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Find a,b,c so that every solution of the preceding linear system also satisfies the quadratic constraint.

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

TOPIC 01

2026 JM02

Checking one parameter value is insufficient when every solution is required.

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

01 / Standard#Your turn

Find a,b,c so that every solution of the preceding linear system also satisfies the quadratic constraint.

ax2+by2+cz2=3ax^2+by^2+cz^2=3

Official paper · jm02-2026 · 5(b)(ii) · PDF 7

Official original and suggested answers ↗ · Suggested answer PDF page 11

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Substitute the full parameterised solution.
Hint 2
The resulting polynomial must be an identity for all real t.
Worked solution
  1. Expand after substitution.

    a(1+t)2+b(1−2t)2+ct2=(a+4b+c)t2+(2a−4b)t+(a+b)a(1+t)^2+b(1-2t)^2+ct^2=(a+4b+c)t^2+(2a-4b)t+(a+b)
  2. Match coefficients with the constant 3 and solve.

    a+4b+c=0,2a−4b=0,a+b=3  ⟹  (a,b,c)=(2,1,−6)a+4b+c=0,\quad2a-4b=0,\quad a+b=3\implies(a,b,c)=(2,1,-6)

a=2, b=1, c=−6.

Checks and common pitfalls: Checking one parameter value is insufficient when every solution is required.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • Expand after substitution.
    a(1+t)2+b(1−2t)2+ct2=(a+4b+c)t2+(2a−4b)t+(a+b)a(1+t)^2+b(1-2t)^2+ct^2=(a+4b+c)t^2+(2a-4b)t+(a+b)
  • Match coefficients with the constant 3 and solve.
    a+4b+c=0,2a−4b=0,a+b=3  ⟹  (a,b,c)=(2,1,−6)a+4b+c=0,\quad2a-4b=0,\quad a+b=3\implies(a,b,c)=(2,1,-6)

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Curriculum and source notes ↗