Completing the Square Slides
Algebra 2 & Honors Unit 3: Quadratic Relations
MATH-ALG2-03
Completing the Square
From Geometric Intuition to Algebraic Mastery & the Quadratic Formula
Geometric Area Models
Leading Coefficients \(a \neq 1\)
Honors Extensions
Interactive Lecture & Guided Problem Set Session Goal: Solve any quadratic in form \((x-h)^2 = k\)
Why Complete the Square?
Conceptual Foundation
When Factoring Fails
Not all quadratics have clean integer roots. Completing the square solves every quadratic with exact radicals.
\(x^2 - 6x - 2 = 0 \to \text{Exact}\)
Reveals Vertex Form
Transforms standard form into vertex form \(y = a(x-h)^2 + k\), making max/min optimization and graphing immediate.
\(\text{Vertex } (h, k) \text{ exposed}\)
The Master Key
The famous Quadratic Formula is not an arbitrary rule; it is derived directly by completing the square on \(ax^2 + bx + c = 0\).
\(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\)
Key Takeaway: We turn an expression with \(x^2\) and \(x\) into an expression with a single squared binomial \((x + d)^2\). \(x \text{ appears once!}\)
The Visual Model: Why "Complete" a Square?
Area Representation
Consider the expression: \(x^2 + bx\) (e.g., \(x^2 + 6x\))
\(x^2\) Area: \(x \times x\)
\(\frac{b}{2}x\) Area: \(\frac{b}{2} \times x\)
\(x \frac{b}{2}\) Area: \(x \times \frac{b}{2}\)
? \(\left(\frac{b}{2}\right)^2\)
The missing corner needed to make a full square is \(\left(\frac{b}{2}\right)^2\)!
1 Start with \(x^2 + bx\)
You have a square of area \(x^2\) plus a rectangular area of \(bx\).
2 Split the linear term into two equal wings
Cut \(bx\) into two halves: \(\frac{b}{2}x\) on the right, and \(\frac{b}{2}x\) on the bottom.
3 Fill the corner to "Complete" the Square
Add \(\left(\frac{b}{2}\right)^2\). The entire shape now has side length \(\left(x + \frac{b}{2}\right)\) and area \(\left(x + \frac{b}{2}\right)^2\).
Algebraic Identity: \(x^2 + bx + \left(\frac{b}{2}\right)^2 = \left(x + \frac{b}{2}\right)^2\) Perfect Square Trinomial
The 5-Step Algebraic Protocol (\(a = 1\))
Standard Operating Procedure
01
Isolate Variables
Move constant \(c\) to the opposite side:
\(x^2 + bx = -c\)
02
Find Magic Value
Halve the coefficient \(b\) and square it:
\(k = \left(\frac{b}{2}\right)^2\)
03
Add to BOTH Sides
Keep balance! Never add to just one side:
\(+ \left(\frac{b}{2}\right)^2 \text{ both}\)
04
Factor as Square
Rewrite left side as a binomial square:
\(\left(x + \frac{b}{2}\right)^2\)
05
Square Root & Solve
Extract roots and remember \(\pm\):
\(x = -\frac{b}{2} \pm \sqrt{\dots}\)
Common Pitfall: Forgetting to add \((b/2)^2\) to the right side breaks equation balance!
Example 1
Solve: \(x^2 - 8x - 9 = 0\)
Case: \(a = 1\), even \(b\)
Step 1: Add 9 to isolate variables
\(x^2 - 8x = 9\)
Step 2: Half of \(-8\) is \(-4\); square it: \((-4)^2 = 16\)
\(\left(\frac{-8}{2}\right)^2 = 16\)
Step 3: Add 16 to both sides
\(x^2 - 8x + 16 = 9 + 16\)
Step 4: Factor left side into square
\((x - 4)^2 = 25\)
Step 5: Take square root of both sides
\(x - 4 = \pm 5\)
Final Solutions
Branch 1 (\(+\)):
\(x = 4 + 5 = 9\)
Branch 2 (\(-\)):
\(x = 4 - 5 = -1\)
Solution Set: \(\{ -1, 9 \}\)
Notice: Since roots are rational, this could have been factored! Next, let's look at irrational roots. Check: \((9)^2 - 8(9) - 9 = 81 - 72 - 9 = 0 \checkmark\)
Your Turn 1
Solve: \(x^2 + 6x - 16 = 0\)
2 Minutes Practice
Follow the Steps on Your Paper:
- 1 Isolate terms with \(x\): \(x^2 + 6x = \text{\_\_\_}\)
- 2 Calculate the magic number: \(\left(\frac{6}{2}\right)^2 = \text{\_\_\_}\)
- 3 Add to both sides & factor left side: \((x + \text{\_\_\_})^2\)
- 4 Take square roots: \(x + \text{\_\_\_} = \pm \text{\_\_\_}\)
Hint: What is half of positive \(6\)?
