If $x^4+y^4+x^2 y^2=21$ and $x^2+y^2-x y=7$, then what is the value of $\frac{x}{y}+\frac{y}{x}$ ?
Answer & explanation
Correct answer: option 4
x4 + x2y2 + y4 = (x2 – xy + y2) (x2 + xy + y2)
If $x^4+y^4+x^2 y^2=21$
$x^2+y^2-x y=7$
Then, $x^2+y^2+x y$ = \(\frac{21}{7}\) = 3
what is the value of $\frac{x}{y}+\frac{y}{x}$ = \(\frac{x^2 + y^2}{xy}\)---(A)
$x^2+y^2-x y=7$
$x^2+y^2+x y$ = 3
So from these two equations ,
x2 + y2 = 5
xy = -2
Put in (A)
\(\frac{x^2 + y^2}{xy}\) = -\(\frac{5}{2}\)