Question

I. $225x^2-4=0$
II. $\sqrt{225}y+2=0$

• A. If $x>y$
• B. If $x\ge y$
• C. If $x<y$
• D. If $x\le y$
• E. $x=y$ or the relationship cannot be established.

Answer 1

The correct option is E $x=y$ or the relationship cannot be established.

Equation I:-
$225x^2-4=0$
$\Rightarrow 225x^2=4$
$\Rightarrow x^2=\frac{4}{225}$
$\Rightarrow x=\sqrt{\frac{4}{225}}=\pm \frac{2}{15}$

Equation II:-
$\sqrt{225}y+2=0$
$\sqrt{225}y=-2$
On squaring both sides, we get
$225y^2=4$
$y^2=\frac{4}{225}$
$\Rightarrow y=\pm \frac{2}{15}$
Hence, $x=y$