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(D) For equation $I$: $36 x^{4}+369 x^{2}+900=0$.Let $x^{2} = p$. Then $36 p^{2} + 369 p + 900 = 0$.Dividing by $9$,we get $4 p^{2} + 41 p + 100 = 0$.

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if $x = y$ or relationship between $x$ and $y$ cannot be established.

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(D) For equation $I$: $36 x^{4}+369 x^{2}+900=0$.Let $x^{2} = p$. Then $36 p^{2} + 369 p + 900 = 0$.Dividing by $9$,we get $4 p^{2} + 41 p + 100 = 0$.Using the quadratic formula $p = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$,we get $p = \frac{-41 \pm \sqrt{41^{2} - 4(4)(100)}}{2(4)} = \frac{-41 \pm \sqrt{1681 - 1600}}{8} = \frac{-41 \pm 9}{8}$.$p = \frac{-32}{8} = -4$ or $p = \frac{-50}{8} = -6.25$.Since $x^{2} = -4$ or $x^{2} = -6.25$,$x$ has only imaginary roots.For equation $II$: $144 y^{4}+337 y^{2}+144=0$.Let $y^{2} = q$. Then $144 q^{2} + 337 q + 144 = 0$.Since all coefficients are positive,$q$ must be negative for the equation to hold,leading to imaginary roots for $y$.Since both equations yield imaginary roots for $x$ and $y$,the relationship between $x$ and $y$ cannot be established.

Solve the given two equations and select the correct answer from the given options.
$I.$ $36 x^{4}+369 x^{2}+900=0$
$II.$ $144 y^{4}+337 y^{2}+144=0$

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Solve the given two equations and select the correct answer from the given options.$I.$ $36 x^{4}+369 x^{2}+900=0$$II.$ $144 y^{4}+337 y^{2}+144=0$