Completing the Square

Completing the square is a way of rearranging quadratic equations from the general form ax² + bx + c = 0 to the vertex form a(x – h)² + k = 0. It is written as a(x + m)² + n, such that the left side is a perfect square trinomial.

Comparing this equation with the general form, we get:

• ${m=\dfrac{b}{2a}}$
• ${n=c-\dfrac{b^{2}}{4a}}$

It is a form used to solve quadratic equations that is also useful in graphing and analyzing a quadratic function and finding its minimum or maximum values. 

For example, the quadratic equation x² – 4x + 4 = 0 can also be written as (x – 2)². Although the two equations are equivalent, the form (x – 2)² is sometimes better to work with in some situations.

Deriving Quadratic Equations by Completing the Square

Let us try to understand the concept using the concept of geometry.

In a quadratic equation ax² + bx + c, we will arrange the expression in the form of a perfect square trinomial. Here, x comes twice, which makes it tough to solve. We will simplify the equation further using the concept of geometry as shown below:

We will start by dividing all terms by a, the coefficient of x²

Now, let us divide the equation into a square of side x and a rectangle of length ${\dfrac{b}{a}}$ and breadth x as shown:

Thus by adding ${\left( \dfrac{b}{2a}\right) ^{2}}$to the equation ${x^{2}+\dfrac{b}{a}x+\dfrac{c}{a}=0}$we can complete the square.

Solving Quadratic Equations by Completing the Square with Steps

Step 1: Dividing all terms by a, the coefficient of x²

${x^{2}+\dfrac{b}{a}x+\dfrac{c}{a}=0}$

Step 2: Rewriting the equation such that the constant term ‘c’ is the only term on the right side of the equation

=>${x^{2}+\dfrac{b}{a}x=-\dfrac{c}{a}}$

Step 3:Adding the square of half the coefficient of the x -term, ${\left( \dfrac{b}{2a}\right) ^{2}}$ to both sides of the equation.

Step 4:Factoring the left side as the square of the binomial

=>${x^{2}+\dfrac{b}{a}x+\left( \dfrac{b}{2a}\right) ^{2}=-\dfrac{c}{a}+\left( \dfrac{b}{2a}\right) ^{2}}$

= ${\left( x+\dfrac{b}{2a}\right) ^{2}= -\dfrac{c}{a}+\dfrac{b}{( 2a}) ^{2}}$

This equation is in the vertex form a(x – h)² + k. From here, we can find the vertex

Step 5: Taking the square root of both sides of the equation

=>${\sqrt{\left( x+\dfrac{b}{2a}\right) ^{2}}=\sqrt{-\dfrac{c}{a}+\begin{bmatrix} – \\ 20 \end{bmatrix}^{2}}}$

Step 6: Subtracting the number that remains on the left side of the equation to find x

=>${x+\dfrac{b}{2a}-\dfrac{b}{2a}=\sqrt{-\dfrac{c}{a}+\left( \dfrac{b}{2a}\right) ^{2}-}\dfrac{b}{2a}}$

=>${x=\sqrt{-\dfrac{c}{a}+\left( \dfrac{b}{2a}\right) ^{2}-}\dfrac{b}{2a}}$

By completing the square, let us solve the quadratic equation x² + 4x + 1 = 0.

Step 1: Here, this step is not needed as a = 1

Step 2:Moving the constant term to the right side of the equation

=>x² + 4x = -1

Step 3:: Adding the square of half the coefficient of the x -term to both sides of the equation.

${\left( \dfrac{4}{2}\right) ^{2}=\left( 2\right) ^{2}=4}$

=>x² + 4x + 4 = -1 + 4

Step 4:Factoring the left side as the square of the binomial

=>(x + 2)² = 3

This gives us the vertex of the parabola corresponding to the given equation as (-2, -3)

Step 5: Taking the square root on both sides of the equation

=>x + 2 = ±√3 = ±1.73

Step 6: Subtracting 2 from both sides of the equation

=> x = (-2 + √3, -2 – √3)

Solved Examples

A Shortcut Approach

Solving a quadratic equation of the form a(x + m)² + n, where a = 1

Expanding (x + m)² + n, we get x² + 2mx + m² + n

Now, if we compare a quadratic equation of the form ax² + bx + c with the above equation, we will obtain the value of m and n

Let us use the equation x² + 12x + 32 = 0

Comparing this equation with the equation x² + 2mx + m² + n, we get

2mx = 12x

=>m = 6

m² + n = 32

=> (6)² + n = 32

=> 36 + n = 32

=> n = -4 Thus, we get the same result (x + 6)² – 4 as obtained while solving the quadratic equation in solved example 1. The rest of the steps remains the same while solving the quadratic equation, and we will get the result x = (-4, -8)