Here, \( a = 1 \), \( b = -5 \), \( c = 6 \). - ECD Germany
Optimizing Solutions with Quadratic Equations: The Case of \( a = 1 \), \( b = -5 \), \( c = 6 \)
Optimizing Solutions with Quadratic Equations: The Case of \( a = 1 \), \( b = -5 \), \( c = 6 \)
Solving quadratic equations is a fundamental skill in algebra, with applications across science, engineering, economics, and beyond. In this article, we explore a specific quadratic equation with the coefficients \( a = 1 \), \( b = -5 \), and \( c = 6 \), demonstrating step-by-step how to find its roots and understand its significance.
Understanding the Context
The Quadratic Equation: A Closer Look
Start with the standard form of a quadratic equation:
\[
ax^2 + bx + c = 0
\]
Given:
\( a = 1 \),
\( b = -5 \),
\( c = 6 \)
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Substituting these values, the equation becomes:
\[
x^2 - 5x + 6 = 0
\]
This quadratic equation opens upward because the coefficient of \( x^2 \) is positive (\( a = 1 > 0 \)), meaning it has a minimum point at its vertex.
Solving the Equation: Step-by-Step
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Method 1: Factoring
Since \( a = 1 \), the equation simplifies nicely to a factorable form. We seek two numbers that multiply to \( c = 6 \) and add to \( b = -5 \).
- The pair: \( -2 \) and \( -3 \)
Because:
\( (-2) \ imes (-3) = 6 \)
\( (-2) + (-3) = -5 \)
Thus, we factor:
\[
x^2 - 5x + 6 = (x - 2)(x - 3) = 0
\]
Set each factor equal to zero:
\[
x - 2 = 0 \quad \Rightarrow \quad x = 2
\]
\[
x - 3 = 0 \quad \Rightarrow \quad x = 3
\]
Method 2: Quadratic Formula
For broader understanding, the quadratic formula expresses solutions directly:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
