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About this video
Grade 11 equations with rational exponents, all solved with the substitution (k) method: put k in place of the smaller power, factorise the trinomial, then resubstitute and square, cube or raise both sides to the right power.
What it covers
- Substituting k for the smaller power (x to the power ½, ⅓ or −⅓)
- Checking that k² gives the first term
- Factorising the trinomial in k
- Resubstituting, then squaring, cubing or raising both sides to a power
- Rejecting a negative value for x to the power ½
Extra worked examples
✍️ Extra example: solve x − 5x½ + 6 = 0
- 1Let x½ = k, so x = k².Squaring x to the power ½ gives x — so the equation becomes a trinomial.
- 2k² − 5k + 6 = 0 → (k − 2)(k − 3) = 0Factorise.
- 3x½ = 2 → x = 4, or x½ = 3 → x = 9Resubstitute and square both sides. Check x = 4: 4 − 10 + 6 = 0 ✓
✍️ Extra example: solve x⅔ + x⅓ − 6 = 0
- 1Let x⅓ = k, so x⅔ = k².Check: (x to the power ⅓)² = x to the power ⅔.
- 2k² + k − 6 = 0 → (k + 3)(k − 2) = 0Factorise.
- 3x⅓ = −3 → x = (−3)³ = −27, or x⅓ = 2 → x = 2³ = 8Cube both sides — a cube root can be negative, so both answers stand.
Exercises to try
Work each one out, then click to reveal the answer.
- 1Solve x − 7x½ + 12 = 0.
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(k − 3)(k − 4) = 0 with k = x½ → x = 9 or x = 16 - 2Solve x + x½ − 6 = 0.
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(k + 3)(k − 2) = 0 → x½ = −3 is rejected (it is never negative); x½ = 2 → x = 4 - 3Solve x⅔ − 5x⅓ + 4 = 0.
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(k − 1)(k − 4) = 0 with k = x⅓ → x = 1 or x = 64 - 4Solve 2x⅔ − x⅓ − 1 = 0.
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(2k + 1)(k − 1) = 0 → x⅓ = −½ → x = −⅛, or x⅓ = 1 → x = 1 - 5Solve x−⅔ − 4x−⅓ + 3 = 0.
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(k − 1)(k − 3) = 0 with k = x−⅓ → x = 1, or x = 3−3 = 1/27 (raise both sides to the power −3)
