Q. 85.0( 1 Vote )

i. If the values of a + b and ab are 12 and 32 respectively, find the values of a2 + b2 and (a – b)2.

ii. If the values of (a – b) and ab are 6 and 40 respectively, find the values of a2 + b2 and (a + b)2.

Answer :

(i) Given (a + b) = 12 and ab = 32


a2 + b2 = (a+ b)2 – 2ab


a2 + b2 = (12)2 – 2(32)


a2 + b2 = 144 – 64


a2 + b2 = 80


(a– b)2 = (a+ b)2 – 4ab


(a– b)2 = (12)2 – 4(32)


(a– b)2 = 144 – 128


(a– b)2 = 16


(ii) Given (a – b) = 6 and ab = 40


a2 + b2 = (a– b)2 + 2ab


a2 + b2 = (6)2 + 2(40)


a2 + b2 = 36 + 80


a2 + b2 = 116


(a+ b)2 = (a– b)2 + 4ab


(a+ b)2 = (6)2 + 4(40)


(a+ b)2 = 36 + 160


(a+ b)2 = 196


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