Expanding the Quadratic Expression: x² − x − 6 = 0 — Solving the Equation Step by Step

Solving quadratic equations is a foundational skill in algebra, essential for students and coding enthusiasts alike. One of the most commonly encountered quadratic expressions is x² − x − 6 = 0. Expanding and solving this equation helps strengthen your understanding of factoring, the quadratic formula, and real-world applications. In this article, we’ll walk through the step-by-step expansion, factoring, and solving of the equation x² − x − 6 = 0, providing clear explanations and practical insight.

Understanding the Context

What is the Quadratic Equation?

A standard quadratic equation takes the form:
ax² + bx + c = 0,
where a, b, and c are constants, and a ≠ 0.

In our example:

  • a = 1
  • b = −1
  • c = −6

Expanding the quadratic allows us to solve for x using factoring, completing the square, or the quadratic formula.

Solved: x^2-x-6=0 [Math]

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Solved: x^2-x-6=0 [Math]
Solved: 2x^2-x-6=0 [algebra]
Solved: Solve by factorising: a) x^2+x-6=0 [Math]
Solved: Expand and simplify (x+6)(x-2) [Math]
Solved 2x2+x−6

Key Insights

Step 1: Expanding the Quadratic Expression

Although x² − x − 6 is already expanded, understanding expansion helps when you start with a factored form. Suppose you factor it as:
x² − x − 6 = (x − 3)(x + 2)

Now, expand it using the distributive property:
(x − 3)(x + 2) = x·x + x·2 − 3·x − 3·2 = x² + 2x − 3x − 6 = x² − x − 6

This confirms the factorization is correct.

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Solution: The given sequence is arithmetic with first term $ a = 3 $ and common difference $ d = 5 $. The last term is $ l = 198 $. Using the formula for the $ n $-th term of an arithmetic sequence:
Solution: This is an arithmetic sequence with first term $ a = 2 $, common difference $ d = 5 $, and last term $ l = 197 $. Using the $ n $-th term formula:
Question: What is the greatest common factor of 84 and 108, two key values used in modeling genetic motif repetition frequencies?

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Final Thoughts

Step 2: Solving x² − x − 6 = 0 by Factoring

Set each factor equal to zero:
(x − 3)(x + 2) = 0

For the product to be zero, either factor must be zero:

  1. x − 3 = 0 → x = 3
  2. x + 2 = 0 → x = −2

solutions: x = 3,         x = −2

Step 3: Using the Quadratic Formula (Alternative Method)

For any ax² + bx + c = 0, the quadratic formula gives:
x = [ −b ± √(b² − 4ac) ] / (2a)

Plugging in a = 1, b = −1, c = −6:
x = [ −(−1) ± √( (−1)² − 4·1·(−6) ) ] / (2·1)
x = [ 1 ± √(1 + 24) ] / 2
x = [ 1 ± √25 ] / 2
x = [ 1 ± 5 ] / 2

Thus:

  • x = (1 + 5)/2 = 6/2 = 3
  • x = (1 − 5)/2 = −4/2 = −2