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

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.

Last updated date: 21st Sep 2023
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