• $$\Large 225x^{2}-4=0; \sqrt{225y}+2=0$$
 A) If x>y B) $$\Large If \ x\ge y$$ C) lfx D) If x = y or relation cannot be established

 D) If x = y or relation cannot be established

$$\Large 225x^{2}-4=0$$

=> $$\Large 225x^{2}=4 => x^{2}=\frac{4}{225}$$

Therefore, $$\Large x=\sqrt{\frac{4}{225}}=\pm \frac{2}{15}, \ i.e., \ \frac{2}{15} \ and -\frac{2}{15}$$

and $$\Large \sqrt{225y}+2=0 \ or \sqrt{225y}=-2$$

On squaring both sides, we get

$$\Large \sqrt{\left(225y\right)^{2}} = \left(-2\right)^{2}$$

So the relationship cannot be established because $$\Large y = \frac{4}{225}$$

lies between  $$\Large \frac{2}{15} and -\frac{2}{15}$$

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