Completing the Square

Rewriting a Quadratic Equation as a 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 $(x + p)^2 = q$. This method is particularly useful for quadratic equations that are difficult to factor using regular factorization.

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:

1. 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 $2x^2 + 6x + 3 = 0$

   $$\frac{2x^2 + 6x + 3}{2} = \frac{0}{2}$$

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

2. Move the Constant Term to the Right Side

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

   Example: From the equation $x^2 + 3x + \frac{3}{2} = 0$

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

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

   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:

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

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

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

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

5. Simplify the Right Side

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

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

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

   So the equation becomes:

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

6. Take the Square Root of Both Sides

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

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

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

7. Isolate the Variable

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

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

   For the positive sign:

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

   For the negative sign:

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

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

Each equation below is solved by turning the left side into a perfect square. The first has a leading coefficient of one, and the later examples divide by the coefficient first.

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.

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

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

Step 5: Simplify the right side.

Step 6: Take the square root of both sides.

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

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 $a = 2$

Step 2: Move the constant to the right side

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

Step 4: Factor the left side into a perfect square

Step 5: Simplify the right side

Step 6: Take the square root of both sides

Step 7: Solve for the value of $x$

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

The two checks below cover the steps that are easiest to skip. Each one names the step and the value to verify.

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

2. 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.

3. Final form: The equation will transform into the form $\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$. Example: $x^2 + 6x + 8 = 0$ becomes $\left(x + 3\right)^2 = 1$

Special Cases and Variations

A negative discriminant makes the square root imaginary, while a perfect square leaves a single repeated solution. Each case changes the number of roots you should report.

Not every quadratic equation produces two real solutions.

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:

Add $1$ 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: $3x^2 - 12 = 0$

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

Practice Problems

Completing the square rewrites the equation until the left side becomes a perfect square. Write down each step, because the position of the vertex can be read off the result directly.

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

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

Answer Key

1. $x^2 + 5x + 6 = 0$

   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:

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

      $$\begin{aligned} &\text{The square of this value} \\ &= \left(\frac{5}{2}\right)^2 \\ &= \frac{25}{4} \end{aligned}$$

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

   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:

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

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

   5. Take the square root of both sides:

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

   6. Solve for the value of $x$:

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

      $$\begin{aligned} &x = \frac{1}{2} - \frac{5}{2} = -2 \\ &\text{or} \\ &x = -\frac{1}{2} - \frac{5}{2} = -3 \end{aligned}$$

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

2. $2x^2 + 6x + 3 = 0$

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

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

   2. Move the constant to the right side:

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

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

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

      $$\begin{aligned} &\text{The square of this value} \\ &= \left(\frac{3}{2}\right)^2 \\ &= \frac{9}{4} \end{aligned}$$

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

   4. Factor the left side into a perfect square:

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

   5. Simplify the right side:

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

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

   6. Take the square root of both sides:

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

   7. Solve for the value of $x$:

      $$\begin{aligned} &x = -\frac{3}{2} + \frac{\sqrt{3}}{2} \\ &\text{or} \\ &x = -\frac{3}{2} - \frac{\sqrt{3}}{2} \end{aligned}$$

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

3. $6x^2 + 2x + \frac{1}{6} = 0$

   1. Divide all terms by the coefficient $a = 6$:

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

   2. Move the constant to the right side:

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

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

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

      $$\begin{aligned} &\text{The square of this value} \\ &= \left(\frac{1}{6}\right)^2 \\ &= \frac{1}{36} \end{aligned}$$

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

   4. Factor the left side into a perfect square:

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

   5. Simplify the right side:

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

   6. Take the square root of both sides:

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

   7. Solve for the value of $x$:

      $$x = -\frac{1}{6}$$

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

4. $x^2 - 12x - 15 = 0$

   1. Move the constant to the right side:

      $$x^2 - 12x = 15$$

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

      Half the coefficient of $x$ is $-6$.

      $$\begin{aligned} &\text{The square of this value} \\ &= (-6)^2 \\ &= 36 \end{aligned}$$

      $$x^2 - 12x + 36 = 15 + 36$$

   3. Factor the left side into a perfect square:

      $$(x - 6)^2 = 51$$

   4. Take the square root of both sides:

      $$x - 6 = \pm\sqrt{51}$$

   5. Solve for the value of $x$:

      $$x = 6 \pm \sqrt{51}$$

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

5. $\frac{3}{2}x^2 - 8x - 6 = 0$

   1. Divide all terms by the coefficient $a = \frac{3}{2}$:

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

   2. Move the constant to the right side:

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

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

      Half the coefficient of $x$ is $-\frac83$.

      $$\begin{aligned} &\text{The square of this value} \\ &= \left(-\frac{8}{3}\right)^2 \\ &= \frac{64}{9} \end{aligned}$$

      $$x^2 - \frac{16}{3}x + \frac{64}{9} = 4 + \frac{64}{9}$$

   4. Factor the left side into a perfect square:

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

   5. Simplify the right side:

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

      $$\left(x - \frac{8}{3}\right)^2 = \frac{100}{9}$$

   6. Take the square root of both sides:

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

   7. Solve for the value of $x$:

      $$\begin{aligned} &x = \frac{8}{3} + \frac{10}{3} = \frac{18}{3} = 6 \\ &\text{or} \\ &x = \frac{8}{3} - \frac{10}{3} = -\frac{2}{3} \end{aligned}$$

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