = 20sides: 4½ cm and 20 cm

Explore 1.2 — equations of the form pq = 1

pq = 1 happens in three ways:

Case Condition Note
① exponent zero q = 0 need p ≠ 0
② base one p = 1 any q
③ base −1 p = −1 q must be EVEN
Equation Working Solutions
A: 4(2x²+x−6) = 1 only ① (4 ≠ ±1): 2x²+x−6 = 0 ⇒ (2x−3)(x+2) = 0 x = 3/2, −2 → 2 values
B: (x²−3x+1)⁶ = 1 exponent fixed (6, even) ⇒ base = 1 or −1: x²−3x = 0 ⇒ 0, 3; x²−3x+2 = 0 ⇒ 1, 2 x = 0, 3, 1, 2 → 4 values
C: (x²−3x+1)(2x²+x−6) = 1 ①: 3/2, −2 (base ≠ 0 ✓) · ②: 0, 3 · ③: x=1 gives exponent −3 (odd ✗); x=2 gives exponent 4 (even ✓) x = −2, 0, 3/2, 2, 3 → 5 values
Remember: always check each answer in the original. If the power is odd (e.g. (…)⁵ = 1) the only option is base = 1; if even (…⁶, …⁸) base = ±1.

2. Completing the Square

Idea: rewrite a quadratic so that x appears only once — easier to solve, and gives the vertex.

Key Point 1.1x² + 2dx = (x + d)² − d²
x² − 2dx = (x − d)² − d²
x²5x5x25x5x5
x² + 10x = whole square (x+5)² minus the missing corner 25

To complete the square on x² + bx: halve the x-coefficient → put in bracket → subtract its square.

x² + 10x10 ÷ 2 = 5= (x + 5)² − 5²= (x + 5)² − 25
x² + 8x − 7= (x + 4)² − 4² − 7= (x + 4)² − 23

If a ≠ 1, take a factor out of the first two terms first:

2x² − 12x + 5 = 2(x² − 6x) + 56 ÷ 2 = 3: x² − 6x = (x−3)² − 3²= 2[(x−3)² − 9] + 5= 2(x−3)² − 18 + 5= 2(x − 3)² − 13
Remember: the factor multiplies the −9 too! 2 × (−9) = −18.

Algebraic method — compare coefficients

Worked Ex 1.5express in form p(x−q)² + r

Write 2x² − 12x + 3 as p(x−q)² + r.

p(x−q)² + r = px² − 2pqx + pq² + rx²: 2 = p (1)x: −12 = −2pq (2)const: 3 = pq² + r (3)p = 2 in (2) ⇒ q = 3in (3): 3 = 2(9) + r ⇒ r = −152(x − 3)² − 15
Worked Ex 1.6express in form (ax+b)² + c

Write 4x² + 20x + 5 as (ax+b)² + c.

(ax+b)² + c = a²x² + 2abx + b² + cx²: 4 = a² ⇒ a = ±2x: 20 = 2abconst: 5 = b² + ca = 2: b = 5, c = 5 − 25 = −20(2x + 5)² − 20alternative a = −2, b = −5 gives (−2x−5)² − 20 = same thing
Worked Ex 1.7solve by completing the square (surds)

Solve 5x+2 + 3x−5 = 1, answer in surd form.

5(x−5) + 3(x+2) = (x+2)(x−5)8x − 19 = x² − 3x − 10x² − 11x + 9 = 0(x − 112)² − (112)² + 9 = 0(x − 112)² = 854x − 112 = ±√854 = ±√852x = 12(11 ± √85)
Note: surd form means leave as exact √ — don't convert to decimals.

3. The Quadratic Formula

For ax² + bx + c = 0 (a ≠ 0, a, b, c constants):

Key Point 1.2x = −b ± √b² − 4ac2a

Proof (completing the square on the general equation):

ax² + bx + c = 0÷a: x² + bax + ca = 0(x + b2a)² − (b2a)² + ca = 0(x + b2a)² = b²4a² − ca = b² − 4ac4a²x + b2a = ± √b² − 4ac2ax = −b ± √b² − 4ac2a
Worked Ex 1.8quadratic formula

Solve 6x² − 3x − 2 = 0 to 3 s.f.

a = 6, b = −3, c = −2x = −(−3) ± √(−3)² − 4×6×(−2)2×6x = 3 ± √5712x = 0.879 or −0.379 (3 s.f.)
Careful: put brackets round negative b, c: (−3)² = 9 (not −9), and −4ac with c = −2 becomes +48. Give the accuracy asked: surd / 2 d.p. / 3 s.f.

4. Simultaneous Equations (1 linear + 1 quadratic)

Idea: solutions = points where the two graphs intersect.

