Which equation has the solutions $x=1 \pm \sqrt{5}$ ?
To find the quadratic equation with the roots \(x_1 = 1 + \sqrt{5}\) and \(x_2 = 1 - \sqrt{5}\), we can use the fundamental relationship between the roots of a quadratic equation and its coefficients. A quadratic equation with roots \(r_1\) and \(r_2\) can be written in the form \((x - r_1)(x - r_2) = 0\). This expression expands to \(x^2 - (r_1 + r_2)x + (r_1 r_2) = 0\), where the coefficient of \(x\) is the negative of the sum of the roots, and the constant term is the product of the roots.
First, we calculate the sum of the roots \(S = x_1 + x_2\):
\[S = (1 + \sqrt{5}) + (1 - \sqrt{5}) = 1 + 1 + \sqrt{5} - \sqrt{5} = 2.\]
Next, we calculate the product of the roots \(P = x_1 \cdot x_2\):
\[P = (1 + \sqrt{5})(1 - \sqrt{5}).\]
Applying the difference of squares formula, \((a+b)(a-b) = a^2 - b^2\), we get:
\[P = 1^2 - (\sqrt{5})^2 = 1 - 5 = -4.\]
Now that we have the sum \(S = 2\) and the product \(P = -4\), we substitute these values into the standard quadratic form \(x^2 - Sx + P = 0\):
\[x^2 - (2)x + (-4) = 0\]
\[x^2 - 2x - 4 = 0.\]
Comparing this result to the given options:
The equation \(x^2 - 2x - 4 = 0\) matches option a.
Quadratic equations are fundamental in physics and engineering, particularly when modeling projectile motion or structural stress. For example, if you are calculating the trajectory of a ball thrown into the air, the height over time can be represented by a quadratic equation. Knowing how to relate the roots (the times when the ball is at a certain height) back to the original equation helps in understanding the motion parameters. This skill of reconstructing equations from given solutions is a key building block for solving more complex systems in real-world applications.