Question

State true or false:

The square root of $\frac{x^2}{y^2}+\frac{y^2}{4x^2}-\frac{x}{y}+\frac{y}{2x}-\frac{3}{4}$ is equal to $\frac{x}{y}-\frac{1}{2}-\frac{y}{2x}$.

• A. True
• B. False

Answer 1

The correct option is A True

Let $z^2=\frac{x^2}{y^2}+\frac{y^2}{4x^2}-\frac{x}{y}+\frac{y}{2x}-\frac{3}{4}$

Now, adding and subtracting $1$ in $z^2$, we get:

$z^2={\left(\frac{x}{y}\right)}^2+{\left(\frac{y}{2x}\right)}^2-\frac{x}{y}+\frac{y}{2x}-\frac{3}{4}+1-1\Rightarrow z^2={\left(\frac{x}{y}\right)}^2+{\left(\frac{y}{2x}\right)}^2-1-\frac{x}{y}+\frac{y}{2x}+\frac{1}{4}\Rightarrow z^2={\left(\frac{x}{y}\right)}^2+{\left(\frac{y}{2x}\right)}^2-2\left(\frac{x}{y}\right)\left(\frac{y}{2x}\right)-\frac{x}{y}+\frac{y}{2x}+\frac{1}{4}\left(Replace1by\left(\frac{2x}{y}\right)\left(\frac{y}{2x}\right)\right)\Rightarrow z^2={(\frac{x}{y}-\frac{y}{2x})}^2-(\frac{x}{y}-\frac{y}{2x}-\frac{1}{4})(\because {(a-b)}^2=a^2+b^2-2ab)$

$\Rightarrow z^2={\left(p\right)}^2-(p-\frac{1}{4})(Let\frac{x}{y}-\frac{y}{2x}=p)\Rightarrow z^2=\frac{{4p}^2-4p+1}{4}\Rightarrow z^2={\left(\frac{2p-1}{2}\right)}^2\Rightarrow z=\frac{2p-1}{2}$

$\Rightarrow z=p-\frac{1}{2}\Rightarrow z=\frac{x}{y}-\frac{y}{2x}-\frac{1}{2}(\because p=\frac{x}{y}-\frac{y}{2x})$

Hence, the square root of $\frac{x^2}{y^2}+\frac{y^2}{4x^2}-\frac{x}{y}+\frac{y}{2x}-\frac{3}{4}$ is $\frac{x}{y}-\frac{y}{2x}-\frac{1}{2}$.