The sum of two numbers is 6 times their geometric mean, then the numbers are in the ratio (3+22):(3-22). State true or false.

# The sum of two numbers is 6 times their geometric mean, then the numbers are in the ratio (3+2$\sqrt{2}$):(3-2$\sqrt{2}$). State true or false.

1. A
True
2. B
False

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### Solution:

The above statement is true.
First, let  two numbers be x and y
Geometric Mean = $\sqrt{\mathit{xy}}$
As per the question,
x + y = 6$\sqrt{\mathit{xy}}$…………(i)
$⇒$ = 36 (xy)
By subtracting 4xy from both the side we get,
$⇒$ - 4xy = 36xy - 4xy
$⇒$ x - y = $\sqrt{32}$ $\sqrt{\mathit{xy}}$
$⇒$ x - y = 4$\sqrt{2}$ $\sqrt{\mathit{xy}}$……..(ii)
By adding both equation (i & ii)  we get,
x + y + x - y  = 6$\sqrt{\mathit{xy}}$ + 4$\sqrt{2}$ $\sqrt{\mathit{xy}}$
$⇒$ 2x = (6 + 4$\sqrt{2}$) $\sqrt{\mathit{xy}}$
$⇒$ x = (3 + 2$\sqrt{2}$) $\sqrt{\mathit{xy}}$
Now lets substitute the value of x in our equation (i),
x + y = 6$\sqrt{\mathit{xy}}$
$⇒$ (3 + 2$\sqrt{2}$) $\sqrt{\mathit{xy}}$ + y =  6$\sqrt{\mathit{xy}}$
$⇒$ y = 6$\sqrt{\mathit{xy}}$ - (3 + 2$\sqrt{2}$) $\sqrt{\mathit{xy}}$
$⇒$ y = (3 - 2$\sqrt{2}$) $\sqrt{\mathit{xy}}$
Then, $\frac{x}{y}$ = =
∴ The required Ratio = (3+2$\sqrt{2}$):(3-2$\sqrt{2}$).

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