NCEA Level 2Quadratic equationsAS 91261

Quadratic equations: 7 exam questions solved step by step

What this lesson covers

  • Every quadratic can be solved by factorising, the formula, or completing the square, and each suits different equations.
  • The roots of a quadratic fix its factors, so the equation can be rebuilt from its solutions.
Key idea

7 questions, one plan

  • 1. Rearrange so one side is 0.
  • 2. Factorise, or use the formula.
  • 3. Each root gives a factor.
  • 4. In context, reject with a reason.
Your turn

Question 1 · Achieved

  • Solve 2x² + 7x − 15 = 0
Worked example

Q1: 2x² + 7x − 15 = 0

  • a = 2: use ac.
  • Check both roots.
  1. ac = 2 × −15 = −30
  2. 10 × −3 = −30, 10 + −3 = 7
  3. 2x² + 10x − 3x − 15 = 0split 7x
  4. 2x(x + 5) − 3(x + 5) = 0group in pairs
  5. (2x − 3)(x + 5) = 0common bracket
  6. x = 3/2 or x = −52x = 3
  7. x = 3/2: 9/2 + 21/2 − 15 = 0
  8. x = −5: 50 − 35 − 15 = 0
Your turn

Question 2 · Achieved

  • Form a quadratic equation with roots
  • x = −4 and x = 2/5,
  • with whole-number coefficients.
Worked example

Q2: roots −4 and 2/5

  • Each root gives a factor.
  • Clear the fraction first.
  1. x = −4, so x + 4 = 0+4 both sides
  2. x = 2/5, so 5x = 2× 5
  3. 5x − 2 = 0−2
  4. (x + 4)(5x − 2) = 0multiply the factors
  5. 5x² − 2x + 20x − 8 = 0expand
  6. 5x² + 18x − 8 = 0collect
  7. x = 2/5: 4/5 + 36/5 − 8 = 0
  8. x = −4: 80 − 72 − 8 = 0
Your turn

Question 3 · Merit

  • Solve 3x² = 2(4x + 1)
  • Give your answers to 3 decimal places.
Worked example

Q3: 3x² = 2(4x + 1)

  • Standard form first.
  • Brackets round negatives.
  1. 3x² = 8x + 2expand
  2. 3x² − 8x − 2 = 0−8x, −2 both sides
  3. a = 3, b = −8, c = −2
  4. x = (−b ± √(b² − 4ac))/2a
  5. b² − 4ac = (−8)² − 4(3)(−2)
  6. = 64 + 24 = 88
  7. x = (8 ± √88)/6−b = 8, 2a = 6
  8. x = (8 + √88)/6 = 2.897
  9. x = (8 − √88)/6 = −0.230
Your turn

Question 4 · Merit

  • Solve 3x − 4/x = 4
Worked example

Q4: 3x − 4/x = 4

  • x ≠ 0: multiply by x.
  • Every term, both sides.
  1. 3x² − 4 = 4x× x, every term
  2. 3x² − 4x − 4 = 0−4x
  3. ac = −12: −6 × 2, −6 + 2 = −4
  4. 3x² − 6x + 2x − 4 = 0split −4x
  5. 3x(x − 2) + 2(x − 2) = 0group in pairs
  6. (3x + 2)(x − 2) = 0common bracket
  7. x = 2 or x = −2/3
  8. x = 2: 6 − 2 = 4
  9. x = −2/3: −2 + 6 = 4
Your turn

Question 5 · Merit, in context

  • A rugby ball is kicked at Eden Park.
  • Its height after t seconds is
  • h = −4.9t² + 14t + 1 (metres).
  • When does it hit the ground?
Worked example

Q5: when is h = 0?

  • Ground means h = 0.
  • Time cannot be negative.
  1. −4.9t² + 14t + 1 = 0ground: h = 0
  2. 4.9t² − 14t − 1 = 0× −1
  3. a = 4.9, b = −14, c = −1
  4. b² − 4ac = (−14)² − 4(4.9)(−1)
  5. = 196 + 19.6 = 215.6
  6. t = (14 ± √215.6)/9.82a = 9.8
  7. t = 2.927 or t = −0.070
  8. t = −0.070before the kick
  9. It lands after about 2.93 s
Your turn

Question 6 · Excellence, in context

  • A school lawn is 12 m by 8 m.
  • A path x m wide goes all round it.
  • Lawn and path cover 140 m².
  • Find the width of the path.
Worked example

Q6: form the equation

  • The path adds x on each side.
  • State the equation.
  1. outer length = 12 + 2xx at each end
  2. outer width = 8 + 2xx on each side
  3. (12 + 2x)(8 + 2x) = 140length × width
  4. 96 + 24x + 16x + 4x² = 140four products
  5. 4x² + 40x + 96 = 140collect
  6. 4x² + 40x − 44 = 0−140
  7. x² + 10x − 11 = 0÷ 4
Worked example

Q6: solve, reject, check

  • A width is positive.
  • Answer in words, with units.
  1. x² + 10x − 11 = 0
  2. 11 × −1 = −11, 11 + −1 = 10
  3. (x + 11)(x − 1) = 0factorise
  4. x = −11 or x = 1
  5. x = −11a width cannot be < 0
  6. outer: 14 m by 10 m12 + 2, 8 + 2
  7. 14 × 10 = 140 m²
  8. The path is 1 m wide
Your turn

Question 7 · Excellence

  • One root of 3x² + kx − 10 = 0
  • is x = 5/3.
  • Find k and the other root.
Worked example

Q7: a root gives a factor

  • x = 5/3 gives (3x − 5).
  • Match the coefficients.
  1. x = 5/3, so 3x − 5 = 0× 3, then −5
  2. 3x² + kx − 10 = (3x − 5)(x + m)
  3. = 3x² + 3mx − 5x − 5mexpand
  4. constants: −5m = −10, m = 2
  5. x terms: k = 3m − 5match the x terms
  6. k = 3(2) − 5 = 1
Worked example

Q7: the other root, and check

  • The second factor gives it.
  • Check by substituting.
  1. 3x² + x − 10 = (3x − 5)(x + 2)
  2. 3x² + 6x − 5x − 10 = 3x² + x − 10
  3. x + 2 = 0, so x = −2
  4. k = 1, other root x = −2
  5. x = 5/3: 25/3 + 5/3 − 10 = 0
  6. x = −2: 12 − 2 − 10 = 0
Summary

The plan that solved all seven

  • Zero first, factorise or formula, then reason about the roots.
StepHow
Rearrangeeverything on one side, = 0
Factoriseac method when a is not 1
Formulabrackets round negatives
Fractionsmultiply every term first
From rootsx = p/q gives (qx − p)
In contextreject only with a reason

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