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:
\( a = 1 \),
\( b = -5 \),
\( c = 6 \)

Solved 1. In ABC we have A1∈BC,B1∈(AC),C1∈(AB). If | Chegg.com

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Solved 1. In ABC we have A1∈BC,B1∈(AC),C1∈(AB). If | Chegg.com
Solved Q6/F1=A'B'+B'C+ABC+A'C' F2=A′+B′C+A′C F3 =B′+ABC F4 | Chegg.com
Solved a₁ b₁ C1 b₂ C2 C₂ = 6. Find each of the following. | Chegg.com
Solved (b) (2) (6 marks) Let A=(a1,a2),B=(b1,b2) and | Chegg.com
View question - (a+b+c)²=a²+b²+c² Proved that 1/a+1/b+1/c=0

Key Insights

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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Final Thoughts

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:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]