Given that:
- The Product of two numbers is 40.
- The Sum of those numbers is 10.
You need to remember that a Product is the result of a Multiplication, and a Sum is the result of an Addition.
Let be "x" and "y" the two numbers.
Using the data given in the exercise, you can set up this System of Equations:
[tex]\begin{cases}xy=40 \\ x+y=10\end{cases}[/tex]In order to solve the System of Equations, you can use the Substitution Method:
1. Solve for "y" from the second equation:
[tex]\begin{gathered} x+y=10 \\ y=10-x \end{gathered}[/tex]2. Substitute the new equation into the first original equation:
[tex]\begin{gathered} xy=40 \\ x(10-x)=40 \end{gathered}[/tex]Solve for "x":
- Apply the Distributive Property on the left side of the equation:
[tex]\begin{gathered} (x)(10)-(x)(x)=40 \\ 10x-x^2=40 \end{gathered}[/tex]- Notice that you get a Quadratic Equation. Then, you need to rewrite it in this form:
[tex]ax^2+bx+c=0[/tex]Then:
[tex]-x^2+10x-40=0[/tex]- Use the Quadratic Formula:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]Knowing that:
[tex]\begin{gathered} a=-1 \\ b=10 \\ c=-40 \end{gathered}[/tex]You can substitute values and simplify:
[tex]x=\frac{-10\pm\sqrt[]{(-10)^2-4(-1)(-40)}}{2(-1)}[/tex][tex]x=\frac{-10\pm\sqrt[]{-60}}{-2}[/tex]Notice that the number inside the square root is negative. That means that you will get two Complex Solutions.
By definition:
[tex]\sqrt[]{-1}=i[/tex]Therefore, you need to simplify the square root of 60 and multiply it by "i":
[tex]x=\frac{-10\pm i2\sqrt[]{15}}{-2}[/tex]Simplifying you get:
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