IGCSE Maths · Algebra and Graphs · Extended only

Equations II: Quadratics, Fractional Equations and Non-Linear Simultaneous Equations Extended only

This entire note is Extended-only. It brings together three skills you’ve already built separately — factorising, completing the square, and clearing algebraic fraction denominators — and puts them to work solving for x. The quadratic formula itself is given to you in the exam, so the real skill here is knowing which method to reach for, and when.

Why this matters Most real quadratics don’t factorise as neatly as textbook examples — completing the square and the quadratic formula exist precisely for the ones that don’t. Knowing all three methods means you’re never stuck, whichever type of quadratic shows up.

1. Solving Quadratic Equations by Factorisation

If a product of two brackets equals zero, at least one of the brackets must itself equal zero — factorise, then set each bracket equal to zero separately.

Worked example Solve x² + 5x + 6 = 0.
Factorise: (x + 2)(x + 3) = 0
Either x + 2 = 0 or x + 3 = 0
Answer: x = −2 or x = −3
Worked example — a ≠ 1 Solve 2x² − 7x + 3 = 0.
Factorise (split the middle term: ac = 6, numbers −6 and −1): (x − 3)(2x − 1) = 0
Answer: x = 3 or x = 1/2
Try it — Factorisation
  1. Solve x² + 7x + 10 = 0.
  2. Solve x² − 3x − 18 = 0.
  3. Solve 3x² + 11x + 6 = 0.
Show answers
  1. (x+2)(x+5)=0 → x = −2 or x = −5
  2. (x−6)(x+3)=0 → x = 6 or x = −3
  3. ac=18, numbers 9 and 2: 3x²+9x+2x+6=0 → 3x(x+3)+2(x+3)=0 → (x+3)(3x+2)=0 → x = −3 or x = −2/3

2. Solving Quadratic Equations by Completing the Square

Useful whether or not the quadratic factorises nicely — rearrange to isolate the squared bracket, then take the square root of both sides (remembering ± ).

Worked example — a factorisable case, solved this way instead Solve x² + 6x + 5 = 0, by completing the square.
(x + 3)² − 9 + 5 = 0 → (x + 3)² = 4
x + 3 = ±2 → x = −1 or x = −5
Worked example — a case that doesn’t factorise neatly Solve x² − 4x − 1 = 0, giving your answer in surd form.
(x − 2)² − 4 − 1 = 0 → (x − 2)² = 5
x − 2 = ±√5
Answer: x = 2 + √5 or x = 2 − √5
Try it — Completing the Square
  1. Solve x² + 8x + 12 = 0 by completing the square.
  2. Solve x² − 2x − 4 = 0 by completing the square, giving your answer in surd form.
Show answers
  1. (x+4)²−16+12=0 → (x+4)²=4 → x+4=±2 → x = −2 or x = −6
  2. (x−1)²−1−4=0 → (x−1)²=5 → x−1=±√5 → x = 1 + √5 or x = 1 − √5

3. The Quadratic Formula

For any quadratic ax² + bx + c = 0, this formula always works — it’s given in the exam’s List of Formulas, so you don’t need to memorise it, just apply it accurately.

x = [−b ± √(b² − 4ac)] / 2a
Worked example Solve 2x² + 3x − 5 = 0 using the quadratic formula.
a = 2, b = 3, c = −5
Discriminant: b² − 4ac = 9 − 4(2)(−5) = 9 + 40 = 49
x = [−3 ± √49] / 4 = [−3 ± 7] / 4
Answer: x = 1 or x = −2.5
Worked example — a surd-form answer Solve x² + 2x − 4 = 0 using the quadratic formula, giving your answer in surd form.
a = 1, b = 2, c = −4
Discriminant: 4 − 4(1)(−4) = 4 + 16 = 20
x = [−2 ± √20] / 2 = [−2 ± 2√5] / 2
Answer: x = −1 + √5 or x = −1 − √5
Try it — Quadratic Formula
  1. Solve 3x² − 5x − 2 = 0 using the quadratic formula.
  2. Solve x² + 4x − 3 = 0 using the quadratic formula, giving your answer in surd form.
  3. Solve 2x² − 3x − 1 = 0 using the quadratic formula, giving your answer in surd form.
Show answers
  1. a=3,b=−5,c=−2. Disc=25+24=49. x=(5±7)/6 → x = 2 or x = −1/3
  2. a=1,b=4,c=−3. Disc=16+12=28. x=(−4±√28)/2=(−4±2√7)/2 → x = −2 + √7 or x = −2 − √7
  3. a=2,b=−3,c=−1. Disc=9+8=17. x=(3±√17)/4

4. Solving Fractional Equations

Clear the denominators first — multiply every term by whatever’s needed so nothing is left as a fraction, then solve the resulting equation (which is often a quadratic).

