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Complete the Square
to Find Solutions

Transform any quadratic equation into perfect square form to reveal its solutions: unlock the power behind the quadratic formula.

Algebra 2

What Is a Perfect Square Trinomial?

A perfect square trinomial is an expression that factors into the square of a binomial. Recognizing them is the foundation of completing the square.

The key insight: the constant term is always the square of half the coefficient of the linear term. Here, half of 6 is 3, and \( 3^2 = 9 \).

The Geometry of Completing the Square

\( x^2 \) \( \tfrac{b}{2} \cdot x \) \( x \cdot \tfrac{b}{2} \) \( \left(\tfrac{b}{2}\right)^{\!2} \) \( x \) \( \tfrac{b}{2} \) \( x \) \( \tfrac{b}{2} \) \( \left(x + \tfrac{b}{2}\right)^{\!2} \) \( x + \tfrac{b}{2} \) \( x + \tfrac{b}{2} \) Add the missing corner to complete the square

Why Complete the Square?

Solve Equations
Works on every quadratic, even when factoring fails.
Derive the Formula
The quadratic formula is derived by completing the square on \( ax^2 + bx + c = 0 \).
Vertex Form
Reveals the vertex \( (h, k) \) of a parabola for graphing.

The Core Transformation

Completing the square converts a quadratic equation into an equivalent equation with a perfect square on one side:

Once in this form, take the square root of both sides:

Factoring vs. Completing the Square

Factoring Completing the Square
When it works Only "nice" integer roots Always
Result form \( (x - r_1)(x - r_2) = 0 \) \( (x - p)^2 = q \)
Handles irrational roots? No Yes
Handles complex roots? No Yes (\( q < 0 \))
Gives vertex form? No Yes

How to Complete the Square

1
Isolate the quadratic and linear terms on one side. Move the constant to the other side.
2
Make the leading coefficient 1. If \( a \neq 1 \), divide every term by \( a \).
3
Find the magic number: take half the coefficient of \( x \), then square it: \( \left(\frac{b}{2}\right)^2 \).
4
Add that number to both sides to keep the equation balanced.
5
Factor the left side as a perfect square: \( (x + \frac{b}{2})^2 \).
6
Solve by taking the square root of both sides. Don't forget \( \pm \)!

Example 1: Simple Case

Solve \( x^2 + 6x + 2 = 0 \)

1
Move constant: \( x^2 + 6x = -2 \)
2
Leading coefficient is already 1. ✓
3
Half of 6 is 3; square it: \( 3^2 = \) 9
4
Add 9 to both sides: \( x^2 + 6x + 9 = -2 + 9 = 7 \)
5
Factor: \( (x + 3)^2 = 7 \)
6
Solve: \( x = -3 \pm \sqrt{7} \)

Visualizing \( x^2 + 6x + 2 = 0 \)

\( 0 \) \( 1 \) \( -1 \) \( -2 \) \( -3 \) \( -4 \) \( -5 \) \( -6 \) \( 4 \) \( -6 \) vertex \( (-3, -7) \) \( -3-\sqrt{7} \) \( -3+\sqrt{7} \) \( x \) \( y \)

Example 2: When \( a \neq 1 \)

Solve \( 2x^2 + 12x + 7 = 0 \)

1
Move constant: \( 2x^2 + 12x = -7 \)
2
Divide by 2: \( x^2 + 6x = -\frac{7}{2} \)
3
Half of 6 is 3; square: \( 3^2 = 9 \)
4
Add 9 to both sides: \( x^2 + 6x + 9 = -\frac{7}{2} + 9 = \frac{11}{2} \)
5
Factor: \( (x + 3)^2 = \frac{11}{2} \)
6
Solve: \( x = -3 \pm \sqrt{\frac{11}{2}} = -3 \pm \frac{\sqrt{22}}{2} \)

Example 3: Odd Coefficients

Solve \( x^2 + 5x + 3 = 0 \)

1
Move constant: \( x^2 + 5x = -3 \)
2
Leading coefficient is 1. ✓
3
Half of 5 is \( \frac{5}{2} \); square: \( \left(\frac{5}{2}\right)^2 = \frac{25}{4} \)
4
Add \( \frac{25}{4} \) to both sides: \( x^2 + 5x + \frac{25}{4} = -3 + \frac{25}{4} = \frac{13}{4} \)
5
Factor: \( \left(x + \frac{5}{2}\right)^2 = \frac{13}{4} \)
6
Solve: \( x = -\frac{5}{2} \pm \frac{\sqrt{13}}{2} = \frac{-5 \pm \sqrt{13}}{2} \)

Deriving the Quadratic Formula

Start with the general equation \( ax^2 + bx + c = 0 \) and complete the square:

1
\( x^2 + \frac{b}{a}x = -\frac{c}{a} \)
2
\( x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = \frac{b^2}{4a^2} - \frac{c}{a} \)
3
\( \left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2} \)
4
\( x + \frac{b}{2a} = \pm\frac{\sqrt{b^2 - 4ac}}{2a} \)

The Discriminant Connection

In the completed-square form \( (x - p)^2 = q \), the value of \( q \) determines the nature of the solutions, just like the discriminant \( b^2 - 4ac \).

\( q > 0 \)
Two distinct real solutions: \( x = p \pm \sqrt{q} \)
\( q = 0 \)
One repeated real solution: \( x = p \)
\( q < 0 \)
Two complex solutions: \( x = p \pm i\sqrt{|q|} \)

Common Mistakes to Avoid

✕ Forgetting both sides
When you add \( \left(\frac{b}{2}\right)^2 \) to the left, you must add it to the right too. The equation must stay balanced.
✕ Skipping the division
If \( a \neq 1 \), you must divide all terms by \( a \) first. Completing the square on \( 2x^2 + 12x \) directly gives wrong results.
✕ Dropping the \( \pm \)
\( \sqrt{q} \) has two values. Writing only the positive root loses half the solutions.

Summary: The Complete Process

Step Action Result
1 Move \( c \) to the right \( ax^2 + bx = -c \)
2 Divide by \( a \) \( x^2 + \frac{b}{a}x = -\frac{c}{a} \)
3 Add \( \left(\frac{b}{2a}\right)^2 \) to both sides Left side becomes a perfect square
4 Factor left, simplify right \( \left(x + \frac{b}{2a}\right)^2 = q \)
5 Square root both sides \( x = -\frac{b}{2a} \pm \sqrt{q} \)

Question 1 of 5

To complete the square on \( x^2 + 10x \), what number must you add?

Question 2 of 5

What is the completed-square form of \( x^2 - 8x + 10 = 0 \)?

Question 3 of 5

Solve \( x^2 + 4x - 1 = 0 \) by completing the square. What are the solutions?

Question 4 of 5

When completing the square on \( 3x^2 + 18x + 5 = 0 \), what is the first step?

Question 5 of 5

If \( (x - 3)^2 = -4 \), how many real solutions does the equation have?