(−1, −3)(3, 5)y = x² − 4y = 2x − 1
y = x² − 4 and y = 2x − 1 meet at 2 points
  1. Make x or y the subject of the linear equation.
  2. Substitute into the quadratic ⇒ quadratic in one variable.
  3. Solve it.
  4. Put each answer back into the linear equation to get the partner value.
y = x² − 4 ……(1)y = 2x − 1 ……(2)2x − 1 = x² − 4x² − 2x − 3 = 0(x + 1)(x − 3) = 0x = −1 ⇒ y = −2−1 = −3x = 3 ⇒ y = 6−1 = 5x = −1, y = −3 and x = 3, y = 5

General quadratic in x and y: ax² + bxy + cy² + dx + ey + f = 0 (at least one of a, b, c ≠ 0) — graph is a conic. Same method works.

Worked Ex 1.9linear + conic

Solve 2x + 2y = 7 …(1) and x² − 4y² = 8 …(2)

Method 1 (x = …)x = 7 − 2y2(7−2y2)² − 4y² = 849 − 28y + 4y²4 − 4y² = 849 − 28y + 4y² − 16y² = 3212y² + 28y − 17 = 0(6y + 17)(2y − 1) = 0y = −176 ⇒ x = 193y = 12 ⇒ x = 3
Method 2 (sub 2y)2y = 7 − 2xx² − (7 − 2x)² = 8x² − 49 + 28x − 4x² = 83x² − 28x + 57 = 0(3x − 19)(x − 3) = 0x = 193 or 3~(4y² = (2y)² made this quicker)

Solutions: x = 19/3, y = −17/6 and x = 3, y = 1/2

Common mistake: always substitute back into the linear equation. Using the quadratic can give false extra pairs — pair each x with its y.

5. Equations Quadratic in a Function of x

Spot a hidden quadratic: let y = (the repeated function), solve for y, then solve back for x.

Looks like Let
x⁴, x² y = x²
x, √x y = √x
9x, 3x y = 3x
x⁶, x³  or  1/x⁴, 1/x² y = x³  /  y = 1/x²
Worked Ex 1.10quartic

Solve 4x⁴ − 37x² + 9 = 0

Method 1let y = x²4y² − 37y + 9 = 0(4y − 1)(y − 9) = 0y = 14 or 9x² = 14 or x² = 9x = ±12 or ±3
Method 2factorise directly:(4x² − 1)(x² − 9) = 0x² = 14 or 9x = ±12 or ±3
Worked Ex 1.11with a square root

Solve x − 4√x − 12 = 0

let y = √xy² − 4y − 12 = 0(y − 6)(y + 2) = 0y = 6 or y = −2√x = 6 ⇒ x = 36√x = −2 has no solution: √x is never negative
Worked Ex 1.12exponential

Solve 3(9x) − 28(3x) + 9 = 0

9x = (3x)², let y = 3x3y² − 28y + 9 = 0(3y − 1)(y − 9) = 0y = 13 or 93x = 13 = 3−1 or 3x = 9 = 3²x = −1 or 2
Reject check: after finding y, test it can occur: x² ≥ 0 and √x ≥ 0 so negative y is out; 3x > 0 so y ≤ 0 is out. ("Real solutions" only.)

6. Max & Min Values — Sketching Parabolas

f(x) = ax² + bx + c graph = parabola. The turning point is the vertex (a stationary / turning point: gradient 0). Every parabola has a line of symmetry through the vertex.

a > 0 : ∪ shape — vertex = MIN
a > 0 ⇒ minimum point
a < 0 : ∩ shape — vertex = MAX
a < 0 ⇒ maximum point
A good sketch shows: ① general shape (from sign of a) ② axis intercepts (y: x=0 → c; x: solve f(x)=0) ③ coordinates of the vertex. (Show some or all, depending on the question.)
focusrays ∥ axis
parabola rotated about its axis = paraboloid (satellite dish): parallel rays reflect through the focus

Did you know? Satellite dishes are paraboloids — a receiver at the focus picks up everything entering the dish.

Method 1 — symmetry (Worked Ex 1.13)

Worked Ex 1.13f(x) = x² − 3x − 4

a) axes crossing points b) sketch & vertex

x = 0 ⇒ y = −4y = 0: x² − 3x − 4 = 0(x + 1)(x − 4) = 0 ⇒ x = −1, 4crossing points: (0,−4) (−1,0) (4,0)axis of symmetry halfway between −1 and 4: x = 32y = (32)² − 3(32) − 4 = −254a > 0 ⇒ U-shape, minimum (32, −254)
−14−4(3/2, −25/4)x = 3/2y = x² − 3x − 4
symmetry line through the midpoint of the roots; vertex is the minimum