Worked example — simple case Solve (x + 1)/(x − 1) = 2.
Multiply both sides by (x − 1): x + 1 = 2(x − 1)
x + 1 = 2x − 2 → x = 3
Worked example — leads to a quadratic Solve 2/(x + 2) + 3/(2x − 1) = 1.
Multiply every term by (x + 2)(2x − 1):
2(2x − 1) + 3(x + 2) = (x + 2)(2x − 1)
Left side: 4x − 2 + 3x + 6 = 7x + 4
Right side: 2x² + 3x − 2
So: 7x + 4 = 2x² + 3x − 2 → 0 = 2x² − 4x − 6 → 0 = x² − 2x − 3
Factorise: (x − 3)(x + 1) = 0 → x = 3 or x = −1
Always check neither solution makes a denominator zero — here, both are fine.
Worked example — a surd-form answer Solve x/(x + 2) = 3/(x − 6), giving your answer in surd form.
Cross-multiply: x(x − 6) = 3(x + 2)
x² − 6x = 3x + 6 → x² − 9x − 6 = 0
Use the quadratic formula: a=1, b=−9, c=−6. Discriminant = 81 + 24 = 105
Answer: x = (9 + √105)/2 or x = (9 − √105)/2
Try it — Fractional Equations
  1. Solve (x + 4)/(x − 2) = 3.
  2. Solve 3/(x + 1) + 2/(x − 3) = 1.
  3. Solve x/(x + 1) = 2/(x − 4).
Show answers
  1. x+4=3(x−2) → x+4=3x−6 → 10=2x → x=5
  2. Multiply by (x+1)(x−3): 3(x−3)+2(x+1)=(x+1)(x−3) → 3x−9+2x+2=x²−2x−3 → 5x−7=x²−2x−3 → 0=x²−7x+4. Discriminant=49−16=33. x=(7±√33)/2
  3. Cross-multiply: x(x−4)=2(x+1) → x²−4x=2x+2 → x²−6x−2=0. Discriminant=36+8=44. x=(6±√44)/2=3±√11

5. Simultaneous Equations — One Linear, One Non-Linear

Use substitution: rearrange the linear equation to make one letter the subject, substitute into the non-linear equation, and solve the resulting quadratic. This always gives up to two pairs of solutions.

Worked example — a curve and a line Solve simultaneously: y = x + 2   and   y = x² − 4.
Substitute: x + 2 = x² − 4
0 = x² − x − 6 → (x − 3)(x + 2) = 0 → x = 3 or x = −2
When x = 3: y = 3 + 2 = 5
When x = −2: y = −2 + 2 = 0
Answer: (3, 5) and (−2, 0)
Worked example — a circle-type equation Solve simultaneously: y = x − 1   and   x² + y² = 25.
Substitute: x² + (x − 1)² = 25
x² + x² − 2x + 1 = 25 → 2x² − 2x − 24 = 0 → x² − x − 12 = 0
(x − 4)(x + 3) = 0 → x = 4 or x = −3
When x = 4: y = 3. When x = −3: y = −4.
Answer: (4, 3) and (−3, −4)
Try it — Linear and Non-Linear Simultaneous
  1. Solve simultaneously: y = x + 1 and y = x² − 2x + 1.
  2. Solve simultaneously: y = 2x − 1 and y = x² − 4x + 5.
  3. Solve simultaneously: y = x + 3 and x² + y² = 9.
Show answers
  1. x+1=x²−2x+1 → 0=x²−3x → x(x−3)=0 → x=0,y=1 or x=3,y=4
  2. 2x−1=x²−4x+5 → 0=x²−6x+6. Discriminant=36−24=12. x=(6±√12)/2=3±√3. This one gives surd answers — a good check that not every non-linear simultaneous system factorises neatly.
  3. (x+3)²… substitute y=x+3: x²+(x+3)²=9 → x²+x²+6x+9=9 → 2x²+6x=0 → 2x(x+3)=0 → x=0,y=3 or x=−3,y=0
Common mistakes to watch for
  • Stopping after finding one solution to a quadratic — always check for a second value from the other bracket, or the other sign in ± .
  • Forgetting the ± when taking a square root, in both completing the square and the quadratic formula.
  • Substituting values incorrectly into the quadratic formula — write out a, b, and c clearly first, especially watching signs when c is negative.
  • Not checking fractional equation solutions against the original denominators — a solution that makes a denominator zero must be rejected.
  • In non-linear simultaneous equations, forgetting to find the matching y-value (or x-value) for each solution — every x needs its own y, not just one shared value.

