- Rewrite the equation \(x^2 - 5x = -4\) in standard form as \(x^2 - 5x + 4 = 0\).
- Factor the quadratic expression into \((x - 1)(x - 4) = 0\).
- Apply the zero-product property to find the roots \(x = 1\) and \(x = 4\).
- Order the roots as \(\boxed{x = 1, x = 4}\).
Explanation
- Rewriting in Standard Form
To solve the quadratic equation \(x^2 - 5x = -4\), we first rewrite the equation in the standard form \(ax^2 + bx + c = 0\). We do this by adding 4 to both sides of the equation, which gives us:
\[x^2 - 5x + 4 = 0\]
This is a standard quadratic equation where \(a = 1\), \(b = -5\), and \(c = 4\). We need to find the values of \(x\) that satisfy this equation.
- Factoring the Quadratic Expression
To solve the quadratic equation \(x^2 - 5x + 4 = 0\), we look for factors of the constant term (4) that sum to the coefficient of the middle term (-5). We are looking for two numbers \(p\) and \(q\) such that:
\[p \cdot q = 4\]
\[p + q = -5\]
The numbers -1 and -4 satisfy both conditions since \((-1) \times (-4) = 4\) and \((-1) + (-4) = -5\). Therefore, we can write the quadratic expression as a product of two binomials:
\[(x - 1)(x - 4) = 0\]
- Finding the Solutions
According to the zero-product property, the product of two factors is zero if at least one of the factors is zero. Thus, we set each factor equal to zero:
\[x - 1 = 0 \quad \text{or} \quad x - 4 = 0\]
Solving for \(x\) in each case, we find:
\[x = 1 \quad \text{or} \quad x = 4\]
Since we are asked to enter the smaller value first, we list the solutions in increasing order as \(x = 1\) and \(x = 4\).
Examples
Understanding quadratic equations is essential for modeling real-world phenomena, such as the trajectory of a projectile. If a ball is thrown into the air, its height over time can often be described by a quadratic equation. Solving for the roots helps us determine when the object is at a specific height or when it hits the ground. This skill is foundational in fields ranging from physics and engineering to economics and data analysis.