x1+x2=20 x1*x2=1

You have a system of equations:

1. \( x_1 + x_2 = 20 \)
2. \( x_1 \cdot x_2 = 1 \)

You can solve this system by substituting one equation into the other. Let's solve for \( x_2 \) from the first equation:

\[ x_2 = 20 - x_1 \]

Now, substitute this expression for \( x_2 \) into the second equation:

\[ x_1 \cdot (20 - x_1) = 1 \]

Expanding this gives:

\[ 20x_1 - x_1^2 = 1 \]

Rearranging the equation, we get:

\[ -x_1^2 + 20x_1 - 1 = 0 \]

Multiplying through by -1 to make the leading coefficient positive:

\[ x_1^2 - 20x_1 + 1 = 0 \]

Now you can solve this quadratic equation using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 1 \), \( b = -20 \), and \( c = 1 \):

\[ x_1 = \frac{20 \pm \sqrt{(-20)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]

Calculating the discriminant:

\[ \sqrt{400 - 4} = \sqrt{396} = 2\sqrt{99} \]

Now, substituting back into the formula gives:

\[ x_1 = \frac{20 \pm 2\sqrt{99}}{2} = 10 \pm \sqrt{99} \]

So, we have:

\[ x_1 = 10 + \sqrt{99} \quad \text{or} \quad x_1 = 10 - \sqrt{99} \]

Now, substituting these back to find \( x_2 \):

1. If \( x_1 = 10 + \sqrt{99} \):

\[ x_2 = 20 - (10 + \sqrt{99}) = 10 - \sqrt{99} \]

2. If \( x_1 = 10 - \sqrt{99} \):

\[ x_2 = 20 - (10 - \sqrt{99}) = 10 + \sqrt{99} \]

Thus, the solutions are:

\[ (x_1, x_2) = (10 + \sqrt{99}, 10 - \sqrt{99}) \quad \text{or} \quad (10 - \sqrt{99}, 10 + \sqrt{99}) \]