NCEA Level 3Algebra with constants and surdsAS 91577

Algebra with k: quadratics, discriminants and surd equations

What this lesson covers

  • The exam uses unknown constants so the graphing calculator cannot do the algebra.
  • Squaring both sides can create false solutions, so every solution must be checked in the original equation.
Key idea

Why the exam uses k

  • Solve for x in terms of k.
  • A calculator cannot solve with k in it.
  • 2025: too much reliance on the calculator.
Worked example

2x² − 3kx + k² = 0

  • Treat k as a number.
  • The discriminant is k².
  • Check by substituting.
  1. a = 2, b = −3k, c = k²k is a number
  2. b² − 4ac = (−3k)² − 4(2)(k²)
  3. = 9k² − 8k² = k²
  4. x = (3k ± √(k²))/4the formula
  5. = (3k ± k)/4± covers the sign
  6. x = 4k/4 or x = 2k/4
  7. x = k or x = k/2
  8. 2k² − 3k(k) + k² = 0x = k
Worked example

Completing the square with k

  • Spot (z − k)².
  • The root of −4 is 2i.
  • Keep the ±.
  1. z² − 2kz + k² + 4 = 0
  2. (z − k)² + 4 = 0first three terms
  3. (z − k)² = −4
  4. z − k = ±√−4 = ±2i√−4 = 2i
  5. z = k ± 2i
  6. (k + 2i − k)² = (2i)² = −4
Worked example

The discriminant with k

  • Find b² − 4ac in k.
  • Then = 0, ≥ 0 or < 0.
  • k ≠ 0: still a quadratic.
  1. a = k, b = 6, c = k
  2. b² − 4ac = 36 − 4k²
  3. 36 − 4k² = 0equal roots
  4. k² = 9, so k = ±3
  5. 36 − 4k² ≥ 0, so k² ≤ 9real roots
  6. −3 ≤ k ≤ 3, k ≠ 0else not quadratic
  7. 36 − 4k² < 0, so k² > 9non-real roots
  8. k < −3 or k > 3
Worked example

Read what is asked

  • Asked: x in terms of k.
  • Not k in terms of x.
  1. asked: x in terms of k
  2. x = k or x = k/2answers it
  3. k² − 3xk + 2x² = 0a quadratic in k
  4. (k − x)(k − 2x) = 0factorise
  5. k = x or k = 2xwrong way round
Worked example

Two surds: the 2025 error

  • Not term by term.
  • (a − b)² is not a² − b².
  1. √(2x) − √(x + 1) = 1
  2. 2x − (x + 1) = 1squared each term
  3. x = 2
  4. √4 − √3 ≈ 0.27not 1, so wrong
  5. (a − b)² = a² − 2ab + b²not a² − b²
Worked example

Isolate one surd, then square

  • Get one surd alone.
  • Expand the square in full.
  1. √(2x) = 1 + √(x + 1)isolate one surd
  2. 2x = (1 + √(x + 1))²square both sides
  3. = 1 + 2√(x + 1) + x + 1expand in full
  4. 2x = x + 2 + 2√(x + 1)collect
  5. x − 2 = 2√(x + 1)take x + 2 off
Worked example

Square again, then solve

  • Square again.
  • Two answers to check.
  1. x − 2 = 2√(x + 1)
  2. (x − 2)² = 4(x + 1)square again
  3. x² − 4x + 4 = 4x + 4expand
  4. x² − 8x = 0
  5. x(x − 8) = 0factorise
  6. x = 0 or x = 8both to check
Worked example

Check both in the original

  • x = 8 works.
  • x = 0 gives −1: reject it.
  • Say why you reject it.
  1. √(2x) − √(x + 1) = 1
  2. x = 8: √16 − √9substitute
  3. = 4 − 3 = 1
  4. x = 0: √0 − √1 = −1−1 is not 1
  5. x = 8
Key idea

Where false answers come from

  • 1 and −1 have the same square.
  • A square root has two values.
  • Include or reject each one.
Key idea

Set out every line

  • Keep k as a letter to the end.
  • Keep surds exact.
  • Cross out working you reject.
Key idea

Where the marks are

  • Achieved: formula, square, discriminant.
  • Merit: surd equations, checked.
  • Excellence: harder surds, new strategies.
Summary

What to take away

  • Treat k as a number, isolate before squaring, and check every answer.
Formulawith k as a number
Complete squarethen square root, ±
Discriminantequation or inequality
Read itx in terms of k
Two surdsisolate one, then square
Every answercheck in the original
Your turn

Your turn: four questions

  • Pause here and answer each one.
  • Question 2: expect complex roots.
  • Question 4: check both answers.
GivenFind
1x² + 2kx − 8k² = 0x in terms of k
2z² + 6z + k = 0, k > 9z in terms of k
3x² + kx + 16 = 0k, for equal roots
4√(x + 4) − √(2x + 1) = 1x
Show answer

1. Discriminant 36k², so x = 2k or x = −4k. 2. (z + 3)² = 9 − k, which is negative, so z = −3 ± i√(k − 9). 3. k² − 64 = 0, so k = ±8. 4. Isolate and square twice: x² − 12x = 0. x = 0 checks; x = 12 gives −1, so reject it. x = 0.

Worked example

The answers: questions 1 to 3

  1. 1. b² − 4ac = 4k² + 32k² = 36k²
  2. x = (−2k ± 6k)/2
  3. x = 2k or x = −4k
  4. 2. (z + 3)² − 9 + k = 0complete the square
  5. (z + 3)² = 9 − k < 0k > 9
  6. z + 3 = ±i√(k − 9)
  7. z = −3 ± i√(k − 9)
  8. 3. b² − 4ac = k² − 64 = 0equal roots
  9. k² = 64, so k = ±8
Worked example

The answers: question 4

  1. 4. √(x + 4) = 1 + √(2x + 1)isolate
  2. x + 4 = 1 + 2√(2x + 1) + 2x + 1square, expand
  3. 2 − x = 2√(2x + 1)tidy up
  4. 4 − 4x + x² = 8x + 4square again
  5. x² − 12x = 0, x = 0 or 12factorise
  6. x = 12: √16 − √25 = −1−1 is not 1
  7. x = 0: √4 − √1 = 1
  8. x = 0

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