The sum of two numbers is 6 times their geometric means,
Question:

The sum of two numbers is 6 times their geometric means, show that the numbers are in the ratio

$(3+2 \sqrt{2}):(3-2 \sqrt{2})$

Solution:

Let the two numbers be $a$ and $b$.

Let the geometric mean between them be $G$.

We have:

$a+b=6 G$

But, $G=\sqrt{a b}$

$\therefore a+b=6 \sqrt{a b}$

$\Rightarrow(a+b)^{2}=(6 \sqrt{a b})^{2}$

$\Rightarrow a^{2}+2 a b+b^{2}=36 a b$

$\Rightarrow a^{2}-34 a b+b^{2}=0$

Using the quadratic formula:

$\Rightarrow a=\frac{-(-34 b) \pm \sqrt{(-34 b)^{2}-4 \times 1 \times b^{2}}}{2 \times 1}$

$\Rightarrow a=\frac{34 b \pm b \sqrt{1156-4}}{2}$

$\Rightarrow a=\frac{b(34 \pm \sqrt{1152})}{2}$

$\Rightarrow \frac{a}{b}=\frac{34 \pm 24 \sqrt{2}}{2}$

$\Rightarrow \frac{a}{b}=17+12 \sqrt{2} \quad[\because a$ and $b$ are positive numbers $]$

$\Rightarrow \frac{a}{b}=3+8+2 \times 3 \times 2 \sqrt{2}$

$\Rightarrow \frac{a}{b}=(3+2 \sqrt{2})^{2}$

$\Rightarrow \frac{a}{b}=\frac{(3+2 \sqrt{2})^{2}(3-2 \sqrt{2})}{(3-2 \sqrt{2})}$

$\Rightarrow \frac{a}{b}=\frac{(3+2 \sqrt{2})(9-8)}{(3-2 \sqrt{2})}$

$\Rightarrow \frac{a}{b}=\frac{(3+2 \sqrt{2})}{(3-2 \sqrt{2})}$

$\Rightarrow a: b=(3+2 \sqrt{2}):(3-2 \sqrt{2})$

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