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  • The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio (3+ 2 square root 2): (3−2 square root 2).

28.   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 ) 

Answers (1)

best_answer

Let there be two numbers a and b

geometric mean =\sqrt{ab}

According to the given condition,

a+b=6\sqrt{ab}

(a+b)^2=36(ab).............................................................(1)

Also,(a-b)^2=(a+b)^2-4ab=36ab-4ab=32ab

(a-b)=\sqrt{32}\sqrt{ab}

 (a-b)=4\sqrt{2}\sqrt{ab}.......................................................(2)

From (1) and (2), we get

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

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

Putting the value of 'a' in (1),

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

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

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

\frac{a}{b}=\frac{(3+2\sqrt{2})}{(3-2\sqrt{2})}

Thus, the ratio is ( 3+ 2 \sqrt 2 ) : ( 3 - 2 \sqrt 2 )

Posted by

seema garhwal

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