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.