Method 2 — completing the square

x² − 3x − 4 = (x − 32)² − (32)² − 4= (x − 32)² − 254(x − 3/2)² ≥ 0, smallest value 0 when x = 3/2 ⇒ min value −25/4
Key Point 1.3f(x) = a(x − h)² + k ⇒ axis of symmetry x = h = −b2a
a > 0 : minimum at (h, k)   |   a < 0 : maximum at (h, k)
Tip: Give vertex coordinates as fractions, not rounded decimals.
Worked Ex 1.14sketch y = 16x − 7 − 4x²

Complete the square (a is negative!):

16x − 7 − 4x² = 9 − 4(x − 2)²check: −4(x²−4x+4) + 9 = −4x²+16x−7 ✓(x−2)² ≥ 0 and it is subtracted from 9 ⇒ greatest when x = 2max value 9: vertex (2, 9), axis x = 2x = 0 ⇒ y = −7y = 0: (x−2)² = 94 ⇒ x − 2 = ±32x = 3½ or ½
max (2, 9)½3½−7x = 2
upside-down parabola: maximum at the vertex

Sketch checklist: shape from a → intercepts (c, roots) → axis halfway between roots → vertex y by substituting x = −b/2a (or completing the square).

7. Quadratic Inequalities

Linear review: solve like an equation, BUT

Key Point 1.4multiply / divide both sides by a NEGATIVE ⇒ REVERSE the inequality sign
2(x+7) < −42x + 14 < −42x < −18x < −9
11 − 2x ≥ 5−2x ≥ −6÷(−2) and flip:x ≤ 3

Quadratic inequality method

  1. Rearrange so one side is 0.
  2. Solve the equation to get the critical values (x-intercepts).
  3. Sketch the graph: just shape + x-intercepts (no vertex / y-intercept needed).
  4. > 0 ⇒ take the part above the x-axis; < 0 ⇒ the part below. (≥, ≤ include the critical values.)
Worked Ex 1.15x² − 5x − 14 > 0
x² − 5x − 14 = 0(x + 2)(x − 7) = 0 ⇒ x = −2, 7want curve ABOVE axis (+):x < −2 or x > 7
Worked Ex 1.162x² + 3x ≤ 27
2x² + 3x − 27 ≤ 0(2x + 9)(x − 3) = 0 ⇒ x = −4½, 3want curve ZERO or BELOW axis (−):−4½ ≤ x ≤ 3
−27++−
x² − 5x − 14 > 0: open circles, two pieces
−4½3++−
2x² + 3x − 27 ≤ 0: closed circles, one piece
Writing the answer: outside the roots ⇒ two separate parts joined by "or": x < −2 or x > 7. Between the roots ⇒ one double inequality: −4½ ≤ x ≤ 3. Never write −2 > x > 7 (impossible).

Explore 1.3 — Ivan's error (inequality with x on the bottom)

Solve 2x − 4x ≥ 7. Ivan: ×x ⇒ 2x − 4 ≥ 7x ⇒ x ≤ −45. Anika tests x = −1: 2(−1)−4−1 = 6, which is not ≥ 7, so Ivan is wrong.

The error: multiplying by x is only allowed (without flipping) when x > 0. If x is negative the sign must reverse — and we don't know the sign of x.

Correct approaches (also note x ≠ 0):

Multiply by x² (always +)x(2x − 4) ≥ 7x²0 ≥ 5x² + 4xx(5x + 4) ≤ 0−45 ≤ x ≤ 0but x ≠ 0 ⇒ −45 ≤ x < 0
Move all to one side2x−4x − 7 ≥ 0−5x − 4x ≥ 05x + 4x ≤ 0 (×−1, flip!)same: −45 ≤ x < 0
y = 7−4/5y = (2x−4)/xy = 2
curve is on/above y = 7 only for −4/5 ≤ x < 0 (x = 0 excluded: asymptote)

8. The Discriminant — Number of Roots

Roots of f(x) = the solutions of f(x) = 0. The part of the formula under the square root is the discriminant.

Key Point 1.5discriminant = b² − 4ac
Equation b² − 4ac Result
x² + 2x − 8 = 0 4 + 32 = 36 (>0) x = 2 or −4 ⇒ two distinct real roots
x² + 6x + 9 = 0 36 − 36 = 0 x = −3 (twice) ⇒ two equal roots
x² + 2x + 6 = 0 4 − 24 = −20 (<0) √(−20) impossible ⇒ no real roots
b² − 4ac Nature of roots Curve y = ax²+bx+c
> 0 two distinct real roots cuts x-axis at 2 points
= 0 two equal (1 repeated) real roots touches x-axis at 1 point
< 0 no real roots entirely above or below the x-axis
b² − 4ac > 0
2 roots
b² − 4ac = 0
1 repeated root
b² − 4ac < 0
no real roots

(Shown for a > 0; for a < 0 flip the parabola upside-down — same rules.)

