The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio.

$$(3+2\sqrt{2}): (3-2\sqrt{2})$$

Asked by Abhisek | 1 year ago |  76

##### Solution :-

Consider the two numbers be a and b.

Then, G.M. = $$\sqrt{ab}$$.

According to the conditions given in the question,

$$a+b=6\sqrt{ab}$$ .........(1)

$$(a+b)^2=36$$

$$(a-b)^2= (a+b)^2-4ab$$

$$=36ab-4ab=32ab$$

$$a-b=\sqrt{32}\sqrt{ab}$$

$$=4\sqrt{2}\sqrt{ab}$$ ..........(2)

$$2a= (6+4\sqrt{2})\sqrt{ab}$$

$$a= (3+2\sqrt{2})\sqrt{ab}$$

Substitute $$a= (3+2\sqrt{2})\sqrt{ab}$$ in equation (1) .

$$b= 6\sqrt{ab}-(3+2\sqrt{2})\sqrt{ab}$$

b=$$(3-2\sqrt{2})\sqrt{ab}$$

Divide a by b .

$$\dfrac{a}{b}=\dfrac{(3+2\sqrt{2})\sqrt{ab}}{(3-2\sqrt{2})\sqrt{ab}}$$

$$\dfrac{(3+2\sqrt{2})}{(3-2\sqrt{2})}$$

Therefore, it is proved that the numbers are in the ratio.

Answered by Abhisek | 1 year ago

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