the arithmetic mean of two numbers is 15, and their geometric mean is 9.
find the numbers
(n + m)/2 = 15
so
n + m = 30
m = (30-n)
(n m )^.5= 9
n m = 81
n(30-n) = 81
-n^2 + 30 n = 81
n^2 -30 + 81 = 0
(n-3)(n-27) = 0
3 and 27
To find the two numbers, let's assume the first number is "x" and the second number is "y".
1. Given that the arithmetic mean of the two numbers is 15, we can write the equation:
(x + y) / 2 = 15
x + y = 30 (multiplying both sides by 2)
2. Now, let's consider the geometric mean, which is the square root of the product of the two numbers:
√(x * y) = 9
x * y = 81 (squaring both sides)
3. We have two equations: x + y = 30 and x * y = 81.
To solve these equations, we can use substitution or elimination:
Method 1: Substitution
- Rearrange the first equation to express one variable in terms of the other (for example, x = 30 - y).
- Substitute this expression for the variable in the second equation, and then solve for the remaining variable.
- Once you find the value for one variable, substitute it back into the first equation to find the other variable.
Method 2: Elimination
- Multiply the first equation by a value to get the sum of the coefficients equal to the sum of the coefficients in the second equation.
- Subtract one equation from the other to eliminate one variable and solve for the remaining variable.
Let's use the substitution method to solve these equations:
From equation 1, we can express x in terms of y:
x = 30 - y
Replace x in equation 2:
(30 - y) * y = 81
Expanding the equation: 30y - y^2 = 81
Rearrange the equation to isolate y^2:
y^2 - 30y + 81 = 0
Now, we can solve this quadratic equation. Factoring may not work in this case, so we can use the quadratic formula:
y = (-(-30) ± √((-30)^2 - 4(1)(81))) / (2(1))
Simplifying:
y = (30 ± √(900 - 324)) / 2
y = (30 ± √576) / 2
y = (30 ± 24) / 2
We have two possible values for y:
1. When y = (30 + 24) / 2 = 54 / 2 = 27
2. When y = (30 - 24) / 2 = 6 / 2 = 3
Now, substitute each value of y back into the equation x = 30 - y:
For y = 27:
x = 30 - 27 = 3
For y = 3:
x = 30 - 3 = 27
Therefore, the two numbers are 3 and 27.
To find the two numbers, let's call them "x" and "y." We are given two pieces of information about these numbers:
1. The arithmetic mean of the two numbers is 15.
The arithmetic mean is obtained by adding the two numbers and dividing the sum by 2. So, we have the equation: (x + y) / 2 = 15.
2. The geometric mean of the two numbers is 9.
The geometric mean is obtained by taking the square root of the product of the two numbers. So, we have the equation: √(x * y) = 9.
Now, we have a system of two equations with two variables. We can solve this system to find the values of "x" and "y."
Let's solve the first equation for one variable and substitute it into the second equation:
(x + y) / 2 = 15
x + y = 30
y = 30 - x
Now, substitute the value of "y" in the second equation:
√(x * (30 - x)) = 9
Squaring both sides of the equation, we get:
x * (30 - x) = 81
Expanding and rearranging the equation, we get:
30x - x^2 = 81
x^2 - 30x + 81 = 0
Now, we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. In this case, the equation factors nicely:
(x - 9)(x - 9) = 0
This yields a double root of x = 9.
Since y = 30 - x, we can substitute x = 9 into this equation to find y:
y = 30 - 9
y = 21
Therefore, the two numbers are 9 and 21.