Completing the Square

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

Remember that a perfect square trinomial follows the pattern $x^2 + 2px + p^2 = (x + p)^2$. We use this pattern to transform quadratic equations into a more solvable form.

Why Use This Method?

Not all quadratic equations are easily factored. For example, the equation $x^2 + 4x + 2 = 0$ cannot be easily factored using rational numbers because there are no two numbers that multiply to give 2 and add up to 4.

In such cases, the completing the square method becomes an effective choice for finding the roots of the equation.

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 x² 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 $2x^2 + 6x + 3 = 0$

   $$2x^2 + 6x + 3 = 0 \div 2$$

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

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 Square of Half the Coefficient of x 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

   Perform calculations on the right side to get a simpler form.

   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. Solve for the Value of x

   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:

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

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

Complete Solution Examples

Equation with Coefficient of x² = 1

Let's 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 $x$.

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

Equation with Coefficient of x² ≠ 1

Let's 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 $x$

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

Important Points in Completing the Square

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

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:

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: $3x^2 - 12 = 0$

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

Practice Problems

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:

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

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

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

   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 $x$:

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

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

   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:

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

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

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

   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 $x$:

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

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

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:

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

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

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

   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 $x$:

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

   Therefore, this equation has one (double) root, which is $x = -\frac{1}{6}$.

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:

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

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

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

   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 $x$:

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

   Therefore, 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:

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

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

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

   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 $x$:

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

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