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.
(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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