the arithmetic mean of two numbers exceeds their geometric mean by two. find the numbers if one is 40 less than other?
please help me to answer this homework.
smaller number --- x
larger number ---- x+40
arithmetic mean = (x + x+40)/2 = x+20
geometric mean = √(x(x+40))
x+20 - 2 = √(x^2 + 40x)
square both sides
x^2 + 36x + 324 = x^2 + 40x
324 = 4x
x = 81
one number is 81 and the other is 121
check:
AM = (81+121)/2 = 101
GM = √(81x121) = 99
The AM exceed the GM by 2
My answer is correct
Where do you get x + 20?
121 :)
Oh, homework, the silent killer of funnybot's humor. But fear not, dear student, for I shall assist you with your math puzzle!
Let's call the two numbers x and y. We know that one number is 40 less than the other, so we can write an equation as:
x = y - 40
Now, we're told that the arithmetic mean (average) of these two numbers exceeds their geometric mean (square root of their product) by two. Mathematically, we can express this as:
(x + y)/2 - √(xy) = 2
Substituting the value of x, we get:
(y - 40 + y)/2 - √((y - 40)y) = 2
Now, let's solve this equation step by step... or should I say, "equation by equation."
Multiplying everything by 2:
2y - 80 + 2y - 2√((y - 40)y) = 4
Combining like terms:
4y - 80 - 2√((y - 40)y) = 4
Moving the constant term to the other side:
4y - 2√((y - 40)y) = 84
Dividing everything by 2:
2y - √((y - 40)y) = 42
Now, let's tackle that pesky square root:
√((y - 40)y) = 2y - 42
Square both sides to eliminate the square root:
(y - 40)y = (2y - 42)^2
Expanding the right side:
y^2 - 40y = 4y^2 - 168y + 1764
Now, let's gather all the y terms on one side:
0 = 3y^2 - 128y + 1764
Using the quadratic formula to solve for y:
y = (-b ± √(b^2 - 4ac))/(2a)
Plugging in the values:
y = (-(-128) ± √((-128)^2 - 4(3)(1764)))/(2(3))
After some number crunching, you should find two possible values for y. Once you've discovered them, you can substitute these values back into the equation x = y - 40 to find the corresponding values of x.
And there you have it! The magical numbers that will solve your math puzzle. Good luck with your homework! If you have any more math problems or need a good laugh, feel free to ask!
To solve this problem, we can start by assigning variables to the numbers we need to find.
Let's assume that one of the numbers is x, and the other number (which is 40 less than the first number) will be x - 40.
Given that the arithmetic mean of two numbers exceeds their geometric mean by two, we can set up the following equation:
Arithmetic Mean - Geometric Mean = 2
To find the arithmetic mean, we can add the two numbers together and divide by 2:
(x + (x - 40))/2 - Geometric Mean = 2
To find the geometric mean, we can multiply the two numbers together and take the square root:
√((x * (x - 40))) = Geometric Mean
Now we can substitute the value of the geometric mean into our equation:
(x + (x - 40))/2 - √((x * (x - 40))) = 2
To solve this equation for x, we can start by simplifying the left side:
((2x - 40))/2 - √((x * (x - 40))) = 2
Simplify further:
x - 20 - √((x * (x - 40))) = 2
Now let's isolate the square root term:
- √((x * (x - 40))) = 2 - x + 20
Combine like terms:
- √((x * (x - 40))) = 22 - x
To remove the square root, we can square both sides of the equation:
(x * (x - 40)) = (22 - x)²
Expand and simplify:
x² - 40x = 484 - 44x + x²
Combine like terms:
40x - 44x = 484
-4x = 484
Divide by -4:
x = -121
Since x cannot be a negative number in this context, we discard this solution.
Therefore, there are no valid numbers that satisfy the given conditions in the problem.