Real roots condition: b² − 4ac ≥ 0 (two distinct or equal).
Worked Ex 1.17two equal roots

Find k so that 4x² + kx + 1 = 0 has equal roots.

b² − 4ac = 0k² − 4(4)(1) = 0k² = 16k = −4 or 4
Worked Ex 1.18rearrange first

Find k so that x² − 5x + 9 = k(5 − x) has equal roots.

x² − 5x + 9 − 5k + kx = 0x² + (k−5)x + (9−5k) = 0 (form ax²+bx+c=0)(k−5)² − 4(1)(9−5k) = 0k² − 10k + 25 − 36 + 20k = 0k² + 10k − 11 = 0(k + 11)(k − 1) = 0k = −11 or 1
Worked Ex 1.19two distinct roots ⇒ inequality in k

Find k so that kx² − 2kx + 8 = 0 has two distinct roots.

b² − 4ac > 0(−2k)² − 4(k)(8) > 04k² − 32k > 04k(k − 8) > 0critical values k = 0, 8k < 0 or k > 8
0 8k < 0 or k > 8
Ex 1.19
−4 0k < −4 or k > 0
Ex 1.21 (sec 9)
−8 8−8 < k < 8
Ex 1.22 (sec 9)
Pattern: "equal roots" ⇒ solve b² − 4ac = 0 (an equation in k). "distinct / no real / real roots" ⇒ solve an inequality in k with the same sign-graph method as section 7.

9. Intersection of a Line and a Quadratic Curve

Solve the two equations simultaneously → get a quadratic → use its discriminant.

cuts at 2 points
touches at 1 point
⇒ tangent
no intersection
b² − 4ac (of resulting quadratic) Roots Line and curve
> 0 two distinct real 2 intersection points
= 0 equal (repeated) 1 point — line is a tangent
< 0 none no intersection
Worked Ex 1.20tangent

Find k so that y = x + k is a tangent to y = x² + 5x + 2.

x² + 5x + 2 = x + kx² + 4x + (2 − k) = 0tangent ⇒ b² − 4ac = 04² − 4(1)(2 − k) = 016 − 8 + 4k = 04k = −8k = −2
Worked Ex 1.21two distinct points

Find k so that y = kx − 1 meets y = x² − 2x at two distinct points.

x² − 2x = kx − 1x² − (k+2)x + 1 = 0need b² − 4ac > 0(k+2)² − 4 > 0k² + 4k > 0k(k + 4) > 0critical values −4, 0k < −4 or k > 0
Worked Ex 1.22general conic xy = 8

Find k so that the line 2x + y = k does not meet the curve xy = 8.

y = k − 2x ⇒ x(k − 2x) = 82x² − kx + 8 = 0no intersection ⇒ b² − 4ac < 0(−k)² − 4(2)(8) < 0k² − 64 < 0(k + 8)(k − 8) < 0−8 < k < 8
Remember: this works for any quadratic curve (xy = 8, x² + y² = 25 …) — substitute the line, then discriminant. Past papers (9709 P11) also ask: coordinates of A, B, mid-point of AB, length AB, perpendicular bisector — find x from the quadratic, y from the line.

10. Full Chapter Cheat-Sheet

✓ Solving quadratics
  • Factorise (get = 0 first; common factor first; never ÷ by an x-expression) · complete the square · formula x = −b ± √b² − 4ac2a
  • x² + 2dx = (x+d)² − d²  |  a x² + bx + c = a(x + b2a)² + c − b²4a
  • Check answers; reject invalid ones (negative lengths, √x = −2, 3x = −1…)
✓ Simultaneous & hidden quadratics
  • Linear → x or y = …; substitute into the quadratic; solve; substitute back into the linear.
  • Let y = x², √x, 3x, x³, 1/x² … solve for y, then for x.
  • pq = 1 ⇒ q = 0 (p ≠ 0)  or  p = 1  or  p = −1 with q even.
✓ Graph, max/min, inequalities
  • f(x) = a(x − h)² + k: axis x = h = −b/2a; a > 0 min (h,k); a < 0 max (h,k).
  • Sketch: shape · intercepts · vertex.
  • Inequalities: = 0 → critical values → sketch → above (>0) / below (<0). Outside ⇒ "or"; between ⇒ one double inequality.
  • Multiply/divide by a negative ⇒ flip sign. Don't multiply by x unless sign known — use x² or one side = 0.
✓ Discriminant & line–curve
b² − 4ac Roots Curve / line + curve
> 0 2 distinct real cuts x-axis twice / 2 intersections
= 0 2 equal real touches / tangent
< 0 none misses x-axis / no intersection
  • Real roots ⇔ b² − 4ac ≥ 0. Equal roots ⇒ solve = 0; otherwise solve an inequality in k.