• \( \Large 225x^{2}-4=0; \sqrt{225y}+2=0 \)

    A) If x>y

      B) \( \Large If \ x\ge y \)

    C) lfx

      D) If x = y or relation cannot be established

    Correct Answer:
      D) If x = y or relation cannot be established

    Description for Correct answer

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

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

    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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