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Completing the Square

Completing the Square

Completing the square is a method for solving quadratic equations by converting the equation from the form $ax^2 + bx + c = 0$ to the form . This method is particularly useful for quadratic equations that are difficult to factor using regular factorization.

Completing the SquareSolve quadratic equations by completing the square and rewriting each expression as a perfect-square trinomial.

On this page

Completing the Square
Quadratics That Are Hard to Factor
Steps for Completing the Square
Solving Quadratic Equations by Completing the Square
- Equation with Leading Coefficient One
- Equation with Leading Coefficient Not One
Checks for Common Errors
Special Cases and Variations
- When the Discriminant is Negative
- For Incomplete Quadratic Equations
Practice Problems
- Answer Key

Comments

(x + p)^2 = q

A perfect square trinomial follows the pattern $x^2 + 2px + p^2 = (x + p)^2$ . Once the left side is a square, solve for the roots by taking the square root of both sides. When the middle term is $2px$ , add $p^2$ to complete the square.

Quadratics That Are Hard to Factor

Factoring $x^2 + 4x + 2 = 0$ over the rational numbers would require two rational numbers with product $2$ and sum $4$ . No such pair exists.

Completing the square produces an equivalent equation and its roots without requiring that rational pair.

Steps for Completing the Square

Here are the steps to solve a quadratic equation $ax^2 + bx + c = 0$ using the completing the square method:

Ensure the coefficient of $x^2$ is $1$

If the coefficient $a$ of $x^2$ is not $1$ , divide the entire equation by the value of $a$ .

Example: For the equation $2x^2 + 6x + 3 = 0$
$\frac{2x^2 + 6x + 3}{2} = \frac{0}{2}$
Move the Constant Term to the Right Side

Move the constant term to the right side of the equation.

Example: From the equation $x^2 + 3x + \frac{3}{2} = 0$
Add the Completing-Square Term to Both Sides

Add $\left(\frac{b}{2}\right)^2$ to both sides of the equation. This value is the square of half the coefficient of $x$ .

Example: For the equation
Factor the Left Side into a Perfect Square

The left side now has the form $x^2 + bx + \left(\frac{b}{2}\right)^2$ , which can be factored as .
Simplify the Right Side

Put the fractions over a common denominator and combine their numerators.

Example: For $\left(x + \frac{3}{2}\right)^2 = -\frac{3}{2} + \frac{9}{4}$
Take the Square Root of Both Sides

To eliminate the square, take the square root of both sides.

Example: From the equation $\left(x + \frac{3}{2}\right)^2 = \frac{3}{4}$
Isolate the Variable

Isolate the variable $x$ to find the roots of the equation.

Example: From $x + \frac{3}{2} = \pm\frac{\sqrt{3}}{2}$

The roots of the equation are $x = -\frac{3}{2} + \frac{\sqrt{3}}{2}$ and $x = -\frac{3}{2} - \frac{\sqrt{3}}{2}$ .

Solving Quadratic Equations by Completing the Square

Equation with Leading Coefficient One

Solve the equation: $x^2 + 5x + 6 = 0$

Step $1$ : The coefficient $a = 1$ , so we proceed to the next step.

Step $2$ : Move the constant to the right side.

x^2 + 5x = -6

Step $3$ : Add the square of half the coefficient of $x$ to both sides.

\text{Half the coefficient of } x = \frac{5}{2}

Step $4$ : Factor the left side into a perfect square.

\left(x + \frac{5}{2}\right)^2 = -6 + \frac{25}{4}

Step $5$ : Simplify the right side.

-6 + \frac{25}{4} = \frac{-24 + 25}{4} = \frac{1}{4}

Step $6$ : Take the square root of both sides.

x + \frac{5}{2} = \pm\frac{1}{2}

Step $7$ : Solve for the value of $x$ .

x + \frac{5}{2} = \frac{1}{2} \quad \text{or} \quad x + \frac{5}{2} = -\frac{1}{2}

The roots of the equation are $x = -2$ and $x = -3$ .

