- Rearrange the equation into standard form: \(2x^2 - 4x - 6 = 0\).
- Simplify by dividing by 2 to get \(x^2 - 2x - 3 = 0\).
- Factor the expression to get \((x - 3)(x + 1) = 0\).
- Identify the roots \(x = 3\) and \(x = -1\), which corresponds to the option \(\boxed{(-1, 3)}\).
Explanation
- Standardizing the Equation
The problem asks us to find the solutions to the quadratic equation \(2x^2 - 3x - 2 = x + 4\). To solve this, we first aim to bring the equation into the standard quadratic form, which is \(ax^2 + bx + c = 0\). We achieve this by moving all terms to one side of the equality:
\[2x^2 - 3x - x - 2 - 4 = 0\]
Combining the like terms, we get:
\[2x^2 - 4x - 6 = 0\]
- Solving for x
Now, we can simplify the equation by dividing all coefficients by their greatest common divisor, which is \(2\):
\[x^2 - 2x - 3 = 0\]
To solve this quadratic equation, we look for two numbers that multiply to \(-3\) and add to \(-2\). These numbers are \(-3\) and \(1\). Thus, we can factor the quadratic expression as:
\[(x - 3)(x + 1) = 0\]
By the zero-product property, the solutions are found by setting each factor equal to zero:
\(x - 3 = 0 \implies x = 3\)
\(x + 1 = 0 \implies x = -1\)
So, the solutions to the equation are \(x = 3\) and \(x = -1\).
- Concluding the Solution
Comparing our findings with the provided options, we see that the set \((-1, 3)\) contains our calculated solutions \(x = -1\) and \(x = 3\). Therefore, the correct choice among the options is \((-1, 3)\). Note that since the solutions are \(-1\) and \(3\), we consider these as the correct values, matching the first provided option.
Examples
Quadratic equations are essential in physics and engineering, such as determining the trajectory of a projectile. If you want to know when a ball thrown into the air will hit the ground, you would set up a quadratic equation where the height is zero. Solving this equation helps you find the specific time instances, much like solving for the variable \(x\) in our current problem, allowing for precise predictions of motion.