the arithmetic mean of two numbers exceeds the geometric mean by 3/2 and geometric mean excceeds the harmonic mean by 6/5. find the numbers.
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sir this question answer 2 and 12
answer 3 and 12
so, if you have the answer, what are your thoughts on how to get it?
To find the numbers, let's assign them variables. Let the first number be 'a' and the second number be 'b'.
We know that the arithmetic mean of two numbers exceeds the geometric mean by 3/2. The arithmetic mean is given by (a + b) / 2, and the geometric mean is given by √(ab). So, we can write the first equation as:
(a + b) / 2 = √(ab) + 3/2
Next, we know that the geometric mean exceeds the harmonic mean by 6/5. The harmonic mean is given by 2ab / (a + b), so we can write the second equation as:
√(ab) = 2ab / (a + b) + 6/5
Now, we have a system of two equations. To solve it, we need to eliminate one variable. Let's solve the first equation for 'a':
(a + b) / 2 = √(ab) + 3/2
Multiply both sides of the equation by 2:
a + b = 2√(ab) + 3
Now, let's square both sides to eliminate the square root:
(a + b)^2 = (2√(ab) + 3)^2
a^2 + 2ab + b^2 = 4ab + 12√(ab) + 9
We can rewrite this equation as:
a^2 + b^2 - 2ab - 4ab - 12√(ab) - 9 = 0
Similarly, let's solve the second equation for 'a':
√(ab) = 2ab / (a + b) + 6/5
Square both sides:
ab = (2ab / (a + b) + 6/5)^2
Simplify the equation:
ab = 4a^2b^2 / (a + b)^2 + 24ab / (a + b) + 36/25
Multiply every term by (a + b)^2 to eliminate the denominators:
ab(a + b)^2 = 4a^2b^2 + 24ab(a + b) + 36(a + b)^2 / 25
Expand the equations and simplify:
a^3 + 2a^2b + ab^2 + b^3 - 4a^2b - 4ab^2 - 12√(ab)(a + b) - 9(a + b)^2 = 0
Now, we have a system of two equations:
a^2 + b^2 - 2ab - 4ab - 12√(ab) - 9 = 0
a^3 + 2a^2b + ab^2 + b^3 - 4a^2b - 4ab^2 - 12√(ab)(a + b) - 9(a + b)^2 = 0
To solve these equations, we can use numerical methods or approximation methods. Unfortunately, there is no simple closed-form solution for this particular problem.