## Precalculus (6th Edition) Blitzer

The numbers are $8\ \text{and}\ 12$.
Let us assume $x$ and $y$ be the two numbers such that: \begin{align} & x+y=20\ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{I} \right) \\ & xy=96\ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{II} \right) \end{align} From equation (I): $y=20-x$ (III) Substitute $y=20-x$ in equation (II): \begin{align} & x\left( 20-x \right)=96 \\ & 20x-{{x}^{2}}=96 \\ & {{x}^{2}}-8x+96=0 \\ & \left( x-8 \right)\left( x-12 \right)=0 \end{align} Now, $x=8,12$ Substitute $x=8$ in equation (III): \begin{align} & y=20-2 \\ & y=12 \\ \end{align} Substitute $x$ in equation (III) Thus, the numbers are 8 and 12.