If the arithmetic mean of two numbers is 15 and their geometric mean is 9 find the two numbers
(a+b)/2 = 15
√(ab) = 9
so,
ab = 81
The only factors of 81 are (1,81),(3,27),(9,9)
Clearly the only possible candidates are 3 and 27, which you can easily verify.
If you want to do the algebra, you have
a+b = 30
a^2+2ab+b^2 = 900
a^2 + 2*81 + (81/a)^2 = 900
a^2 + 6561/a^2 = 738
a^4-738a^2+6561 = 0
(a^2-9)(a^2-729) = 0
a = 3 or 27
To find the two numbers, let's assume the numbers are x and y.
We are given two pieces of information:
1. The arithmetic mean of the two numbers is 15.
The arithmetic mean is the sum of the two numbers divided by 2.
So, we can write the equation: (x + y) / 2 = 15.
2. The geometric mean of the two numbers is 9.
The geometric mean is the square root of the product of the two numbers.
So, we can write the equation: √(xy) = 9.
Let's solve these equations simultaneously to find the values of x and y.
First, let's manipulate the arithmetic mean equation to get rid of the fraction:
x + y = 30.
Now, let's square both sides of the geometric mean equation:
(√(xy))^2 = 9^2,
xy = 81.
Now, we have a system of equations:
x + y = 30,
xy = 81.
We can use substitution or elimination method to solve these equations.
Let's use substitution method:
From the first equation, we can write y = 30 - x.
Substituting this value of y in the second equation:
x(30 - x) = 81.
Expanding and rearranging the equation:
30x - x^2 = 81,
x^2 - 30x + 81 = 0.
Now, we can solve this quadratic equation to find the values of x.
Using factoring or quadratic formula, we get:
(x - 9)(x - 9) = 0.
So, x = 9.
Now, substituting the value of x in the first equation:
9 + y = 30,
y = 30 -9,
y = 21.
Therefore, the two numbers are 9 and 21.