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Real Pairs with a²+b²=25 and 3(a+b)-ab=15 | IOQM

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

Question
The number of all real pairs (a,b)(a, b) satisfying a2+b2=25a^2 + b^2 = 25 and 3(a+b)−ab=153(a + b) - ab = 15.
Solution
Answer: 4
Step 1: Combine the equations by subtracting twice the second from the first:
(a2+b2)−2[3(a+b)−ab]=25−30=−5.(a^2 + b^2) - 2\big[3(a + b) - ab\big] = 25 - 30 = -5.
 Step 2: Simplify the left side.
a2+b2+2ab−6(a+b)=−5  ⟹  (a+b)2−6(a+b)+5=0.a^2 + b^2 + 2ab - 6(a + b) = -5 \implies (a + b)^2 - 6(a + b) + 5 = 0.
 Step 3: Let s=a+bs = a + b. Then s2−6s+5=0s^2 - 6s + 5 = 0, so (s−1)(s−5)=0(s - 1)(s - 5) = 0, giving s=1s = 1 or s=5s = 5. Step 4: Case a+b=5a + b = 5. From 3(a+b)−ab=153(a + b) - ab = 15: 15−ab=1515 - ab = 15, so ab=0ab = 0. With a+b=5a + b = 5 and ab=0ab = 0, (a,b)=(5,0)(a, b) = (5, 0) or (0,5)(0, 5). Step 5: Case a+b=1a + b = 1. Then 3−ab=153 - ab = 15, so ab=−12ab = -12. With a+b=1a + b = 1 and ab=−12ab = -12, aa and bb are roots of t2−t−12=0t^2 - t - 12 = 0, that is t=4t = 4 or t=−3t = -3. So (a,b)=(4,−3)(a, b) = (4, -3) or (−3,4)(-3, 4). Answer: (a,b)=(5,0), (0,5), (4,−3), (−3,4)(a, b) = (5, 0),\ (0, 5),\ (4, -3),\ (-3, 4).
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