Question

The geometric mean of two numbers is \[6\] and their arithmetic mean is \[6.5\]. The numbers are
A. \[(3,12)\]
B. \[(4,9)\]
C. \[(2,18)\]
D. \[(7,6)\]

Answer 1

Hint: Use the formula related to G.M and A.M to get an equation related to the terms.

Formula Used: GM = \[\sqrt {xy}\]
and AM = \[\dfrac{{x + y}}{2}\]
where $x$ and $y$ are number

Complete step by step solution: Let us assume that x and y are the two required numbers.
The geometric mean of two numbers is equal to six.
\[\sqrt {xy} = 6\]
\[ = > xy = 36\] --(1)
Additionally, it is assumed that the identical two values' arithmetic mean is equal to \[6.5\].
\[\dfrac{{x + y}}{2} = 6.5\]
\[ = > x + y = 13\]
\[ = > x = 13 - y\] --(2)
Find the required numbers from the equations (1) and (2)
\[(13 - y)y = 36\]
\[ = > 13y - {y^2} = 36\]
\[ = > {y^2} - 13y + 36 = 0\]
The roots for the quadratic equation are,
\[y = \dfrac{{13 \pm \sqrt {{{13}^3} - 4 \times 36} }}{2}\]
\[y = \dfrac{{13 \pm \sqrt {25} }}{2}\]
\[y = \dfrac{{13 \pm 5}}{2}\]
\[ = > y = \{ 4,9\} \]
Substitute the value of y as four,
\[x = 13 - 4\]
\[ = > x = 9\]
Substitute the value of y as nine,
\[x = 13 - 9\]
\[ = > x = 4\]
So, the two numbers that are required are \[4\] and \[9\].

Therefore, the correct option is B.

Note: As we have the quadratic equation we should take care of the sign.