NCEA Level 2The discriminantAS 91261

The discriminant: how many roots, and what kind

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

Before you solve, count the roots

  • Inside the quadratic formula is b² − 4ac.
  • It is called the discriminant.
  • Its sign decides how many real roots there are.
Key idea

Three cases, one curve

  • Positive: the curve crosses the axis twice.
  • Zero: it just touches, once.
  • Negative: it never meets the axis.
Three cases, one curve
Key idea

The rule

  • Work out b² − 4ac.
  • Positive, zero or negative.
  • Two roots, one repeated root, or none.
DiscriminantRootsGraph
b² − 4ac > 0two real rootscrosses twice
b² − 4ac = 0one repeated roottouches once
b² − 4ac < 0no real rootsnever meets
Key idea

Why it works

  • The formula has ± √(b² − 4ac).
  • Positive gives two answers; zero gives one.
  • Negative has no real square root.
Worked example

Two quick examples

  • Write a, b and c first.
  • Then state the nature.
  1. x² − 6x + 9: a = 1, b = −6, c = 9
  2. Δ = (−6)² − 4(1)(9)substitute
  3. = 36 − 36 = 0one repeated root
  4. x² + 2x + 5: a = 1, b = 2, c = 5
  5. Δ = 2² − 4(1)(5)substitute
  6. = 4 − 20 = −16no real roots
Worked example

Watch the sign when c is negative

  • c negative: −4ac adds.
  • Bracket negative values.
  1. 2x² + 3x − 2 = 0
  2. a = 2, b = 3, c = −2c is negative
  3. Δ = 3² − 4(2)(−2)substitute
  4. −4(2)(−2) = +16minus times minus
  5. Δ = 9 + 16 = 25two real roots
Worked example

Rational roots: a perfect square

  • Perfect square: rational.
  • The equation factorises.
  1. Δ = 25 = 5²a perfect square
  2. x = (−3 ± √25)/4−b = −3, 2a = 4
  3. = (−3 ± 5)/4√25 = 5
  4. x = (−3 + 5)/4 = 1/2
  5. x = (−3 − 5)/4 = −2
  6. (2x − 1)(x + 2) = 0it factorises
Worked example

Irrational roots: not a square

  • Not a square: irrational.
  • Leave the root as a surd.
  1. x² − 4x + 1: a = 1, b = −4, c = 1
  2. Δ = (−4)² − 4(1)(1)substitute
  3. = 16 − 4 = 12not a perfect square
  4. x = (4 ± √12)/2−b = 4, 2a = 2
  5. √12 = √4 × √3 = 2√3
  6. x = (4 ± 2√3)/2
  7. = 2 ± √3irrational
Worked example

Find k for equal roots

  • Equal roots: Δ = 0.
  • Give both values of k.
  1. x² + kx + 9 = 0
  2. a = 1, b = k, c = 9
  3. Δ = 0equal roots
  4. k² − 4(1)(9) = 0substitute
  5. k² − 36 = 0
  6. k² = 36add 36
  7. k = 6 or k = −6both square roots
Worked example

Find k for no real roots

  • No real roots: Δ < 0.
  • The answer is a range.
  1. kx² − 4x + 1 = 0
  2. a = k, b = −4, c = 1
  3. Δ < 0no real roots
  4. (−4)² − 4(k)(1) < 0substitute
  5. 16 − 4k < 0
  6. 16 < 4kadd 4k
  7. 4 < k, so k > 4divide by 4
Key idea

The words tell you the condition

  • "Equal" or "repeated": Δ = 0.
  • "Two real" or "distinct": Δ > 0.
  • "No real": Δ < 0. "Real": Δ ≥ 0.
Key idea

Where the marks are

  • Achieved: calculate Δ and state the nature.
  • Merit: use Δ to find an unknown.
  • Excellence: link Δ to a real problem.
Summary

What to take away

  • The sign of b² − 4ac counts the roots.
Δ = b² − 4acfrom ax² + bx + c = 0
Δ > 0two real roots
Δ = 0one repeated root
Δ < 0no real roots
Δ a perfect squarethe roots are rational
Unknown kset up Δ = 0, > 0 or < 0
Your turn

Your turn: four questions

  • Pause here and answer each one.
  • Question 4 has a range of answers.
  • Show the discriminant each time.
Question
1Describe the roots of 3x² − 5x + 1 = 0.
2Describe the roots of x² − 10x + 25 = 0.
3Find k so x² + 6x + k = 0 has equal roots.
4For which k does 2x² + kx + 8 = 0 have no real roots?
Show answer

1. Δ = 25 − 12 = 13 > 0 and not a square: two irrational roots. 2. Δ = 100 − 100 = 0: one repeated root, x = 5. 3. 36 − 4k = 0, so k = 9. 4. k² − 64 < 0, so −8 < k < 8.

Worked example

The answers: questions 1 and 2

  1. 1. a = 3, b = −5, c = 1
  2. Δ = (−5)² − 4(3)(1)substitute
  3. = 25 − 12 = 13
  4. 13 > 0, not a squaretwo irrational roots
  5. 2. a = 1, b = −10, c = 25
  6. Δ = (−10)² − 4(1)(25)substitute
  7. = 100 − 100 = 0one repeated root
  8. (x − 5)² = 0, so x = 5factorise
Worked example

The answers: questions 3 and 4

  1. 3. 6² − 4(1)(k) = 0equal roots: Δ = 0
  2. 36 − 4k = 0
  3. 4k = 36, so k = 9
  4. 4. k² − 4(2)(8) < 0no real roots: Δ < 0
  5. k² − 64 < 0
  6. k² < 64add 64
  7. −8 < k < 8k between −8 and 8

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