### Instruction: If $x,$ $y,$ and $k$ are positive real numbers such that \[3=k^2\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\right)+k\left(\dfrac{x}{y}+\dfrac{y}{x}\right),\]find the maximum possible value of $k.$.### Response: This is a tricky problem because it involves a quadratic equation in $k$ and two other variables $x$ and $y.$
I notice that the equation is symmetric in $x$ and $y,$ so I can assume without loss of generality that $x \geq y.$
Then I can write $x = cy$ for some constant $c \geq 1.$
Substituting this into the equation, I get \[3 = k^2(c^4 + 1/c^4) + k(c + 1/c).\]
Now I have a quadratic equation in $k$ with coefficients depending on $c.$
To find the maximum possible value of $k,$ I need to minimize the value of the discriminant of this quadratic equation, which is \[b^2 - 4ac = (c + 1/c)^2 - 4(c^4 + 1/c^4).\]
I notice that this is a difference of two squares, so I can factor it as \[(c + 1/c + 2)(c + 1/c - 2)(c^2 + 1/c^2 - 2).\]
Since $c \geq 1,$ the first factor is always positive, so I can ignore it.
The second factor is also always positive, since $c + 1/c \geq 2$ by AM-GM inequality.
The third factor can be negative, zero, or positive, depending on the value of $c.$
I want to minimize the absolute value of the discriminant, so I want to find the value of $c$ that makes the third factor as close to zero as possible.
I see that the third factor is zero when $c^2 + 1/c^2 = 2,$ which implies that $c = 1.$
Therefore, the minimum value of the discriminant occurs when $c = 1,$ which means that $x = y.$
Plugging this back into the original equation, I get \[3 = k^2(2) + k(2),\] which simplifies to \[k^2 + k - 3/2 = 0.\]
To find the maximum possible value of $k,$ I use the quadratic formula and choose the positive root, which is \[k = \dfrac{-1 + \sqrt{1 + 6}}{2} = \dfrac{\sqrt{7} - 1}{2}.\]