If the [arithmetic mean](https://artofproblemsolving.com/wiki/index.php/Arithmetic_mean) of two [numbers](https://artofproblemsolving.com/wiki/index.php/Number) is $6$ and their [geometric mean](https://artofproblemsolving.com/wiki/index.php/Geometric_mean) is $10$, then an [equation](https://artofproblemsolving.com/wiki/index.php/Equation) with the given two numbers as [roots](https://artofproblemsolving.com/wiki/index.php/Root) is:
$\text{(A)} \ x^2 + 12x + 100 = 0 ~~ \text{(B)} \ x^2 + 6x + 100 = 0 ~~ \text{(C)} \ x^2 - 12x - 10 = 0$
$\text{(D)} \ x^2 - 12x + 100 = 0 \qquad \text{(E)} \ x^2 - 6x + 100 = 0$
Let the numbers be $\eta$ and $\zeta$.
\[\dfrac{\eta+\zeta}{2}=6\Rightarrow \eta+\zeta=12\].
\[\sqrt{\eta\zeta}=10\Rightarrow \eta\zeta=100\].
The [monic](https://artofproblemsolving.com/wiki/index.php/Monic_polynomial) [quadratic](https://artofproblemsolving.com/wiki/index.php/Quadratic) with roots $\eta$ and $\zeta$ is $x^2-(\eta+\zeta)x+\eta\zeta$. Therefore, an equation with $\eta$ and $\zeta$ as roots is $x^2 - 12x + 100 = 0\Rightarrow \text{(D)}$