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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.

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True

False

answer is A.

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Detailed Solution

The above statement is true.
First, let two numbers be x and y
Geometric Mean =

\sqrt{xy}

As per the question,
x + y = 6

\sqrt{xy}

…………(i)

\Rightarrow

{(x + y)}^{2}

= 36 (xy)
By subtracting 4xy from both the side we get,

\Rightarrow

{(x + y)}^{2}

- 4xy = 36xy - 4xy

\Rightarrow

x - y =

\sqrt{32}

\sqrt{xy}

\Rightarrow

x - y = 4

\sqrt{2}

\sqrt{xy}

……..(ii)
By adding both equation (i & ii) we get,
x + y + x - y = 6

\sqrt{xy}

+ 4

\sqrt{2}

\sqrt{xy}

\Rightarrow

2x = (6 + 4

\sqrt{2}

)

\sqrt{xy}

\Rightarrow

x = (3 + 2

\sqrt{2}

)

\sqrt{xy}

Now lets substitute the value of x in our equation (i),
x + y = 6

\sqrt{xy}

\Rightarrow

(3 + 2

\sqrt{2}

)

\sqrt{xy}

+ y = 6

\sqrt{xy}

\Rightarrow

y = 6

\sqrt{xy}

- (3 + 2

\sqrt{2}

)

\sqrt{xy}

\Rightarrow

y = (3 - 2

\sqrt{2}

)

\sqrt{xy}

Then,

\frac{x}{y}

\frac{(3 + 2 \sqrt{2}) \sqrt{xy}}{(3 - 2 \sqrt{2}) \sqrt{xy}}

\frac{(3 + 2 \sqrt{2})}{(3 - 2 \sqrt{2})}

∴ The required Ratio = (3+2

\sqrt{2}

):(3-2

\sqrt{2}

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