The sum of two numbers is 6 times their geometric means, show that the numbers are in the ratio ( 3+2root2 : 3 - 2root2).

Let the numbers be a and b respectively.Then G.M=sqrt(ab) 

a+b=6sqrt(ab)

Square on both sides and make a quadratic in (a/b)

Arrange the terms and divide throughout by b^2.

Solve this eqn. and rationalize the numerator and then u will get the answer.

 Pooja your answer is wrong.

Let number be a and ar^2 

So a+ ar^2 = 6ar

So r =3+/-2sqrt(2) 

But you are not asking for r. You are asking for ratio of numbers = r^2 = (3+/-2sqrt(2))^2

OK ....dont rationalize the numerator of the quadratic..

RATIONALIZE THE DENOMINATOR OF YOUR GIVEN ANSWER ....YOU WILL GET IT ....AND BTW I DONT POST WRONG ANSWERS....

 Let A and G be the arithmetic and geometric mean for a, b respectively.

Thn given that, a+b = 6G

                         (a+b)/2 = 3G

                          A = 3G 

 Let  take a quadratic equation in x whose roots are a and b. Then, x^2 - (a+b)x + ab = 0

                                                                                                                        x^2 - 2Ax + G^2 = 0

Using quadratic formula, we get   x = [2A +- root(4A^2 - 4G^2)] / 2

                                                             x = A +- sqrt(A^2 - G^2)

Since a and b are the roots of the then, a = A + sqrt(A^2 - G^2)  & b = A - sqrt(A^2 -G^2)

Then,

Now, putting A = 3G in the above equation we get,

Hence proved.....