Find 2 numbers whose arithmetic mean exceeds their geometric mean by 2and whose harmonic mean is 1/5 of the larger number
To solve this problem, let's assume the two numbers we need to find are "a" and "b", where "a" is the larger number.
Given that the arithmetic mean exceeds the geometric mean by 2, we can write the following equation:
(a + b)/2 - √(a * b) = 2
Similarly, the harmonic mean being 1/5 of the larger number can be expressed as:
2/(1/a + 1/b) = a/5
Now, let's solve these equations step by step to find the values of "a" and "b".
1. Solve the first equation:
(a + b)/2 - √(a * b) = 2
Rearranging, we get:
a + b - 2√(a * b) = 4
2√(a * b) = a + b - 4
Squaring both sides:
4(a * b) = (a + b - 4)^2
4ab = a^2 + 2ab + b^2 - 8a - 8b + 16
Simplifying:
a^2 + b^2 - 2ab - 8a - 8b + 16 = 0
a^2 + b^2 - 2ab - 8(a + b - 2) + 16 = 0
a^2 + b^2 - 2ab - 8a - 8b + 32 = 0
2. Solve the second equation:
2/(1/a + 1/b) = a/5
Multiply both sides by a * b:
2 * ab/(a + b) = a^2/5
Cross-multiply:
10ab = a^2(a + b)
10ab = a^3 + a^2b
3. Combine the equations:
We have two equations:
a^2 + b^2 - 2ab - 8a - 8b + 32 = 0 (Equation 1)
10ab = a^3 + a^2b (Equation 2)
We'll solve these equations using a numerical method such as substitution or graphical analysis.
Let's use the substitution method:
From Equation 2, we get:
10ab = a^3 + a^2b
a^3 + (a^2 - 10ab) = 0
a^3 + a^2(1 - 10b) = 0
Simplifying:
a^3 - 10ab + a^2 = 0
a^2(a - 10b + 1) = 0
Now, two cases can occur:
Case 1: a^2 = 0
If a^2 = 0, it means a = 0. But in our case, a can't be 0, so we discard this possibility.
Case 2: (a - 10b + 1) = 0
If a - 10b + 1 = 0, we get a = 10b - 1.
Substituting this value of "a" in Equation 1, we can solve for "b".
(a^2 + b^2 - 2ab - 8a - 8b + 32 = 0) (using a = 10b - 1)
(100b^2 - 20b + 1 + b^2 - 2(10b - 1)b - 8(10b - 1) - 8b + 32 = 0)
(101b^2 - 92b + 9 = 0)
Solve for "b" using this quadratic equation.
Once you find the value of "b", substitute it back into the equation a = 10b -1 to find "a".
And that's how you can find the two numbers "a" and "b" using a step-by-step approach.