Solve the equation: $2 x(x-5)+3(x+5)=0$
We are tasked with solving the quadratic equation \(2x(x - 5) + 3(x + 5) = 0\). To begin, we need to simplify the expression by expanding the terms. Distributing the terms, we get \(2x \cdot x - 2x \cdot 5 + 3 \cdot x + 3 \cdot 5 = 0\), which simplifies to \(2x^2 - 10x + 3x + 15 = 0\). Combining the like terms \(-10x\) and \(3x\), the equation becomes \(2x^2 - 7x + 15 = 0\).
The equation is now in the standard quadratic form \(ax^2 + bx + c = 0\), where \(a = 2\), \(b = -7\), and \(c = 15\). To find the values of \(x\), we can use the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Substituting our coefficients into the formula, we have: \[x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(15)}}{2(2)}\] Calculating the terms inside the formula gives us \(x = \frac{7 \pm \sqrt{49 - 120}}{4}\). The discriminant is \(49 - 120 = -71\). Since the discriminant is negative, the roots of this equation are complex numbers. Specifically, \(x = \frac{7 \pm i\sqrt{71}}{4}\).
Quadratic equations are fundamental in modeling real-world phenomena. For instance, if you were launching a small model rocket, the path of the rocket could be modeled by a quadratic equation relating time to height. If the trajectory resulted in a negative discriminant in a specific model, it would physically mean the rocket never reaches a certain height or ground level, depending on the constraints. Mastering these equations allows engineers and scientists to predict outcomes in physics, economics, and various fields of engineering.