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The sum of the two numbers is 6 times their geometric mean. Show that the two numbers are in the ratio $\left( 3+2\sqrt{2} \right):\left( 3-2\sqrt{2} \right)$.

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Hint: Consider two numbers and give each number a variable, say a and b. For these two numbers, the geometric mean is given by $\sqrt{ab}$. In the question, we are given that the sum of the two numbers is 6 times the geometric mean of the two numbers. Use this to find the value of the ratio of these two numbers.

Complete step-by-step answer:

Before proceeding with the question, we must know the formula that will be required to solve this question. For any two numbers a and b, the geometric mean of these two numbers is given by,
$\sqrt{ab}$ . . . . . . . . . . . . (1)
In this question, it is given that the sum of the two numbers is 6 times their geometric mean and we are required to find the ratio of these two numbers.
Let us assume that these two numbers are a and b. From (1), the geometric mean of these two numbers is equal to $\sqrt{ab}$. Since it is given that the sum of the two numbers is 6 times their geometric mean, we can write,
$\begin{align}
  & a+b=6\sqrt{ab} \\
 & \Rightarrow \dfrac{a+b}{\sqrt{ab}}=6 \\
 & \Rightarrow \dfrac{a}{\sqrt{ab}}+\dfrac{b}{\sqrt{ab}}=6 \\
 & \Rightarrow \sqrt{\dfrac{a}{b}}+\sqrt{\dfrac{b}{a}}=6 \\
 & \Rightarrow \sqrt{\dfrac{a}{b}}+\dfrac{1}{\sqrt{\dfrac{a}{b}}}=6 \\
\end{align}$
Let us substitute $\sqrt{\dfrac{a}{b}}=t$. So, we get,
$\begin{align}
  & t+\dfrac{1}{t}=6 \\
 & \Rightarrow \dfrac{{{t}^{2}}+1}{t}=6 \\
 & \Rightarrow {{t}^{2}}+1=6t \\
 & \Rightarrow {{t}^{2}}-6t+1=0 \\
\end{align}$
To solve this equation, we will use quadratic formula from which, the roots of the quadratic equation \[a{{x}^{2}}+bx+c=0\] are given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. So, for the above equation, we can say,
$\begin{align}
  & t=\dfrac{-\left( -6 \right)\pm \sqrt{{{\left( -6 \right)}^{2}}-4.1.1}}{2.1} \\
 & \Rightarrow t=\dfrac{6\pm \sqrt{36-4}}{2} \\
 & \Rightarrow t=\dfrac{6\pm \sqrt{32}}{2} \\
 & \Rightarrow t=\dfrac{6\pm 4\sqrt{2}}{2} \\
 & \Rightarrow t=3\pm 2\sqrt{2} \\
\end{align}$
 Since $\sqrt{\dfrac{a}{b}}=t$, we can write,
\[\begin{align}
  & \sqrt{\dfrac{a}{b}}=3\pm 2\sqrt{2} \\
 & \Rightarrow \sqrt{\dfrac{a}{b}}=3+2\sqrt{2},\sqrt{\dfrac{a}{b}}=3-2\sqrt{2} \\
\end{align}\]
Let us consider \[\sqrt{\dfrac{a}{b}}=3+2\sqrt{2}\]. This can be also written as,
\[\begin{align}
  & \sqrt{\dfrac{a}{b}}={{\left( \sqrt{2} \right)}^{2}}+{{1}^{2}}+2.1.\sqrt{2} \\
 & \Rightarrow \sqrt{\dfrac{a}{b}}={{\left( \sqrt{2}+1 \right)}^{2}} \\
 & \Rightarrow \sqrt{\dfrac{a}{b}}={{\left( \sqrt{2}+1 \right)}^{2}} \\
 & \Rightarrow \sqrt{\dfrac{a}{b}}=\dfrac{{{\left( \sqrt{2}+1 \right)}^{2}}}{\left( \sqrt{2}-1 \right)\left( \sqrt{2}+1 \right)} \\
 & \Rightarrow \sqrt{\dfrac{a}{b}}=\dfrac{\left( \sqrt{2}+1 \right)}{\left( \sqrt{2}-1 \right)} \\
\end{align}\]
Squaring both the sides, we get,
\[\begin{align}
  & {{\left( \sqrt{\dfrac{a}{b}} \right)}^{2}}=\dfrac{{{\left( \sqrt{2}+1 \right)}^{2}}}{{{\left( \sqrt{2}-1 \right)}^{2}}} \\
 & \Rightarrow \dfrac{a}{b}=\dfrac{{{\left( \sqrt{2} \right)}^{2}}+{{1}^{2}}+2.1.\sqrt{2}}{{{\left( \sqrt{2} \right)}^{2}}+{{1}^{2}}-2.1.\sqrt{2}} \\
 & \Rightarrow \dfrac{a}{b}=\dfrac{2+1+2\sqrt{2}}{2+1-2\sqrt{2}} \\
 & \Rightarrow \dfrac{a}{b}=\dfrac{3+2\sqrt{2}}{3-2\sqrt{2}} \\
\end{align}\]
Hence, we have proved that the ratio of the two numbers is \[\dfrac{3+2\sqrt{2}}{3-2\sqrt{2}}\].

Note: There is a possibility that one may write $t$ as our answer in a hurry to solve the question. But since $t=\sqrt{\dfrac{a}{b}}$ and we are required to find the $\dfrac{a}{b}$ in the question, we have to perform squaring and then answer the question.

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