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The stated equation is quadratic and has exactly one real root. Find a.

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

TOPIC 01

2023 JM01

a=0 gives a linear equation and violates the stated quadratic condition.

Try it. Leave your reasoning visible.

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

01 / Standard#Your turn

The stated equation is quadratic and has exactly one real root. Find a.

4a2x2+2(a+3)x+9=04a^2x^2+2(a+3)x+9=0

Official paper · jm01-2023 · I.5 · PDF 2

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

Skills and prerequisite lessons
  1. Option A3/53/5
  2. −1 or 3/2
  3. Option C3/23/2
  4. −3/7 or 3/5
  5. Any real number

Working and explanation

BUILD THE REASONING

Hint 1
Quadratic means a≠0.
Hint 2
A repeated real root requires discriminant zero.
Worked solution
  1. Set the discriminant equal to zero.

    4(a+3)2−144a2=0  ⟺  (a+3)2=36a24(a+3)^2-144a^2=0\iff(a+3)^2=36a^2
  2. Solve the two linear equations; both satisfy a≠0.

    a+3=±6a  ⟹  a=3/5 or −3/7a+3=\pm6a\implies a=3/5\text{ or }-3/7

D: a=−3/7 or 3/5.

Checks and common pitfalls: a=0 gives a linear equation and violates the stated quadratic condition.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • Set the discriminant equal to zero.
    4(a+3)2−144a2=0  ⟺  (a+3)2=36a24(a+3)^2-144a^2=0\iff(a+3)^2=36a^2
  • Solve the two linear equations; both satisfy a≠0.
    a+3=±6a  ⟹  a=3/5 or −3/7a+3=\pm6a\implies a=3/5\text{ or }-3/7

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Curriculum and source notes ↗