Solution Verification
\(x^2 + 6x = 16\)
\(x^2 + 6x + \mathbf{9} = 16 + \mathbf{9}\)
\((x + 3)^2 = 25\)
\(x + 3 = \pm 5 \implies x = -3 \pm 5\)
\(x = 2 \quad \text{or} \quad x = -8\)
Self-Check: Did you remember that the binomial inside is \((x + 3)\), not \((x + 9)\)? Ready for radicals!
Example 2
Solve: \(x^2 + 5x - 3 = 0\)
Odd \(b\) \(\to\) Working with Fractions
1. Isolate: Move constant to right
\(x^2 + 5x = 3\)
2. Find Magic Term: Keep fraction!
\(\left(\frac{5}{2}\right)^2 = \frac{25}{4}\)
3. Add to Both: Common denom \(\frac{12}{4} + \frac{25}{4}\)
\(x^2 + 5x + \frac{25}{4} = \frac{37}{4}\)
4. Factor as Square:
\(\left(x + \frac{5}{2}\right)^2 = \frac{37}{4}\)
5. Square Root: \(\sqrt{37/4} = \frac{\sqrt{37}}{2}\)
\(x + \frac{5}{2} = \pm \frac{\sqrt{37}}{2}\)
Exact Radical Form
Combined Single Fraction:
\(x = \frac{-5 \pm \sqrt{37}}{2}\)
Pro Tip for Odd Numbers:
Never turn fractions into rounded decimals! Keep \(\frac{b}{2}\) as an improper fraction. Squaring gives a denominator of 4 every time!
Notice how the denominator in the square root is always \(\sqrt{4} = 2\)! \(\Delta = 37\) (Non-perfect square \(\implies\) irrational roots)
Your Turn 2
Solve: \(x^2 - 3x - 5 = 0\)
2 Minutes Practice
Step-by-Step Prompts:
1. Move constant:
\(x^2 - 3x = 5\)
2. Halve and square:
\(\left(\frac{-3}{2}\right)^2 = \text{\_\_\_}\)
3. Add \(\frac{9}{4}\) to both sides:
\(5 + \frac{9}{4} = \frac{20}{4} + \frac{9}{4} = \text{\_\_\_}\)
Keep exact radical form. Do not use decimals.
Walkthrough & Result
\(x^2 - 3x + \mathbf{\frac{9}{4}} = 5 + \mathbf{\frac{9}{4}}\)
\(\left(x - \frac{3}{2}\right)^2 = \frac{29}{4}\)
\(x - \frac{3}{2} = \pm \frac{\sqrt{29}}{2}\)
Final Exact Solution
\(x = \frac{3 \pm \sqrt{29}}{2}\)
Notice the sign inside the parentheses: because \(b = -3\), the binomial is \((x - 3/2)\). Now: What happens when \(a \neq 1\)?
Conquering Leading Coefficients: When \(a \neq 1\)
Critical Strategy
The Golden Rule: Completing the square REQUIRES \(a = 1\)
You cannot complete the square on \(ax^2 + bx\). You must eliminate or factor out the leading coefficient \(a\) first!
\(a \neq 1 \implies \div a\)
Approach A Best for Equations (\(= 0\))
Divide Entire Equation by \(a\)
Divide every single term on both sides by \(a\). This creates an equivalent equation with leading coefficient 1 immediately.
\(2x^2 + 8x - 10 = 0 \xrightarrow{\div 2} x^2 + 4x - 5 = 0\)
Approach B Essential for Vertex Form
Factor \(a\) Out of \(x\)-Terms Only
Keep \(a\) outside parentheses. Remember: when adding inside, you are really adding \(a \times \text{value}\) to the whole expression!
\(y = 2(x^2 + 4x \mathbf{+ 4}) - 10 \mathbf{- 8}\)
Let's master Approach A for solving equations first, then apply Approach B for vertex form. Step-by-step next \(\to\)
Example 3
Solve: \(2x^2 - 8x + 3 = 0\)
Case: \(a = 2\)
1. Divide all terms by 2:
\(x^2 - 4x + \frac{3}{2} = 0\)
2. Isolate variable terms:
\(x^2 - 4x = -\frac{3}{2}\)
3. Find magic value: \(\left(\frac{-4}{2}\right)^2\)
\((-2)^2 = \mathbf{4}\)
4. Add 4 to both sides: \(-\frac{3}{2} + \frac{8}{2}\)
\(x^2 - 4x + 4 = \frac{5}{2}\)
5. Factor & Square Root:
\((x - 2)^2 = \frac{5}{2} \implies x - 2 = \pm\sqrt{\frac{5}{2}}\)
Rationalizing the Denominator
Rationalize \(\sqrt{\frac{5}{2}}\):
\(\frac{\sqrt{5}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{10}}{2}\)
Exact Solutions
\(x = 2 \pm \frac{\sqrt{10}}{2} = \frac{4 \pm \sqrt{10}}{2}\)
Honors Standard: Always rationalize radicals in denominators before reporting final answers! Both answers are real & irrational
Your Turn 3
Solve: \(3x^2 - 12x + 6 = 0\)
2 Minutes Practice
Student Milestones:
Step 1: Divide by 3
\(x^2 - 4x + 2 = 0\)
Step 2: Move 2 & complete square
\(x^2 - 4x + \text{\_\_\_} = -2 + \text{\_\_\_}\)
Step 3: Solve for \(x\)
\((x - 2)^2 = \text{\_\_\_} \implies x = \text{\_\_\_}\)
Check: Does 3 divide nicely into every coefficient? Yes!
