Question:
If (x2 - xy + y2) = 8 and (x4 + x2y2 + y4) = 16, then find the value of (x2 + xy + y2).
If (x2 - xy + y2) = 8 and (x4 + x2y2 + y4) = 16, then find the value of (x2 + xy + y2).
Updated On: Sep 10, 2024
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The Correct Option is C
Solution and Explanation
Given, (x2 + x2y2 + y4) = 16
={(x2 + y2)2 - (xy)2} = 16
Since, a2 - b2 = (a - b)(a + b)
Therefore, (x2 + y2 + xy)(x2 + y2 - xy) = 16
Putting the value of (x2 + y2 - xy), we get
8(x2 + y2 + xy) = 16
Or, (x2 + y2 + xy) = \(\frac{16}{8}\) = 2
So, the correct option is (C) : 2.
={(x2 + y2)2 - (xy)2} = 16
Since, a2 - b2 = (a - b)(a + b)
Therefore, (x2 + y2 + xy)(x2 + y2 - xy) = 16
Putting the value of (x2 + y2 - xy), we get
8(x2 + y2 + xy) = 16
Or, (x2 + y2 + xy) = \(\frac{16}{8}\) = 2
So, the correct option is (C) : 2.
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