ICSE Class 8 Answered
If 64^x2+y2 =72 nd xy=2 then find (4^x2 ×y/2)^2
Asked by Takkarsakshi2003 | 19 Nov, 2019, 10:41: PM
Expert Answer
The question should be like this
If 64x2 + y2 =72 nd xy = 2 then find (4x + y/2)2
(4x + y/2)2 = (4x)2 + 2 × 4x × y/2 + y2/4
= 16x2 + 4xy + y2/4
= (64x2 + 16xy + y2)/4
= (64x2 + y2 + 16xy)/4
= (72 + 32)/4
= 26
(4x + y/2)2 = 26
Answered by Sneha shidid | 20 Nov, 2019, 09:36: AM
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