Question

The sum of two numbers is six times their geometric mean. Show that the numbers are in the ratio3+2√2:3-2√2

Answer 1

$\Large{\textbf{\underline{\underline{According\;to\;the\;Question}}}}$

Assume

Numbers be x and y

Then,

x + y = 6√xy

$\rm \: \: \: \: \: \: \: \: \: \dfrac{x+y}{2\sqrt{xy}}=\dfrac{3}{1}$

By Componendo and dividendo

$\rm \: \: \: \: \: \: \: \: \: \dfrac{x+y+2\sqrt{xy}}{x+y-2\sqrt{xy}}=\dfrac{3+1}{3-1}=\dfrac{2}{1}$

$\rm \: \: \: \: \: \: \: \: \: (\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}})^2=\dfrac{2}{1}$

$\rm \: \: \: \: \: \: \: \: \: \dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}=(\dfrac{2}{1})^2$

$\rm \: \: \: \: \: \: \: \: \: \dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\sqrt{\dfrac{2}{1}}$

Again,

By Componendo and dividendo

$\rm \: \: \: \: \: \: \: \: \: \dfrac{(\sqrt{x}+\sqrt{y})+(\sqrt{x}-\sqrt{y})}{(\sqrt{x}+\sqrt{y})+(\sqrt{x}-\sqrt{y})}=\dfrac{\sqrt{2}+1}{\sqrt{2}-1}$

$\rm \: \: \: \: \: \: \: \: \: \dfrac{\sqrt{x}}{\sqrt{y}}=\dfrac{\sqrt{2}+1}{\sqrt{2}-1}$

Squarring Both sides :-

$\rm \: \: \: \: \: \: \: \: \: (\dfrac{\sqrt{x}}{\sqrt{y}})^2=\dfrac{(\sqrt{2}+1)^2}{(\sqrt{2}-1)^2}$

$\rm \: \: \: \: \: \: \: \: \: \dfrac{x}{y}=\dfrac{3+2\sqrt{2}}{3-2\sqrt{2}}$

Hence,

x : y = (3 + 2√2) : (3 - 2√2)

Answer 2

Answer:

Solve this question by using componendo and dividendo.

According to componendo and dividendo,

if a/b = c/d

then,

a+b/a-b = c+d/c-d

You may refer to the above answer for complete detailed solution.