Solving by completing the square

Completing the square involves collecting the \( x^2\) and \( x\) terms together in a bracket which can be quicker than using the quadratic formula to solve a quadratic equation.

The method of completing the square can be streamlined by using the formula:

\[ x^2 + bx = \left(x - \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 \]

This formula allows you to rewrite the quadratic equation into a form that is easier to solve while keeping the equation balanced.

1. Steps to Solve by Completing the Square:

To solve a quadratic equation using this formula, follow these steps:

1. Ensure the coefficient of \( x^2 \) is 1. If it is not, divide the entire equation by \( a \).
2. Apply the formula: \[ x^2 + bx = \left(x - \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 \]
3. Substitute this into the original equation and simplify.
4. Solve for \( x \) by isolating it using square roots and simplifying.

2. Example: Solve \( x^2 + 6x + 5 = 0 \):

Step 1: Ensure the coefficient of \( x^2 \) is 1:

The coefficient of \( x^2 \) is already 1, so no adjustment is needed.

Step 2: Apply the formula:

For \( x^2 + 6x \), the formula gives: \[ x^2 + 6x = \left(x - \frac{6}{2}\right)^2 - \left(\frac{6}{2}\right)^2 \] \[ x^2 + 6x = (x - 3)^2 - 9 \]

Step 3: Substitute into the equation:

Replace \( x^2 + 6x \) in the original equation \( x^2 + 6x + 5 = 0 \): \[ (x - 3)^2 - 9 + 5 = 0 \]

Simplify: \[ (x - 3)^2 - 4 = 0 \]

Step 4: Solve for \( x \):

Isolate the perfect square: \[ (x - 3)^2 = 4 \]

Take the square root of both sides: \[ x - 3 = \pm \sqrt{4} \] \[ x - 3 = \pm 2 \]

Isolate \( x \): \[ x = 3 + 2 \quad \text{or} \quad x = 3 - 2 \] \[ x = 5 \quad \text{or} \quad x = 1 \]

3. Special Cases and Notes:

• If the coefficient of \( x^2 \) is not 1, divide the entire equation by \( a \) before applying the formula.
• The method is particularly useful because it avoids the need to manually calculate \( \left(\frac{b}{2}\right)^2 \) separately.
• If the square root results in a negative number, the roots are complex.