# The sum of two numbers is $6$ times their geometric mean.Show that the ratio of the numbers is $3+2\sqrt 2\::\:3-2\sqrt 2$

$\begin{array}{1 1}3+2\sqrt 2: 3-2\sqrt 2 \\ 4+2\sqrt 2: 2-2\sqrt 2 \\3+3\sqrt 2: 3-3\sqrt 2 \\ 3+2\sqrt 3: 3-2\sqrt 3\end{array}$

Toolbox:
• The G.M. between two numbers $a$ and $b$ = $\sqrt {ab}$
• $(a+b)^2=a^2+b^2+2ab$
• The solutions of the quadratic equation $ax^2+bx+c=0$ are $\large\frac{-b\pm\sqrt {b^2-4ac}}{2a}$
• $(a+b)(a-b)=a^2-b^2$
Let the two numbers be $a$ and $b$.
It is given that the sum of the two numbers $=6\times$ their G.M.
$\Rightarrow\:a+b=6.\sqrt {ab}$
Squaring on both the sides
$(a+b)^2=36ab$
$\Rightarrow\:a^2+b^2+2ab=36ab$
$\Rightarrow\:a^2+b^2=34ab$
Dividing both the sides by $ab$ we get
$\large\frac{a^2+b^2}{ab}$$=34 \Rightarrow\:\large\frac{a}{b}+\frac{b}{a}$$=34$
Step 2
Let $\large\frac{a}{b}$$=x \therefore \:\large\frac{b}{a}=\frac{1}{x} Substituting the value of x \Rightarrow\:x+\large\frac{1}{x}$$=34$
$\Rightarrow\:\large\frac{x^2+1}{x}$$=34 \Rightarrow\:x^2+1=34x \Rightarrow\:x^2-34x+1=0 We know that the solutions of the quadratic equation ax^2+bx+c=0 are \large\frac{-b\pm\sqrt {b^2-4ac}}{2a} In the above equation a=1,\:b=-34\:\:and\:\:c=1 \Rightarrow\:x=\large\frac{-(-34)\pm\sqrt {(-34)^2-4\times 1\times 1}}{2\times 1} \Rightarrow\:x=\large\frac{34\pm \sqrt {1152}}{2} \Rightarrow\:x=17\pm\sqrt {288} \Rightarrow\:x=17\pm 12\sqrt 2 i.e.,\:\:\large\frac{a}{b}$$=17+12\sqrt 2$
Step 3
But $\large\frac{3+2\sqrt 2}{3-2\sqrt 2}=\large\frac{3+2\sqrt 2}{3-2\sqrt 2}\times\large\frac{3+2\sqrt 2}{3+2\sqrt 2}$
(Rationalising the denominator)
$(3+2\sqrt 2)(3-\sqrt 2)=3^2-(2\sqrt 2)^2=9-8=1$
$=\large\frac{9+8+12\sqrt 2}{9-8}$$=17+12\sqrt 2 Hence \large\frac{a}{b}$$=17+12\sqrt 2=\large\frac{3+2\sqrt 2}{3-2\sqrt 2}$
$i.e.,$ The ratio of the two numbers =$a:b=3+2\sqrt 2\::\:3-2\sqrt 2$
Hence proved.