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Find 2ab from a⁴+a²b²+b⁴=900 and a²+ab+b²=45 | IOQM

JEE Maths question with a full step-by-step solution.

Question
Let aa and bb be real numbers such that a4+a2b2+b4=900a^4 + a^2 b^2 + b^4 = 900 and a2+ab+b2=45a^2 + ab + b^2 = 45. Find the value of 2ab2ab.
Solution
Answer: 25
Step 1: Record the given relations.
a2+ab+b2=45...(i),a4+a2b2+b4=900.a^2 + ab + b^2 = 45 \quad \text{...(i)}, \qquad a^4 + a^2 b^2 + b^4 = 900.
 Step 2: Factor the quartic, using a4+a2b2+b4=(a2+ab+b2)(a2−ab+b2)a^4 + a^2 b^2 + b^4 = (a^2 + ab + b^2)(a^2 - ab + b^2):
(a2+ab+b2)(a2−ab+b2)=900.(a^2 + ab + b^2)(a^2 - ab + b^2) = 900.
 Step 3: Substitute (i):
45(a2−ab+b2)=900  ⟹  a2−ab+b2=20...(ii).45(a^2 - ab + b^2) = 900 \implies a^2 - ab + b^2 = 20 \quad \text{...(ii)}.
 Step 4: Subtract (ii) from (i):
(a2+ab+b2)−(a2−ab+b2)=45−20  ⟹  2ab=25.(a^2 + ab + b^2) - (a^2 - ab + b^2) = 45 - 20 \implies 2ab = 25.
 Answer: 2ab=252ab = 25.
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