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Find the complete real solution set.

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

TOPIC 01

2023 JM01

Reject t=−6 because t is a principal square root.

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

Find the complete real solution set.

x2−3x+4x2−3x=12x^2-3x+4\sqrt{x^2-3x}=12

Official paper · jm01-2023 · I.4 · PDF 2

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

Skills and prerequisite lessons
  1. Option A−1{-1}
  2. Option B2,−6{2,-6}
  3. Option C−1,4{-1,4}
  4. Option D4{4}
  5. Option E3{3}

Working and explanation

BUILD THE REASONING

Hint 1
Set t=√(x²−3x), with t≥0.
Hint 2
Solve for t before returning to x.
Worked solution
  1. Factor the equation in t.

    t2+4t−12=(t−2)(t+6)=0  ⟹  t=2t^2+4t-12=(t-2)(t+6)=0\implies t=2
  2. Solve the remaining quadratic and check its radicand.

    x2−3x=4  ⟺  (x−4)(x+1)=0x^2-3x=4\iff(x-4)(x+1)=0

C: {−1,4}.

Checks and common pitfalls: Reject t=−6 because t is a principal square root.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • Factor the equation in t.
    t2+4t−12=(t−2)(t+6)=0  ⟹  t=2t^2+4t-12=(t-2)(t+6)=0\implies t=2
  • Solve the remaining quadratic and check its radicand.
    x2−3x=4  ⟺  (x−4)(x+1)=0x^2-3x=4\iff(x-4)(x+1)=0

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Curriculum and source notes ↗