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The reciprocal numbers 1/α and 1/β are the roots; evaluate the exponential product.

Read the idea, work independently, then explain what changed.

TOPIC 01

2025 JM01

α and β themselves are not the roots of the printed quadratic.

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Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Standard#Your turn

The reciprocal numbers 1/α and 1/β are the roots; evaluate the exponential product.

2x2+2x−1=0,2α+12β+12x^2+2x-1=0,\quad 2^{\alpha+1}2^{\beta+1}

Official paper · jm01-2025 · I.2 · PDF 2

Official original and suggested answers ↗ · Suggested answer PDF page 5

Skills and prerequisite lessons
  1. Option A11
  2. Option B22
  3. Option C44
  4. Option D1616
  5. Option E116\frac1{16}

Working and explanation

BUILD THE REASONING

Hint 1
Use Vieta for the reciprocal roots.
Hint 2
Their sum divided by their product equals α+β.
Worked solution
  1. Recover the sum of α and β.

    1α+1β=−1,1αβ=−12  ⟹  α+β=2\frac1\alpha+\frac1\beta=-1,\quad\frac1{\alpha\beta}=-\frac12\implies\alpha+\beta=2
  2. Add exponents with the same base.

    2α+12β+1=2α+β+2=162^{\alpha+1}2^{\beta+1}=2^{\alpha+\beta+2}=16

D: 16.

Checks and common pitfalls: α and β themselves are not the roots of the printed quadratic.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • Recover the sum of α and β.
    1α+1β=−1,1αβ=−12  ⟹  α+β=2\frac1\alpha+\frac1\beta=-1,\quad\frac1{\alpha\beta}=-\frac12\implies\alpha+\beta=2
  • Add exponents with the same base.
    2α+12β+1=2α+β+2=162^{\alpha+1}2^{\beta+1}=2^{\alpha+\beta+2}=16

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Curriculum and source notes ↗