Equation with Leading Coefficient Not One

Solve the equation: $2x^2 + 6x + 3 = 0$

Step $1$ : Divide all terms by the coefficient $a = 2$

x^2 + 3x + \frac{3}{2} = 0

Step $2$ : Move the constant to the right side

x^2 + 3x = -\frac{3}{2}

Step $3$ : Add the square of half the coefficient of $x$ to both sides

\text{Half the coefficient of } x = \frac{3}{2}

Step $4$ : Factor the left side into a perfect square

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

Step $5$ : Simplify the right side

-\frac{3}{2} + \frac{9}{4} = \frac{-6 + 9}{4} = \frac{3}{4}

Step $6$ : Take the square root of both sides

x + \frac{3}{2} = \pm\frac{\sqrt{3}}{2}

Step $7$ : Solve for the value of $x$

x + \frac{3}{2} = \frac{\sqrt{3}}{2} \quad \text{or} \quad x + \frac{3}{2} = -\frac{\sqrt{3}}{2}

The roots of the equation are $x = -\frac{3}{2} + \frac{\sqrt{3}}{2}$ and $x = -\frac{3}{2} - \frac{\sqrt{3}}{2}$ .

Checks for Common Errors

For equations with coefficient of $x^2$ not equal to $1$ : Always divide the entire equation by the coefficient $a$ first. Example: $3x^2 + 6x + 2 = 0$ becomes $x^2 + 2x + \frac{2}{3} = 0$
Constant to be added: Always add the square of half the coefficient of $x$ to both sides. Example: For $x^2 + 8x = 5$ , add $\left(\frac{8}{2}\right)^2 = 16$ to both sides.
Final form: The equation will transform into the form $\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$ . Example: becomes

Special Cases and Variations

When the Discriminant is Negative

If $b^2 - 4ac < 0$ , then the equation has no real roots.

Concrete example: $x^2 + 2x + 2 = 0$

Completing the square:

x^2 + 2x + 1 = -2 + 1 \quad \text{(adding 1 to both sides)}

Since no real number has a square of $-1$ , this equation has no real roots.

For Incomplete Quadratic Equations

For equations of the form $ax^2 + c = 0$ , we don't need to complete the square.

Concrete example: $3x^2 - 12 = 0$

3x^2 = 12

The roots of the equation are $x = 2$ and $x = -2$ .

Practice Problems

Solve the following quadratic equations using the completing the square method:

$x^2 + 5x + 6 = 0$
$2x^2 + 6x + 3 = 0$
$6x^2 + 2x + \frac{1}{6} = 0$
$x^2 - 12x - 15 = 0$
$\frac{3}{2}x^2 - 8x - 6 = 0$

Answer Key

$x^2 + 5x + 6 = 0$
1. Move the constant to the right side:
  
  $x^2 + 5x = -6$
2. Add the square of half the coefficient of $x$ to both sides:
  $\text{Half the coefficient of } x = \frac{5}{2}$
3. Factor the left side into a perfect square:
  
  $\left(x + \frac{5}{2}\right)^2 = -6 + \frac{25}{4}$
4. Simplify the right side:
  $-6 + \frac{25}{4} = \frac{-24 + 25}{4} = \frac{1}{4}$
5. Take the square root of both sides:
  
  $x + \frac{5}{2} = \pm\frac{1}{2}$
6. Solve for the value of $x$ :
  $x + \frac{5}{2} = \frac{1}{2} \quad \text{or} \quad x + \frac{5}{2} = -\frac{1}{2}$
Therefore, the roots of the equation are $x = -2$ and $x = -3$ .
$2x^2 + 6x + 3 = 0$
1. Divide all terms by the coefficient $a = 2$ :
$6x^2 + 2x + \frac{1}{6} = 0$
1. Divide all terms by the coefficient :
$x^2 - 12x - 15 = 0$
1. Move the constant to the right side:
  
  $x^2 - 12x = 15$
$\frac{3}{2}x^2 - 8x - 6 = 0$
1. Divide all terms by the coefficient :

Nonreal Roots of Quadratic Equations

Types of Quadratic Equation Roots

Published: Apr 19, 2025. Updated: Sep 1, 2026.

x^2 + 3x + \frac{3}{2} = 0

x^2 + 3x = -\frac{3}{2}

x^2 + 3x = -\frac{3}{2}

Half of the coefficient of $x$ is $\frac{3}{2}$

The square of this value: $\left(\frac{3}{2}\right)^2 = \frac{9}{4}$

Add to both sides:

x^2 + 3x + \frac{9}{4} = -\frac{3}{2} + \frac{9}{4}

\left(x + \frac{b}{2}\right)^2

Example: From the equation $x^2 + 3x + \frac{9}{4} = -\frac{3}{2} + \frac{9}{4}$

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

-\frac{3}{2} + \frac{9}{4} = \frac{-6 + 9}{4} = \frac{3}{4}

So the equation becomes:

\left(x + \frac{3}{2}\right)^2 = \frac{3}{4}

x + \frac{3}{2} = \pm\sqrt{\frac{3}{4}} = \pm\frac{\sqrt{3}}{2}

For the positive sign:

x = \frac{\sqrt{3}}{2} - \frac{3}{2} = -\frac{3}{2} + \frac{\sqrt{3}}{2}

For the negative sign:

x = -\frac{\sqrt{3}}{2} - \frac{3}{2} = -\frac{3}{2} - \frac{\sqrt{3}}{2}

x^2 + 6x + 8 = 0

\left(x + 3\right)^2 = 1

x^2 + 5x + \frac{25}{4} = -6 + \frac{25}{4}

\left(x + \frac{5}{2}\right)^2 = \frac{1}{4}

x = \frac{1}{2} - \frac{5}{2} = -2 \quad \text{or} \quad x = -\frac{1}{2} - \frac{5}{2} = -3

x^2 + 3x + \frac{3}{2} = 0

Move the constant to the right side:

x^2 + 3x = -\frac{3}{2}

Add the square of half the coefficient of $x$ to both sides:

\text{Half the coefficient of } x = \frac{3}{2}

Factor the left side into a perfect square:

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

Simplify the right side:

-\frac{3}{2} + \frac{9}{4} = \frac{-6 + 9}{4} = \frac{3}{4}

Take the square root of both sides:

x + \frac{3}{2} = \pm\frac{\sqrt{3}}{2}

Solve for the value of $x$ :

x = -\frac{3}{2} + \frac{\sqrt{3}}{2} \quad \text{or} \quad x = -\frac{3}{2} - \frac{\sqrt{3}}{2}

Therefore, the roots of the equation are $x = -\frac{3}{2} + \frac{\sqrt{3}}{2}$ and $x = -\frac{3}{2} - \frac{\sqrt{3}}{2}$ .

a = 6

x^2 + \frac{1}{3}x + \frac{1}{36} = 0

Move the constant to the right side:

x^2 + \frac{1}{3}x = -\frac{1}{36}

Add the square of half the coefficient of $x$ to both sides:

\text{Half the coefficient of } x = \frac{1}{6}

Factor the left side into a perfect square:

\left(x + \frac{1}{6}\right)^2 = -\frac{1}{36} + \frac{1}{36}

Simplify the right side:

\left(x + \frac{1}{6}\right)^2 = 0

Take the square root of both sides:

x + \frac{1}{6} = 0

Solve for the value of $x$ :

x = -\frac{1}{6}

This equation has the repeated root $x = -\frac{1}{6}$ .

Add the square of half the coefficient of $x$ to both sides:

\text{Half the coefficient of } x = -6

Factor the left side into a perfect square:

(x - 6)^2 = 51

Take the square root of both sides:

x - 6 = \pm\sqrt{51}

Solve for the value of $x$ :

x = 6 \pm \sqrt{51}

The roots of the equation are $x = 6 + \sqrt{51}$ and $x = 6 - \sqrt{51}$ .

a = \frac{3}{2}

x^2 - \frac{16}{3}x - 4 = 0

Move the constant to the right side:

x^2 - \frac{16}{3}x = 4

Add the square of half the coefficient of $x$ to both sides:

\text{Half the coefficient of } x = -\frac{8}{3}

Factor the left side into a perfect square:

\left(x - \frac{8}{3}\right)^2 = 4 + \frac{64}{9}

Simplify the right side:

4 + \frac{64}{9} = \frac{36 + 64}{9} = \frac{100}{9}

Take the square root of both sides:

x - \frac{8}{3} = \pm\frac{10}{3}

Solve for the value of $x$ :

x = \frac{8}{3} + \frac{10}{3} = \frac{18}{3} = 6 \quad \text{or} \quad x = \frac{8}{3} - \frac{10}{3} = -\frac{2}{3}

The roots of the equation are $x = 6$ and $x = -\frac{2}{3}$ .