Advanced Completing the Square Problems (a ≠ 1) with Solutions

📚 Completing the Square: When a ≠ 1

Completing the square is a technique used to rewrite a quadratic expression in the form $ax^2 + bx + c$ into the form $a(x + h)^2 + k$. When $a = 1$, the process is relatively straightforward. However, when $a ≠ 1$, extra steps are required to ensure the accuracy of the transformation. This guide will provide a detailed walkthrough of these advanced problems.

📜 History and Background

The method of completing the square has ancient roots, dating back to Babylonian mathematicians who used geometric approaches to solve quadratic equations. The technique was further developed by Greek mathematicians like Euclid. The algebraic formulation we use today evolved over centuries, becoming a fundamental tool in algebra and calculus.

🔑 Key Principles

• ➗ Factor out 'a': The first and most crucial step is to factor out the coefficient 'a' from the $x^2$ and $x$ terms. This isolates the quadratic expression within the parentheses.
• 🧮 Complete the Square Inside Parentheses: Focus on the expression inside the parentheses. Take half of the coefficient of the $x$ term, square it, and add it inside the parentheses. Remember to subtract the correct amount outside the parentheses to maintain the equation's balance.
• ✍️ Rewrite as a Squared Term: Rewrite the expression inside the parentheses as a squared term, $(x + h)^2$.
• ➕ Simplify: Simplify the constant term outside the parentheses to obtain the final completed square form: $a(x + h)^2 + k$.

⚙️ Step-by-Step Guide

Let's consider a general quadratic expression: $ax^2 + bx + c$. Here’s how to complete the square when $a ≠ 1$:

1. Factor out 'a' from the $x^2$ and $x$ terms: $a(x^2 + \frac{b}{a}x) + c$
2. Take half of the coefficient of $x$ inside the parenthesis, square it, and add and subtract it inside the parenthesis: Half of $\frac{b}{a}$ is $\frac{b}{2a}$. Squaring it gives $(\frac{b}{2a})^2 = \frac{b^2}{4a^2}$. So we have: $a(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} - \frac{b^2}{4a^2}) + c$
3. Rewrite the expression inside the parentheses as a perfect square: $a[(x + \frac{b}{2a})^2 - \frac{b^2}{4a^2}] + c$
4. Distribute 'a' to the second term inside the brackets: $a(x + \frac{b}{2a})^2 - a(\frac{b^2}{4a^2}) + c$
5. Simplify: $a(x + \frac{b}{2a})^2 - \frac{b^2}{4a} + c$
6. Combine the constants: $a(x + \frac{b}{2a})^2 + \frac{4ac - b^2}{4a}$

➗ Example 1: $2x^2 + 8x + 6$

1. Factor out 2: $2(x^2 + 4x) + 6$
2. Complete the square inside the parentheses: Half of 4 is 2, and $2^2 = 4$. So, $2(x^2 + 4x + 4 - 4) + 6$
3. Rewrite as a squared term: $2[(x + 2)^2 - 4] + 6$
4. Distribute the 2: $2(x + 2)^2 - 8 + 6$
5. Simplify: $2(x + 2)^2 - 2$

➕ Example 2: $3x^2 - 12x + 5$

1. Factor out 3: $3(x^2 - 4x) + 5$
2. Complete the square inside the parentheses: Half of -4 is -2, and $(-2)^2 = 4$. So, $3(x^2 - 4x + 4 - 4) + 5$
3. Rewrite as a squared term: $3[(x - 2)^2 - 4] + 5$
4. Distribute the 3: $3(x - 2)^2 - 12 + 5$
5. Simplify: $3(x - 2)^2 - 7$

📝 Practice Quiz

Complete the square for the following quadratic expressions:

1. 🧮 $2x^2 + 4x + 1$
2. ➗ $3x^2 - 6x + 2$
3. ➕ $4x^2 + 8x - 3$
4. ➖ $5x^2 - 10x + 1$
5. ✖️ $2x^2 + 12x + 5$

✅ Solutions

1. 🔑 $2(x + 1)^2 - 1$
2. 💡 $3(x - 1)^2 - 1$
3. 🖍️ $4(x + 1)^2 - 7$
4. ✔️ $5(x - 1)^2 - 4$
5. 📐 $2(x + 3)^2 - 13$

💡 Conclusion

Completing the square when $a ≠ 1$ requires careful attention to detail and a systematic approach. By factoring out the leading coefficient, completing the square inside the parentheses, and then simplifying, you can rewrite any quadratic expression in the vertex form. This technique is valuable in various mathematical contexts, including finding the vertex of a parabola, solving quadratic equations, and simplifying expressions in calculus.