The correct answer is:
There are no real numbers that fit these criteria.
Explanation:
Let x and y represent the two numbers we are looking for.
x*y = 100
x+y = 10
Solving for y,
x+y-x = 10-x
y = 10-x
Substituting this into our first equation, we have:
x(10-x) = 100
Using the distributive property,
x*10 - x*x = 100
10x - x² = 100
Writing this in standard form, we need to subtract 100 from each side:
10x - x² - 100 = 100 - 100
10x - x² - 100 = 0
In standard form, this is
-x²+10x-100 = 0.
This is in the form ax²+bx+c; in our equation, a = -1, b = 10 and c = -100.
We will use the quadratic formula for this:
[tex]x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}
\\
\\=\frac{-10\pm \sqrt{10^2-4(-1)(-100)}}{2(-1)}
\\
\\=\frac{-10\pm \sqrt{100-400}}{-2}
\\
\\=\frac{-10\pm \sqrt{-300}}{-2}[/tex]
There is no real square root of -300, so there are no real number solutions to this.
