Question

If AM and GM of $x$ and $y$ are in the ratio $p:q$, then $x:y$ is

• A. $p–\surd (p^2+q^2):p+\surd (p^2+q^2)$
• B. $p+\surd (p^2–q^2):p–\surd (p^2–q^2)$
• C. $p:q$
• D. None of these

Answer 1

The correct option is B

$p+\surd (p^2–q^2):p–\surd (p^2–q^2)$

Explanation for the correct option:

Step 1. Find the value of $x:y$:

AM of $x$ and $y=\frac{(x+y)}{2}$

GM $=\surd \left(xy\right)$

$\therefore$AM/GM $=\frac{(x+y)}{2\surd \left(xy\right)}$

$=\frac{p}{q}$

Step 2. By componendo dividendo rule, we get

$\frac{x+y+2\surd \left(xy\right)}{x+y–2\surd \left(xy\right)}=\frac{(p+q)}{(p-q)}$

$\Rightarrow$$\frac{{(\surd x+\surd y)}^2}{{(\surd x–\surd y)}^2}=\frac{(p+q)}{(p-q)}$

Take square root on both sides, we get

$\frac{\surd x+\surd y}{\surd x–\surd y}=\frac{\surd (p+q)}{\surd (p-q)}$

Step 3. Again By componendo dividendo rule, we get

$\frac{(\surd x+\surd y)+(\surd x–\surd y)}{(\surd x+\surd y)–(\surd x–\surd y)}=\frac{\surd (p+q)+\surd (p-q)}{\surd (p+q)-\surd (p-q)}$

$\Rightarrow$ $\frac{2\surd x}{2\surd y}=\frac{\surd (p+q)+\surd (p-q)}{\surd (p+q)-\surd (p-q)}$

$\Rightarrow$ $\frac{\surd x}{\surd y}=\frac{\surd (p+q)+\surd (p-q)}{\surd (p+q)-\surd (p-q)}$

Step 4. By Squaring both sides, we get

$\begin{array}{rcl}\frac{x}{y}& =& \frac{p+q+p–q+2\surd (p+q)\surd (p-q)}{(p+q)-(p-q)–2\surd (p+q)\surd (p-q)}\\ & =& \frac{2p+2\surd (p^2-q^2)}{2p–2\surd (p^2-q^2)}\\ & =& \frac{\mathbf{p}\mathbf{}\mathbf{+}\mathbf{}\mathbf{\surd }\mathbf{(}{\mathbf{p}}^{\mathbf{2}}\mathbf{-}{\mathbf{q}}^{\mathbf{2}}\mathbf{)}\mathbf{}}{\mathbf{p}\mathbf{}\mathbf{–}\mathbf{}\mathbf{\surd }\mathbf{(}{\mathbf{p}}^{\mathbf{2}}\mathbf{-}{\mathbf{q}}^{\mathbf{2}}\mathbf{)}}\end{array}$

Hence, Option ‘B’ is Correct.