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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 $\left( 3+2\sqrt{2} \right):\left( 3-2\sqrt{2} \right)$.  Answer Verified
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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