## Precalculus (6th Edition) Blitzer

The two numbers are $5+i\sqrt{15}$ and $5-i\sqrt{15}$.
Consider the two numbers $x$ and $y$ whose sum is $10$ and product is $40$. The sum of the two numbers is, $x+y=10$ The product of the numbers is, $xy=40$ Divide $x$ on both sides. \begin{align} & \frac{xy}{x}=\frac{40}{x} \\ & y=\frac{40}{x} \end{align} Substitute $y=\frac{40}{x}$ in the equation$x+y=10$. \begin{align} & x+\frac{40}{x}=10 \\ & \frac{{{x}^{2}}+40}{x}=10 \\ & {{x}^{2}}+40=10x \\ & {{x}^{2}}-10x+40=0 \end{align} Compare the equation ${{x}^{2}}-10x+40=0$ with $a{{x}^{2}}+bx+c$. \begin{align} & a=1 \\ & b=-10 \\ & c=40 \end{align} Substitute $a=1$, $b=-10$ and $c=40$ in the formula $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. \begin{align} & x=\frac{-\left( -10 \right)\pm \sqrt{{{\left( -10 \right)}^{2}}-4\left( 1 \right)\left( 40 \right)}}{2\left( 1 \right)} \\ & =\frac{10\pm \sqrt{100-160}}{2} \\ & =\frac{10\pm \sqrt{-60}}{2} \end{align} Use the property $\sqrt{-b}=i\sqrt{b}$. \begin{align} & x=\frac{10\pm i\sqrt{60}}{2} \\ & =\frac{10\pm i\sqrt{4\cdot 15}}{2} \\ & =\frac{10\pm 2i\sqrt{15}}{2} \\ & =5\pm i\sqrt{15} \end{align} So, the two numbers are $5+i\sqrt{15}$ and $5-i\sqrt{15}$. Check that the sum of the two complex numbers $5+i\sqrt{15}$ and $5-i\sqrt{15}$ is $10$. \begin{align} & \left( 5+i\sqrt{15} \right)+\left( 5-i\sqrt{15} \right)=\left( 5+5 \right)+\left( \sqrt{15}-\sqrt{15} \right)i \\ & =10+\left( 0 \right)i \\ & =10 \end{align} So, it shows that the complex numbers $5+i\sqrt{15}$ and $5-i\sqrt{15}$ satisfy the condition. Check that the product of two complex numbers $5+i\sqrt{15}$ and $5-i\sqrt{15}$ is $40$. Use the FOIL method. \begin{align} & \left( 5+i\sqrt{15} \right)\left( 5-i\sqrt{15} \right)=25-5i\sqrt{15}+5i\sqrt{15}-{{\left( \sqrt{15} \right)}^{2}}{{i}^{2}} \\ & =25-15{{i}^{2}} \end{align} Replace the value ${{i}^{2}}=-1$. \begin{align} & \left( 5+i\sqrt{15} \right)\left( 5-i\sqrt{15} \right)=25-15\left( -1 \right) \\ & =25+15 \\ & =40 \end{align} So, it shows that the complex numbers $5+i\sqrt{15}$ and $5-i\sqrt{15}$ satisfy the condition. Therefore, the two numbers are $5+i\sqrt{15}$ and $5-i\sqrt{15}$.