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


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  \( \Large \frac{2}{15} and -\frac{2}{15} \)


Part of solved Quadratic Equations questions and answers : >> Elementary Mathematics >> Quadratic Equations








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