Question : If $x^4+x^2 y^2+y^4=21$ and $x^2+xy+y^2=3$, then what is the value of $4xy $?
Option 1: –8
Option 2: 4
Option 3: –4
Option 4: 12
Answer (1)
19th Jan, 2024
Correct Answer: –8
Solution :
We have,
$x^4+x^2 y^2+y^4=21$ and,
$x^2+xy+y^2=3$ -----------------(i)
Use the identity,
$x^4+x^2 y^2+y^4=(x^2-x y+y^2)(x^2+x y+y^2)$
$⇒21=3(x^2-x y+y^2)$
$⇒x^2-x y+y^2=7$ ---------------------(ii)
From (i) and (ii),
$⇒2xy=-4$
$⇒4xy=-8$
Hence, the correct answer is –8.
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