We will solve the quadratic equation
ax2 + bx + c = 0
a
b
c

# Completing the Square Calculator

By Anna Szczepanek, PhD

This completing the square calculator solves any given quadratic equation in the form `ax2 + bx + c = 0` with help from the method known as... "completing the square" 😉 Below, we give a detailed example of how to complete the square and then explain geometrically the steps of completing the square. As a by-product, we will also derive the well-known quadratic formula.

## How does this completing the square calculator work?

You just need to enter the coefficients of the quadratic equation

`ax2 + bx + c = 0`

you want to solve. Remember that `a` can't be zero (otherwise, your equation is linear instead of quadratic).

Our completing the square calculator not only solves the equation, but also shows you the steps of how to complete the square.

Go to the advanced options of this completing the square calculator if you want to:

• allow complex roots (by default, only real roots are displayed).
• increase the precision with which non-integer coefficients are computed.

## How do you complete the square: a step-by-step example

Let's take the equation

` x2 + 6x - 7 = 0 `
and solve it by completing the square. We break the process into several simple steps, so that nobody gets overwhelmed by the completing the square formula:
1. Add `7` to either side of the equation so that the left-hand side contains only terms with `x`:

` x2 + 6x - 7 + 7 = 7 `
`x2 + 6x = 7 `
2. Now it's time to complete the square! Take one half of the coefficient in front of `x` and square it:

• the coefficient in front of `x` is `6`
• one half of `6` is `3`
• after squaring, we get `32 = 9`
1. Add the number computed in Step 2 to both sides of the equation:

`x2 + 6x + 9 = 7 + 9`
`x2 + 6x + 9 = 16`
2. Recall the short multiplication formula, `(p + q)2 = p2 + 2pq + q2`, and note that we may apply it 'backwards' to the left-hand side of our equation (with `p = x` and `q = 3`). We obtain:

`(x + 3)2 = 16`
3. Take the square root of both sides:

`x + 3 = 4`   or   `x + 3 = - 4`

which gives:

`x = 1`   and   `x = - 7`

Hence, we've found the solutions of `x2 + 6x - 7 = 0`.

This means we've determined the points where the parabola `y = x2 + 6x - 7` intersects the x-axis. Maybe you want to calculate the other properties of this parabola?

## Special cases

• What if `a ≠ 1`?

In the example above it is important that the coefficient in front of `x2` is equal to `1`. What do you do if you are asked to solve a quadratic equation where `a ≠ 1`? Just apply one of the most frequently used problem-solving techniques in math, namely... transform the problem to one you've already solved 🙃 What does that mean in our context? Just divide your equation by `a`!

For example: assume we have to solve

`2x2 + 12x - 5 = 0`

by completing the square. We see that `a = 2`. Dividing either side by `2`, we obtain

`x2 + 6x - 2.5 = 0`

and so the coefficient in front of `x2` is equal to `1`. Now just go ahead with the steps we explained in the example above.

• What if `b = 0`?

If `b=0`, then you may skip Steps 2, 3, and 4, and go from `x2 = - c` (Step 1) directly to `x = ±√|c|` (Step 5).

Why?

Because in Step 2 we take `b` and perform some arithmetic operations on it, which gives us the number with which we will 'complete the square.' Note that if `b = 0`, then `(b/2)2 = 0`, and so we would add `0` to both sides of the equations. This is, of course, redundant.

## Quadratic formula via completing the square

Let's go once again through the completing the square steps. This time we apply it to the general equation `x2 + bx + c = 0`

• Note that we've already made sure that `a = 1`

• Move `c` to the right-hand side: `x2 + bx = - c `

• `x2` is the area of the square with side `x`

• `bx` is the area of the rectangle of sides `x` and `b`

• Let the magic happen: Completing the square
(by Krishnavedala - Own work, CC BY-SA 3.0 wikimedia.org / Cropped)

As a consequence, we have

`(x + b/2)2 = -c + b2/4`.

Taking the square root and moving `b/2` to the right-hand side, we get:

• `x = -b/2 ± √(-c + b2/4)` if ` b2/4 > c`; in such a case the equation has two distinct real roots.

• `x = -b/2` if ` b2/4 = c`: the equation has one real root.

• the equation has no real roots if ` b2/4 < c`. If we allow complex roots, they are given by `x = -b/2 ± i √(c - b2/4)`.

Anna Szczepanek, PhD