6. Quick-Fire Challenge Round

True or False?
  1. Every quadratic equation has exactly two solutions.
  2. The quadratic formula only works if the quadratic factorises.
  3. A fractional equation’s solution must always be checked against the original denominators.
  4. Simultaneous equations with one linear and one quadratic equation always give exactly two solution pairs.
Show answers
  1. False — a quadratic can have two solutions, one repeated solution, or no real solutions at all (if the discriminant is negative).
  2. False — the quadratic formula works for every quadratic, factorisable or not; that’s exactly why it’s useful.
  3. True.
  4. False — it depends on the discriminant of the resulting quadratic; there could be two, one (repeated), or zero real solution pairs.
Spot the Error
  1. “Solve x² = 9: x = 3.”
  2. “Using the quadratic formula on x² + 3x + 5 = 0: discriminant = 9 − 20 = −11, so x = (−3 ± √−11)/2.”
  3. “Solve 3/(x−2) = 5: 3 = 5(x−2), so 3 = 5x − 2, giving x = 1.”
Show answers
  1. Taking a square root gives both a positive and negative answer: x = 3 or x = −3, not just x = 3.
  2. The discriminant is negative, which means the equation has no real solutions — writing √−11 as though it were a normal surd is incorrect at this level (this situation means “no real solutions,” not a valid surd answer).
  3. 3 = 5(x−2) expands to 3 = 5x − 10, not 5x − 2. Correct: 13 = 5x → x = 13/5.

7. Mixed Practice — Bringing It All Together

  1. Solve x² + x − 12 = 0 by factorisation.
  2. Solve x² − 6x + 3 = 0 by completing the square, giving your answer in surd form.
  3. Solve 2x² + x − 4 = 0 using the quadratic formula, giving your answer in surd form.
  4. Solve 4/(x − 1) − 1/(x + 2) = 1.
  5. Solve simultaneously: y = x − 2 and y = x² − 6x + 8.
Show answers
  1. (x+4)(x−3)=0 → x = −4 or x = 3
  2. (x−3)²−9+3=0 → (x−3)²=6 → x = 3 ± √6
  3. a=2,b=1,c=−4. Disc=1+32=33. x=(−1±√33)/4
  4. Multiply by (x−1)(x+2): 4(x+2)−(x−1)=(x−1)(x+2) → 4x+8−x+1=x²+x−2 → 3x+9=x²+x−2 → 0=x²−2x−11. Disc=4+44=48. x=(2±√48)/2=1±2√3
  5. x−2=x²−6x+8 → 0=x²−7x+10 → (x−5)(x−2)=0 → x=5,y=3 or x=2,y=0

Practice at Home

A large set again — take it steadily across a few sittings.

Part A — Factorisation
  1. Solve x² + 9x + 14 = 0.
  2. Solve x² − 4x − 21 = 0.
  3. Solve 2x² − x − 6 = 0.
Part B — Completing the square
  1. Solve x² + 10x + 21 = 0 by completing the square.
  2. Solve x² − 6x − 2 = 0 by completing the square, giving your answer in surd form.
Part C — Quadratic formula
  1. Solve 3x² + 2x − 8 = 0 using the quadratic formula.
  2. Solve x² − 5x + 1 = 0 using the quadratic formula, giving your answer in surd form.
  3. Solve 2x² + 7x + 2 = 0 using the quadratic formula, giving your answer in surd form.
Part D — Fractional equations
  1. Solve (2x + 1)/(x + 3) = 2.
  2. Solve 1/(x + 1) + 4/(x − 2) = 1.
  3. Solve x/(x − 3) = 4/(x + 1).
Part E — Non-linear simultaneous equations
  1. Solve simultaneously: y = x + 4 and y = x² + 2x.
  2. Solve simultaneously: y = 3 − x and x² + y² = 17 (careful with the squared terms).
Part F — Applied / challenge
  1. A rectangle has a length 3 cm more than its width, and an area of 40 cm². Let the width be x, form a quadratic equation, and solve it to find the dimensions.
  2. Solve 5/(x − 1) − 2 = 3/(x + 2), showing all your working.
  3. Solve simultaneously: y = 2x + 1 and 2x² + y² = 33 (this one has powers no higher than two, but two different squared terms — approach it the same way).

Key Vocabulary Recap

Discriminant — the value b² − 4ac, found under the square root in the quadratic formula
Quadratic formula — x = [−b ± √(b² − 4ac)] / 2a, given in the exam formula sheet
Fractional equation — an equation containing algebraic fractions
Non-linear equation — an equation where a variable appears to a power higher than 1

Syllabus Reference

This note covers Cambridge IGCSE Mathematics (0580) section E2.5, points 3, 5 and 6 (Extended subject content) — solving fractional equations, simultaneous equations involving one linear and one non-linear equation, and solving quadratic equations by factorisation, completing the square, and the quadratic formula. There is no Core equivalent for any of this content.

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