Solve the equation $0=2 x^2-5 x-3$ algebraically using any method.
The equation to solve is \(2x^2 - 5x - 3 = 0\). This is a quadratic equation in the form \(ax^2 + bx + c = 0\), where \(a = 2\), \(b = -5\), and \(c = -3\). We are looking for values of \(x\) that satisfy this equality. We can solve this by factoring or by using the quadratic formula.
To factor the quadratic expression \(2x^2 - 5x - 3\), we look for two numbers that multiply to \(a \times c = 2 \times (-3) = -6\) and add up to the middle coefficient \(b = -5\). The numbers \(-6\) and \(1\) satisfy these conditions since \((-6) \times 1 = -6\) and \(-6 + 1 = -5\). We can now split the middle term \(-5x\) using these two numbers:
\[2x^2 - 6x + 1x - 3 = 0\]
Now, we factor the expression by grouping:
\[(2x^2 - 6x) + (x - 3) = 0\]
\[2x(x - 3) + 1(x - 3) = 0\]
Factoring out the common binomial \((x - 3)\), we get:
\[(2x + 1)(x - 3) = 0\]
To find the solutions for \(x\), we set each factor equal to zero:
Thus, the roots of the equation are \(x = -\frac{1}{2}\) and \(x = 3\). Looking at the provided options, the first option \(x = -\frac{1}{2}, x = 3\) is the correct one.
Quadratic equations are widely used in physics and engineering to model projectile motion. For example, if you throw a ball, its height \(h\) as a function of time \(t\) can be modeled by a quadratic equation \(h(t) = -gt^2 + v_0t + h_0\). Solving for \(h(t) = 0\) helps determine when the object hits the ground, allowing you to calculate flight time and range. Mastering these algebraic techniques provides the foundation for solving such real-world problems.