In the following questions,two equations numb | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) For equation $I$: $\sqrt{25 x^{2}} - 125 = 0$$\Rightarrow \sqrt{25 x^{2}} = 125$Squaring both sides: $25 x^{2} = 125^{2} = 15625$$\Rightarrow x^{2

Vedclass · Accepted Answer

$x = y$ or the relationship cannot be established.

Vedclass · Answer

(D) For equation $I$: $\sqrt{25 x^{2}} - 125 = 0$$\Rightarrow \sqrt{25 x^{2}} = 125$Squaring both sides: $25 x^{2} = 125^{2} = 15625$$\Rightarrow x^{2} = \frac{15625}{25} = 625$$	herefore x = \pm 25$For equation $II$: $\sqrt{361} y + 95 = 0$$\Rightarrow 19 y + 95 = 0$$\Rightarrow 19 y = -95$$\Rightarrow y = -\frac{95}{19} = -5$Comparing the values: If $x = 25$,then $x > y$ $(25 > -5)$. If $x = -25$,then $x Since we get two different results,the relationship between $x$ and $y$ cannot be established.

In the following questions,two equations numbered $I$ and $II$ are given. You have to solve both the equations and give the answer.
$I$: $\sqrt{25 x^{2}} - 125 = 0$
$II$: $\sqrt{361} y + 95 = 0$

Similar Questions

If $\sqrt{5} = 2.236$,then the value of $\frac{\sqrt{5}}{2} - \frac{10}{\sqrt{5}} + \sqrt{125}$ is:

If $\sqrt{\left(1+\frac{27}{169}\right)}=1+\frac{x}{13},$ then $x$ equals

$\sqrt{\frac{0.324 \times 0.081 \times 4.624}{1.5625 \times 0.0289 \times 72.9 \times 64}} = ?$

If $\sqrt{2} = 1.414$,the square root of $\frac{\sqrt{2}-1}{\sqrt{2}+1}$ is nearest to

$\frac{\sqrt{1296}}{?} = \frac{?}{2.25}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

In the following questions,two equations numbered $I$ and $II$ are given. You have to solve both the equations and give the answer.$I$: $\sqrt{25 x^{2}} - 125 = 0$$II$: $\sqrt{361} y + 95 = 0$