Full Worked Verification
\(x^2 - 4x = -2\)
\(x^2 - 4x + \mathbf{4} = -2 + \mathbf{4}\)
\((x - 2)^2 = 2\)
\(x - 2 = \pm \sqrt{2}\)
Final Solutions
\(x = 2 \pm \sqrt{2}\)
Approximation check: \(2 + 1.414 \approx 3.414\) and \(2 - 1.414 \approx 0.586\) Honors Extensions Ahead \(\to\)
Honors Challenge 1
Non-Real & Complex Solutions (\(i\))
Discriminant < 0
Solve: \(x^2 - 6x + 25 = 0\)
1. Isolate: \(x^2 - 6x = -25\)
2. Add \((-3)^2 = 9\): \(x^2 - 6x + 9 = -25 + 9\)
3. Factor binomial: \((x - 3)^2 = -16\)
4. Square root with \(i\): \(x - 3 = \pm 4i\)
Conjugate Pairs
Because \(\sqrt{-1} = i\), equations with no real roots yield complex conjugate pairs:
\(x = 3 \pm 4i\)
Graph Meaning: The parabola's vertex is at \((3, 16)\) and opens up—never touching the \(x\)-axis!
Completing the square immediately reveals both the vertex \((3, 16)\) and the complex roots! Fundamental Theorem of Algebra \(\checkmark\)
Honors Challenge 2
The Grand Derivation: Quadratic Formula
From Literal \(ax^2 + bx + c = 0\)
1. Divide by \(a\) and move constant: \(x^2 + \frac{b}{a}x = -\frac{c}{a}\)
2. Half of \(\frac{b}{a}\) is \(\frac{b}{2a}\); square it: \(\text{Add } \left(\frac{b}{2a}\right)^2 = \frac{b^2}{4a^2}\)
3. Complete square on left; common denom on right: \(\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}\)
4. Square root both sides (\(\sqrt{4a^2} = 2a\)): \(x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a}\)
5. Subtract \(\frac{b}{2a}\) to isolate \(x\): \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Mystery Solved
The Quadratic Formula isn't magic—it is literally completing the square executed once for all possible coefficients!
Notice the discriminant \(b^2 - 4ac\) inside the radical!
Every high school math student memorizes this formula; Honors students know where it comes from. Q.E.D.
Honors Extension
Converting Standard Form to Vertex Form
\(y = a(x-h)^2 + k\)
Example: Rewrite \(f(x) = 2x^2 - 12x + 23\)
1. Group \(x\)-terms & factor out \(a = 2\):
\(f(x) = 2(x^2 - 6x) + 23\)
2. Complete square INSIDE parentheses:
Half of \(-6\) is \(-3\); squared is \(+9\).
Caution: Adding 9 inside multiplies by 2 outside \(\to\) we added 18!
\(f(x) = 2(x^2 - 6x + \mathbf{9}) + 23 - \mathbf{18}\)
3. Factor binomial and combine constants:
\(f(x) = 2(x - 3)^2 + 5\)
Immediate Graph Features
Vertex \((h, k)\): \((3, 5)\)
Axis of Symmetry: \(x = 3\)
Minimum Value: \(y = 5\)
Direction: Opens Upward (\(a=2 > 0\))
Optimization connection: In business/physics, finding the vertex gives maximum profit or minimum cost. Vertex revealed in seconds!
Quadratic Solver Decision Matrix
Strategic Selection
Method 1
Factoring
Fastest method when coefficients are friendly and rational roots exist.
Best when:
• \(a = 1\) and factors of \(c\) add to \(b\)
• Difference of squares
Speed: Lightning (if it factors)
Method 2
Complete Square
Unbeatable when \(b\) is an even integer or when converting to vertex form.
Best when:
• \(a = 1\) and \(b\) is even
• Finding vertex or circle equations
• Conic sections / Calculus
Power: Structural Insight
Method 3
Quadratic Formula
The brute force calculator when numbers are messy, large, or decimal-heavy.
Best when:
• \(a\) is ugly (e.g., \(7x^2 - 19x + 3 = 0\))
• \(b\) is an awkward odd number
• Only numerical roots needed
Reliability: 100% Universal
Exit Ticket Challenge: Solve \(x^2 + 10x + 29 = 0\) by completing the square!
Answer: \(x = -5 \pm 2i\)
