The **Completing the square method **is one way to solve a quadratic equation. It requires a few more steps than some of the other methods (like the quadratic formula), but it forms a foundation for an integration method of the same name.

**Contents**:

## Completing the Square Method: Solving Quadratic Equations

**Example question:** Solve 2x^{2} + 12x – 18 = 0 by completing the square.

Step 1: Separate the constant term from the other terms by moving it to the right side of the equation. Adding 18 to both sides gives:

2x^{2} + 12x = 18.

Step 2: Remove any coefficient from x^{2}. For this example, we can get rid of the “2” by dividing:

This gives x^{2} + 6x = 9.

Step 3: Halve the x-coefficient, square it, then add the result to both sides. The x-coefficient is 6, so:

*Halving the x-coefficient*: ½ * 6 = 3*Squaring*: 3^{2}= 9*Adding to both sides*: x^{2}+ 6x + 9 = 18.

Step 4: Factor the left side of the equation from Step 3:

Factoring x^{2} + 6x + 9:

- (x + 3(x + 3) = 18
- (x + 3)
^{2}= 18

I skipped a few factoring steps here, but if you’re a bit rusty, you can view the intermediate steps on Symbolab.

Step 5: Solve for x:

- Take the square root of both sides:
- √[(x + 3)
^{2}] = ±√(18) - x + 3 = ±√18

- √[(x + 3)
- Subtract 3 from both sides: x = ±√18 – 3

The solution is x = ±√18 – 3, which simplifies to x = -3±3√2.

## Integration of Fractions

The Completing the Square Method is a method of last resort to deal with expressions that can’t be integrated using the “usual” rules like the power rule. Specifically, you use this method when you have a quadratic polynomial in the denominator and no variables in the numerator.

**Example question:** Find the integral by completing the square:

Step 1: Bring the constant in front and rewrite with dx at the end:

Step 2: Complete the square for the denominator only (following the above procedure for quadratic equations):

That’s it! Now the function is in a form where you can use the usual rules of integration. For this particular function, you can **solve it with u-substitution:**

- Substitute u = x – 1
- Integral Substitution u = 3√3 / √3 v

- Take out the constant
- Use the common integral ∫ 1/(v
^{2}+ 1) dv = arctan (v) - Substitute the original function back in and simplify

Hi. I am pretty sure that your example is worked incorrectly. When you complete the square for 3x^2 -3x+30, you would end up with (x-1/2)^2 rather than (x-1)^2.

You’re right! I made the correction. Thanks for catching that :)