NCEA Level 2The discriminantAS 91261

The discriminant: 7 exam questions solved step by step

What this lesson covers

  • The discriminant b² − 4ac decides how many real roots a quadratic has before it is solved.
  • A perfect-square discriminant means the roots are rational; otherwise they are irrational.
Key idea

7 questions, one tool

  • 1. Get to zero, then read a, b and c.
  • 2. Δ = b² − 4ac
  • 3. Δ > 0 two, Δ = 0 one, Δ < 0 none.
  • 4. Unknown: write the condition first.
Your turn

Question 1 · Achieved

  • Describe the nature of the roots of
  • 2x² − 5x + 4 = 0
Worked example

Q1: 2x² − 5x + 4 = 0

  • Bracket a negative b.
  • The sign of Δ decides.
  1. a = 2, b = −5, c = 4
  2. Δ = b² − 4acthe discriminant
  3. = (−5)² − 4(2)(4)substitute
  4. = 25 − 32(−5)² = +25
  5. = −7
  6. Δ < 0
  7. no real roots
Your turn

Question 2 · Achieved

  • Show that 3x² + 7x + 2 = 0
  • has two rational roots, then find them.
Worked example

Q2: 3x² + 7x + 2 = 0

  • Square Δ: rational roots.
  • √Δ goes in the formula.
  1. a = 3, b = 7, c = 2
  2. Δ = 7² − 4(3)(2)b² − 4ac
  3. = 49 − 24 = 25
  4. Δ > 0, 25 = 5²a perfect square
  5. two rational roots
  6. x = (−7 ± 5)/6formula, √25 = 5
  7. x = −2/6 = −1/3
  8. x = −12/6 = −2
  9. x = −1/3 or x = −2
Your turn

Question 3 · Merit

  • Describe the nature of the roots of
  • 2x(x − 3) = 1 − x
  • Careful: c is not 1.
Worked example

Q3: 2x(x − 3) = 1 − x

  • Zero on one side first.
  • Then read a, b, c.
  1. 2x² − 6x = 1 − xexpand
  2. 2x² − 5x − 1 = 0+x, −1 both sides
  3. a = 2, b = −5, c = −1
  4. Δ = (−5)² − 4(2)(−1)b² − 4ac
  5. = 25 + 8 = 33−4(2)(−1) = +8
  6. Δ > 0
  7. 33 is not a perfect square
  8. two irrational real roots
  9. c = 1: Δ = 25 − 8 = 17not rearranged
Your turn

Question 4 · Merit

  • kx² + 12x + 4 = 0 has equal roots.
  • Find k, then find the root.
Worked example

Q4: equal roots, find k

  • Equal roots: Δ = 0.
  • Then solve with k in.
  1. equal roots: Δ = 0
  2. 12² − 4(k)(4) = 0a = k, b = 12, c = 4
  3. 144 − 16k = 0
  4. 16k = 144+16k both sides
  5. k = 9÷ 16
  6. 9x² + 12x + 4 = 0k = 9
  7. (3x + 2)² = 0perfect square
  8. x = −2/3
  9. 9(4/9) + 12(−2/3) + 4 = 0
Your turn

Question 5 · Merit

  • Find the values of p for which
  • 2x² + px + 18 = 0
  • has two distinct real roots.
Worked example

Q5: two real roots, find p

  • Two roots: Δ > 0.
  • p² > 144 has two parts.
  1. two real roots: Δ > 0
  2. p² − 4(2)(18) > 0a = 2, b = p, c = 18
  3. p² − 144 > 0
  4. (p − 12)(p + 12) > 0difference of squares
  5. zero at p = 12 and p = −12
  6. p > 12 onlyloses p < −12
  7. p = −13: 169 − 144 = 25> 0, two roots
  8. p = 0: 0 − 144 < 0between: none
  9. p < −12 or p > 12
Your turn

Question 6 · Excellence, in context

  • A school fences a rectangular garden
  • with exactly 20 m of fencing.
  • Use the discriminant: can its area be
  • 30 m²? What is the greatest area?
Worked example

Q6: can the area be 30 m²?

  • Form the quadratic first.
  • Δ < 0: no real width.
  1. width = x m
  2. length = 10 − x mhalf of 20 m is 10 m
  3. x(10 − x) = 30area = 30
  4. 10x − x² = 30expand
  5. x² − 10x + 30 = 0all to the right
  6. Δ = (−10)² − 4(1)(30)b² − 4ac
  7. = 100 − 120 = −20
  8. Δ < 0: no real width x
  9. No, 30 m² is impossible
Worked example

Q6: the greatest area

  • Real width: Δ ≥ 0.
  • Answer in words, with units.
  1. x(10 − x) = Aarea A
  2. x² − 10x + A = 0rearrange
  3. real width: Δ ≥ 0
  4. (−10)² − 4(1)(A) ≥ 0b² − 4ac
  5. 100 − 4A ≥ 0
  6. A ≤ 25+4A, ÷ 4
  7. A = 25: (x − 5)² = 0x² − 10x + 25
  8. x = 5, length 10 − 5 = 5
  9. Greatest area 25 m², 5 m by 5 m
Your turn

Question 7 · Excellence

  • Show that x² + kx + k − 1 = 0
  • has real roots for every value of k.
  • For which k are the roots equal?
Worked example

Q7: real roots for every k

  • c is the whole k − 1.
  • A square is never negative.
  1. a = 1, b = k, c = k − 1
  2. Δ = k² − 4(1)(k − 1)b² − 4ac
  3. = k² − 4k + 4expand
  4. = (k − 2)²perfect square
  5. (k − 2)² ≥ 0 for every ka square
  6. Δ ≥ 0: real roots for every k
  7. equal roots: (k − 2)² = 0
  8. k = 2
  9. x² + 2x + 1 = (x + 1)²root x = −1
Summary

The method that solved all seven

  • Zero, a b c, the discriminant, then the condition.
StepHow
Firstget to zero, read a, b, c
Δb² − 4ac, negatives bracketed
Δ > 0two real roots
Δ = 0one repeated root
Δ < 0no real roots
Δ squarethe roots are rational
Unknownwrite the condition, solve

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