Input the geometric mean and harmonic mean of two numbers to instantly find their values.
Let the two numbers be \(x\) and \(y\). You are given their geometric mean (\(G\)) and harmonic mean (\(H\)).
Geometric Mean (G): The square root of the product of the two numbers: \( G = \sqrt{x \cdot y} \)
Harmonic Mean (H): The reciprocal of the average of the reciprocals of the two numbers: \( H = \frac{2xy}{x + y} \)
Using the geometric mean, calculate the product of the two numbers: \( x \cdot y = G^2 \) Substitute the product into the harmonic mean formula to find the sum of the two numbers: \( x + y = \frac{2xy}{H} \) \( x + y = \frac{2G^2}{H} \) Using the sum and product, construct a quadratic equation: \( t^2 - \frac{2G^2}{H}t + G^2 = 0 \) Apply the quadratic formula to find \(t\): \( t = \frac{ \frac{2G^2}{H} \pm \sqrt{( \frac{2G^2}{H})^2 - 4 \cdot G^2}}{2} \) The two solutions correspond to \(x\) and \(y\).
Solution:
1. Calculate the Product \( x \cdot y \):
\( x \cdot y = 12^2 = 144 \)
2. Calculate the Sum \( x + y \):
\( x + y = \frac{2 \cdot 12^2}{7.2} = \frac{2 \cdot 144}{7.2} = 40 \)
3. Form the Quadratic Equation:
\( t^2 - 40t + 144 = 0 \)
4. Solve the Equation:
\( t = \frac{40 \pm \sqrt{40^2 - 4 \cdot 144}}{2} = \frac{40 \pm \sqrt{1600 - 576}}{2} = \frac{40 \pm \sqrt{1024}}{2} \)
\( t_1 = \frac{40 + \sqrt{1024}}{2} = 36 \)
\( t_2 = \frac{40 - \sqrt{1024}}{2} = 4 \)
Result: The two numbers are 36 and 4.
Solution:
1. Calculate the Product \( x \cdot y \):
\( x \cdot y = 6^2 = 36 \)
2. Calculate the Sum \( x + y \):
\( x + y = \frac{2 \cdot 6^2}{\frac{72}{13}} = \frac{2 \cdot 36}{\frac{72}{13}} = \frac{72 \cdot 13}{72} = 13 \)
3. Form the Quadratic Equation:
\( t^2 - 13t + 36 = 0 \)
4. Solve the Equation:
\( t = \frac{13 \pm \sqrt{13^2 - 4 \cdot 36}}{2} = \frac{13 \pm \sqrt{169 - 144}}{2} = \frac{13 \pm \sqrt{25}}{2} \)
\( t_1 = \frac{13 + \sqrt{25}}{2} = 9 \)
\( t_2 = \frac{13 - \sqrt{25}}{2} = 4 \)
Result: The two numbers are 9 and 4.