Completing the Square Calculator
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
7to either side of the equation so that the left-hand side contains only terms with
x2 + 6x - 7 + 7 = 7
x2 + 6x = 7
Now it's time to complete the square! Take one half of the coefficient in front of
xand square it:
- the coefficient in front of
- one half of
- after squaring, we get
32 = 9
Add the number computed in Step 2 to both sides of the equation:
x2 + 6x + 9 = 7 + 9
x2 + 6x + 9 = 16
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 = xand
q = 3). We obtain:
(x + 3)2 = 16
Take the square root of both sides:
x + 3 = 4or
x + 3 = - 4
x = 1and
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?
- 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
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?
b=0, then you may skip Steps 2, 3, and 4, and go from
x2 = - c (Step 1) directly to
x = ±√|c| (Step 5).
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
cto the right-hand side:
x2 + bx = - c
x2is the area of the square with side
bxis the area of the rectangle of sides
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/2if
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).