Completing the Square

🔲 Completing the Square

Part 1 of 5 — Perfect Square Trinomials

Topics in This Part

🔑 Key Concept: "Completing the square" turns any quadratic into a perfect square plus a constant. Before we can complete a square, we need to recognize one — that's Part 1.

What Is a Perfect Square Trinomial?

A perfect square trinomial factors into a binomial squared:

$x^2 + 2ax + a^2 = (x + a)^2$

Notice the pattern:

• The first term is a perfect square: $x^2$
• The last term is a perfect square: $a^2$
• The middle term is twice the product: $2ax$

Examples

🔑 Key Idea: To make $x^2 + bx$ into a perfect square, add $\left(\dfrac{b}{2}\right)^2$. It then factors as $\left(x + \dfrac{b}{2}\right)^2$.

Concept Check 🎯

Find the Missing Constant 🧮

What number must be added to make each a perfect square trinomial? (Use $\left(\dfrac{b}{2}\right)^2$.)

1) $x^2 + 8x + \,?$ 2) $x^2 - 6x + \,?$ 3) $x^2 + 5x + \,?$ (decimal is fine)

Factoring Back to a Square

Once the constant is $\left(\dfrac{b}{2}\right)^2$, the trinomial factors with the same sign as the middle term:

$x^2 + bx + \left(\tfrac{b}{2}\right)^2 = \left(x + \tfrac{b}{2}\right)^2$

Example: $x^2 + 8x + 16 = (x+4)^2$ — half of $8$ is $4$.

Example: $x^2 - 6x + 9 = (x-3)^2$ — half of $-6$ is $-3$.

You now have the one tool the whole method depends on. In Part 2 we use it to complete squares that aren't finished yet.

🔲 Completing the Square

Part 2 of 5 — The Method (leading coefficient = 1)

🔑 The Idea: Take $x^2 + bx + c$, force a perfect square by adding $\left(\dfrac{b}{2}\right)^2$, and balance it by subtracting the same amount.

The Steps

To complete the square on $x^2 + bx + c$:

1. Group the $x$-terms: $(x^2 + bx) + c$
2. Add and subtract $\left(\dfrac{b}{2}\right)^2$ inside: $\left(x^2 + bx + \left(\tfrac{b}{2}\right)^2\right) - \left(\tfrac{b}{2}\right)^2 + c$
3. Factor the perfect square: $\left(x + \dfrac{b}{2}\right)^2 - \left(\dfrac{b}{2}\right)^2 + c$

Worked Example: $x^2 + 6x + 1$

Half of $6$ is $3$, and $3^2 = 9$:

$x^2 + 6x + 1 = (x^2 + 6x + 9) - 9 + 1 = (x+3)^2 - 8$

✅ Check: $(x+3)^2 - 8 = x^2 + 6x + 9 - 8 = x^2 + 6x + 1$ ✓

Worked Example: $x^2 - 10x + 30$

Half of $-10$ is $-5$, and $(-5)^2 = 25$:

$x^2 - 10x + 30 = (x^2 - 10x + 25) - 25 + 30 = (x-5)^2 + 5$

Worked Example: $x^2 + 3x$

Half of $3$ is $\dfrac{3}{2}$, and $\left(\dfrac{3}{2}\right)^2 = \dfrac{9}{4}$:

$x^2 + 3x = \left(x + \tfrac{3}{2}\right)^2 - \tfrac{9}{4}$

💡 Completing the square always produces the form $(x + h)^2 + k$ — that form is the key to both solving (Part 3) and graphing (Part 4).

Concept Check 🎯

Complete the Square 🧮

Rewrite each as $(x+h)^2 + k$. Enter the value of $k$ (the constant outside the square).

1) $x^2 + 8x + 3 = (x+4)^2 + k,\quad k = \,?$ 2) $x^2 - 2x + 6 = (x-1)^2 + k,\quad k = \,?$ 3) $x^2 + 10x = (x+5)^2 + k,\quad k = \,?$

🔲 Completing the Square

Part 3 of 5 — Solving Quadratic Equations

🔑 Why it works: Once a quadratic is written as $(x+h)^2 = \text{number}$, you can take the square root of both sides and solve — including the all-important $\pm$.

Solving by Completing the Square

Example: $x^2 + 6x + 5 = 0$

1. Move the constant: $x^2 + 6x = -5$
2. Add $\left(\frac{6}{2}\right)^2 = 9$ to both sides: $x^2 + 6x + 9 = 4$
3. Factor: $(x+3)^2 = 4$
4. Square root (don't forget $\pm$): $x + 3 = \pm 2$
5. Solve: $x = -3 \pm 2 \Rightarrow x = -1 \text{ or } x = -5$

⚠️ The most common mistake is dropping the $\pm$. A square has two square roots.

When the Leading Coefficient Isn't 1

If $a \ne 1$, divide every term by $a$ first, then complete the square.

Example: $2x^2 + 8x - 10 = 0$

Divide by $2$: $\;x^2 + 4x - 5 = 0$

$x^2 + 4x = 5 \;\Rightarrow\; x^2 + 4x + 4 = 9 \;\Rightarrow\; (x+2)^2 = 9$

$x + 2 = \pm 3 \;\Rightarrow\; x = 1 \text{ or } x = -5$

✅ Check $x=1$: $2(1)+8(1)-10 = 0$ ✓

Order the Steps 🔽

You're solving $x^2 + 6x + 5 = 0$. Choose what happens at each stage.

Solve It 🧮

Solve by completing the square. Enter the larger solution.

1) $x^2 + 4x - 5 = 0 \;\Rightarrow\; x = \,?$ (larger root) 2) $x^2 - 8x + 7 = 0 \;\Rightarrow\; x = \,?$ (larger root)

🔲 Completing the Square

Part 4 of 5 — Vertex Form & Graphing

🔑 Big Payoff: Completing the square rewrites $y = ax^2+bx+c$ as vertex form $y = a(x-h)^2 + k$, which hands you the vertex $(h, k)$ for free.

Vertex Form

$y = a(x - h)^2 + k \quad\Longrightarrow\quad \text{vertex } (h,\, k)$

Watch the sign: $(x - h)$ means the vertex's $x$-coordinate is $+h$.

Example: $y = x^2 + 6x + 5$

Complete the square: $y = (x^2+6x+9) - 9 + 5 = (x+3)^2 - 4$.

Rewrite as $(x - (-3))^2 + (-4)$, so the vertex is $(-3, -4)$.

Example: $y = x^2 - 4x + 1$

Half of $-4$ is $-2$; $(-2)^2 = 4$:

$y = (x^2 - 4x + 4) - 4 + 1 = (x-2)^2 - 3$

Vertex $(2, -3)$, axis of symmetry $x = 2$, minimum value $-3$.

💡 Because the leading coefficient is positive, the parabola opens up, so the vertex is the lowest point.

Read the Vertex 🔽

Find the Vertex 🧮

Complete the square and enter the vertex coordinates.

1) $y = x^2 + 2x - 3$. Vertex $x$-coordinate $= \,?$ 2) Same parabola: vertex $y$-coordinate $= \,?$

🔲 Completing the Square

Part 5 of 5 — Mixed Practice & Mastery Check

You can now (1) recognize perfect squares, (2) complete the square, (3) solve equations, and (4) find a parabola's vertex. Let's put it together.

Quick Reference

⚠️ Remember: divide by $a$ first if the leading coefficient isn't $1$, and never drop the $\pm$ when taking a square root.

Mixed Practice 🎯

Exit Quiz ✅

Answer all three to finish the lesson.