Which equation has the solutions $x=\frac{5+2 \sqrt{7}}{3}$?
We are given the solutions \(x = \frac{5 \pm 2\sqrt{7}}{3}\) for a quadratic equation. Our task is to identify which of the provided options matches the quadratic equation that has these roots. To find the equation, we can reverse the process of solving a quadratic equation using the quadratic formula. Given the solutions \(x_1 = \frac{5 + 2\sqrt{7}}{3}\) and \(x_2 = \frac{5 - 2\sqrt{7}}{3}\), we want to derive the original quadratic equation.
We start with the equation \(x = \frac{5 \pm 2\sqrt{7}}{3}\). To isolate the radical term, we first multiply both sides by 3 to get \(3x = 5 \pm 2\sqrt{7}\). Next, we subtract 5 from both sides, yielding \(3x - 5 = \pm 2\sqrt{7}\). Now, we square both sides of the equation to eliminate the radical:
\[(3x - 5)^2 = (\pm 2\sqrt{7})^2\]
Expanding the left side gives \(9x^2 - 30x + 25\), and calculating the square on the right side gives \(4 \times 7 = 28\). Setting them equal, we have:
\[9x^2 - 30x + 25 = 28\]
Subtracting 28 from both sides, we obtain:
\[9x^2 - 30x - 3 = 0\]
Finally, to simplify the equation, we divide every term by their common factor, which is 3:
\[3x^2 - 10x - 1 = 0\]
This corresponds to option (d).
Understanding how to reconstruct a quadratic equation from its roots is useful in fields like engineering and physics. For example, if you are designing a trajectory for a projectile and you know the points in time when the object hits a certain height, those points are the roots of a quadratic equation. By knowing these times, you can determine the specific equation of motion for the projectile, which helps in calculating its maximum height, velocity, and landing position. This skill allows for reverse-engineering systems to understand the governing dynamics of physical phenomena.