Question

$I.\sqrt{500x}+\sqrt{402}=0$
$II.\sqrt{360y}+(200{)}^{\frac{1}{2}}=0$

• A. $x>y$
• B. $x\ge y$
• C. $x<y$
• D. $x\le y$
• 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
$\sqrt{500x}+\sqrt{402}=0$
$\sqrt{500x}+\sqrt{402}=0$
$\Rightarrow \sqrt{500x}=\sqrt{402}$
On by squaring both sides, we get
$500x^2=402$
$\Rightarrow x=\sqrt{\frac{402}{500}}=\pm 0.897$
Equation II
$\sqrt{360}y+(200{)}^{\frac{1}{2}}=0\Rightarrow (200{)}^{\frac{1}{2}}=-\sqrt{360}y$
On by squaring both sides, we get
$\left(\right(200{)}^{\frac{1}{2}}{)}^2=(-\sqrt{360}y{)}^2\Rightarrow 200=360y^2\therefore y=\sqrt{\frac{200}{360}}=\pm 0.75$
Hence, relationship cannot established