Question

# If the AM of two positive numbers be three times their geometric mean then the ratio of the numbers isThis question has multiple correct optionsA.$3\pm 2\sqrt{2}$B.$\sqrt{2}\pm 1$C$17+12\sqrt{2}$D.${{(3-2\sqrt{2})}^{-2}}$

Hint: We can assume the numbers to be x and y and proceed to form equations so that we can find the desired result. For that we need to remember the formula of AM and GM i.e. arithmetic mean and geometric mean.

The formula for finding arithmetic mean of two numbers ‘a’ and ‘b’ is $\dfrac{a+b}{2}$
And the formula for calculating geometric mean of two numbers ‘a’ and ‘b’ is $\sqrt{a\cdot b}$
Let us assume that the two positive numbers are ‘x’ and ‘y’.
Their arithmetic mean is given by- $AM=\dfrac{x+y}{2}$ …(i)
Their geometric mean is given by- $GM=\sqrt{x\cdot y}$ …(ii)
From the question we are given that, AM of x and y is equal to 3 times GM of x and y which implies
$AM=3\cdot GM$
Substituting the value of AM and GM from equation (i) and (ii) we have,
\begin{align} & \dfrac{x+y}{2}=3\sqrt{xy} \\ & \Rightarrow x+y=6\sqrt{xy} \\ \end{align}
Dividing both sides by $\sqrt{xy}$ we have,
\begin{align} & \dfrac{x+y}{\sqrt{xy}}=6 \\ & \Rightarrow \dfrac{x}{\sqrt{xy}}+\dfrac{y}{\sqrt{xy}}=6 \\ \end{align}
On simplifying we get,
$\sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{y}{x}}=6$ …(iii)
Now our question was asking us to find the ratio of x and y. Therefore we need to find $x:y\Rightarrow \dfrac{x}{y}$ .
Now let us assume $\dfrac{x}{y}=t$ .
Substituting t in place of $\dfrac{x}{y}$ in equation (iii) we have,
$\sqrt{t}+\sqrt{\dfrac{1}{t}}=6$
On simplifying we get,
\begin{align} & \dfrac{\sqrt{t}\cdot \sqrt{t}+\sqrt{1}}{\sqrt{t}}=6 \\ & \Rightarrow \dfrac{t+1}{\sqrt{t}}=6 \\ \end{align}
Multiplying $\sqrt{t}$ both sides we get,
\begin{align} & t+1=6\sqrt{t} \\ & \Rightarrow t-6\sqrt{t}+1=0 \\ \end{align}
Now we need to solve this equation to get ‘t’.
If we assume $t={{p}^{2}}$ and substitute in the above equation we will have a quadratic equation which we can solve easily.
After substituting ‘t’ we get,
${{p}^{2}}-6p+1=0$
We use the formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to find roots of the above quadratic equation where a is coefficient of ${{x}^{2}}$ , b is coefficient of x and c is the constant term.
We have, a=1, b=-6 and c=1
Substituting these values in the above equation we get,
\begin{align} & p=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\ & \Rightarrow p=\dfrac{-(-6)\pm \sqrt{{{(-6)}^{2}}-4\cdot 1\cdot 1}}{2\cdot 1} \\ \end{align}
On simplifying we get,
\begin{align} & p=\dfrac{6\pm \sqrt{36-4}}{2} \\ & \Rightarrow p=\dfrac{6\pm \sqrt{32}}{2} \\ & \Rightarrow p=\dfrac{6\pm 4\sqrt{2}}{2} \\ & \Rightarrow p=3\pm 2\sqrt{2} \\ \end{align}
We had, $t={{p}^{2}}$
Therefore, $t={{(3\pm 2\sqrt{2})}^{2}}$
On squaring we get,
\begin{align} & t={{3}^{2}}+{{(2\sqrt{2})}^{2}}\pm 2\cdot 3\cdot 2\sqrt{12} \\ & \Rightarrow t=9+8\pm 12\sqrt{12} \\ & \Rightarrow t=17\pm 12\sqrt{12} \\ \end{align}
Therefore, $t=\dfrac{x}{y}=17\pm 12\sqrt{12}$
Hence, the answer is option (c).

Note: To find the ratio we may think that we separately need to find x and y but for that we would need two equations to find two variables. We only had one equation so we converted it into one variable and we were able to directly find the ratio without actually knowing the value of x and y.