Algebra with constants: 7 exam questions solved step by step

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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

7 questions, one method

  • 1. Treat k as a number.
  • 2. Formula, square or discriminant.
  • 3. Two surds: isolate one, then square.
  • 4. Check every answer.
Your turn

Question 1 · Achieved

  • Solve for x in terms of k:
  • 3x² + 2kx − k² = 0
  • Use the quadratic formula.
Worked example

Q1: 3x² + 2kx − k² = 0

  • k is just a number.
  • ± covers the sign of k.
  1. a = 3, b = 2k, c = −k²k is a number
  2. b² − 4ac = (2k)² − 4(3)(−k²)
  3. = 4k² + 12k² = 16k²
  4. x = (−2k ± √(16k²))/6the formula
  5. = (−2k ± 4k)/6√(16k²) = 4k
  6. x = 2k/6 or x = −6k/6
  7. x = k/3 or x = −k
  8. 3(k/3)² + 2k(k/3) − k² = 0x = k/3
  9. 3k² − 2k² − k² = 0x = −k
Your turn

Question 2 · Achieved

  • k is real and not zero. Solve
  • z² + 4kz + 5k² = 0
  • by completing the square.
Worked example

Q2: z² + 4kz + 5k² = 0

  • Half of 4k is 2k.
  • The root of −k² is ki.
  1. z² + 4kz + 5k² = 0
  2. (z + 2k)² = z² + 4kz + 4k²half of 4k is 2k
  3. (z + 2k)² − 4k² + 5k² = 0take off 4k²
  4. (z + 2k)² + k² = 0
  5. (z + 2k)² = −k²
  6. z + 2k = ±ki√(−k²) = ki
  7. z = −2k ± kitwo complex roots
  8. (−2k + ki + 2k)² = (ki)² = −k²
Your turn

Question 3 · Merit

  • The quadratic kx² − 4x + (k − 3) = 0
  • has real roots.
  • Find the possible values of k.
  • Careful with the inequality sign.
Worked example

Q3: real roots

  • Real roots: b² − 4ac ≥ 0.
  • ÷ by a negative: flip.
  1. a = k, b = −4, c = k − 3
  2. b² − 4ac = 16 − 4k(k − 3)
  3. = 16 − 4k² + 12kexpand
  4. 16 − 4k² + 12k ≥ 0real roots
  5. k² − 3k − 4 ≤ 0÷ −4: sign flips
  6. (k − 4)(k + 1) ≤ 0factorise
  7. −1 ≤ k ≤ 4between the roots
  8. k ≠ 0else not quadratic
  9. −1 ≤ k ≤ 4, k ≠ 0
Your turn

Question 4 · Merit

  • Solve √(2x − 4) − √(x − 1) = 1
  • Squaring each term will not work.
Worked example

Q4: isolate, then square

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

Q4: square again, then check

  • Square again.
  • Check in the original.
  1. x − 4 = 2√(x − 1)
  2. (x − 4)² = 4(x − 1)square again
  3. x² − 8x + 16 = 4x − 4expand
  4. x² − 12x + 20 = 0
  5. (x − 2)(x − 10) = 0factorise
  6. x = 2 or x = 10
  7. x = 10: √16 − √9 = 4 − 3 = 1
  8. x = 2: √0 − √1 = −1−1 is not 1
  9. x = 10
Your turn

Question 5 · Merit, in context

  • A rectangle is (x + 2k) m long and
  • (x − k) m wide, where k > 0.
  • Its area is 10k² m².
  • Find x and the sides in terms of k.
Worked example

Q5: the rectangle

  • Area = length × width.
  • A width is never negative.
  1. (x + 2k)(x − k) = 10k²length × width
  2. x² + kx − 2k² = 10k²expand
  3. x² + kx − 12k² = 0
  4. b² − 4ac = k² + 48k² = 49k²
  5. x = (−k ± 7k)/2√(49k²) = 7k
  6. x = 3k or x = −4k
  7. x = −4kwidth −5k < 0
  8. (5k)(2k) = 10k²x = 3k
  9. x = 3k: 5k m by 2k m
Your turn

Question 6 · Excellence

  • k is a non-zero constant. Solve
  • kx² − (k² + 1)x + k = 0
  • Careful: x in terms of k.
Worked example

Q6: x in terms of k

  • A perfect square inside.
  • Answer for x, not k.
  1. a = k, b = −(k² + 1), c = k
  2. b² − 4ac = (k² + 1)² − 4k²
  3. = k⁴ + 2k² + 1 − 4k²expand
  4. = k⁴ − 2k² + 1 = (k² − 1)²
  5. x = (k² + 1 ± (k² − 1))/(2k)
  6. x = 2k²/(2k) or x = 2/(2k)
  7. x = k or x = 1/k
  8. k = x or k = 1/xwrong way round
  9. x = 1/k: 1/k − k − 1/k + k = 0
Your turn

Question 7 · Excellence

  • k > 0. Solve for x in terms of k:
  • √(2x − 5k) − √(x + k) = √k
Worked example

Q7: isolate and square

  • Isolate one surd.
  • Keep √k exact.
  1. √(2x − 5k) = √k + √(x + k)isolate
  2. 2x − 5k = k + 2√(k(x + k)) + x + k
  3. 2x − 5k = x + 2k + 2√(k(x + k))
  4. x − 7k = 2√(k(x + k))take x + 2k off
Worked example

Q7: square again, then check

  • Square again.
  • Check both in the original.
  1. x − 7k = 2√(k(x + k))
  2. x² − 14kx + 49k² = 4k(x + k)
  3. x² − 18kx + 45k² = 0collect
  4. (x − 3k)(x − 15k) = 03k × 15k = 45k²
  5. x = 3k or x = 15k
  6. x = 15k: √(25k) − √(16k)
  7. = 5√k − 4√k = √k
  8. x = 3k: √k − √(4k) = −√knot √k
  9. x = 15k
Summary

The method that solved all seven

  • Treat k as a number, isolate before squaring, check every root.
StepHow
Formulak stays a letter to the end
Squarehalf the middle, then ±
Discriminantflip the sign if ÷ by −
Two surdsisolate, square, repeat
Checkevery root, in the original
Read itx in terms of k

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