# Completing The Square Method

**Completing the square method** is one of the methods to find the roots of the given quadratic equation. A polynomial equation with degree equal to two is known as a **quadratic equation**. ‘Quad’ means four but ‘Quadratic’ means ‘to make square’. A quadratic equation in its standard form is represented as:

ax^{2 }+ bx + c = 0, where a,b and c are real numbers such that a ≠ 0 and x is a variable.

Since the degree of the above-written equation is two; it will have two roots or solutions. The roots of polynomials are the values of x which satisfy the equation. There are several methods to find the roots of a quadratic equation. One of them is by completing the square.

## Completing the Square for Quadratic Equation

We have our quadratic equation as;

ax^{2 }+ bx + c = 0

The roots of the quadratic equation can also be evaluated by using the quadratic formula. For simplification, let us take a=1. The equation hence becomes

x^{2 }+ bx + c = 0

If we wanted to represent a quadratic equation using geometry, one way would be representing the terms of the expression in the L.H.S. of the equation by using geometric figures such as squares, rectangles etc. If we take a square with the side equal to x units, its area would be equal to x^{2} square units. This area will hence represent the first term of the expression. Similarly, a rectangle with its two sides as x units and b units will have the area equal to bx square units. And let us take a square with area equal to c square units to represent the last term of the expression. In the figure below, we have the geometrical equivalent of the expression x^{2}, bx and c.

**Geometrical equivalent of x ^{2},bx and c**

This method is known as completing the square. Let us complete some squares. If we break the rectangle representing bx into two equal parts cutting vertically, we will have two figures with an area of each equal to b/2 x square units. The figures are arranged accordingly in the second figure, below. Now, we have

x^{2 }+ bx + c = 0

**Rearranging the figures**

But our square is not complete yet. To complete the square, one square of side b/2 units is needed. This final part of the main square can be taken from the square with the area c square units. Cutting it out and putting it at the place, it results in below figure.

**Completing the Square**

The square is finally complete. The area of the square is equal to

(x+b/2)^{2} square units

The remaining area is equal to

(c-b^{2}/4) square units

All this time, we were rearranging the same figures that we had initially. It would hence be correct to say that

x^{2 }+ bx + c = (x+b/2)^{2 }+ (c-b^{2}/4)

This method is known as completing the square method. We have achieved it geometrically. We know that x^{2 }+ bx + c = 0. So,

(x+b/2)^{2 }+ (c-b^{2}/4) = 0

⇒ (x+b/2)^{2 } = -(c-b^{2}/4)

All the terms in the R.H.S. of the above equation are known. That’s why it is very easy to determine the roots. Let us look at some examples for better understanding.

**Steps for Completing the square method**

Suppose ax^{2 }+ bx + c = 0 is the given quadratic equation. Then follow the given steps to solve it by completing square method.

- Write the equation in the form, such that c is on the right side.
- If a is not equal to 1, then divide the complete equation by a, such that co-efficient of x
^{2 }is 1. - Now add the square of half of the co-efficient of term-x, (b/2a)
^{2}, on both the sides. - Factorize the left side of the equation as the square of the binomial term.
- Take the square root on both the sides
- Solve for variable x and find the roots.

### Examples

**Example 1: x ^{2 }+ 4x – 5b = 4; c = -5**

So, (x+4/2)^{2} = -(-5-4^{2}/4)

⇒ (x+2)^{2} = 9

⇒ (x+2) = ±√9 , where ± is read as ‘plus minus’

⇒ (x + 2 ) = ± 3

⇒ x = 1 , -5

**Example 2: 3x ^{2 }-5x+2 = 0**

The given equation is not in the form to which we apply method of completing squares, i.e. coefficient of x^{2}is not 1. To make it 1, we need to divide the whole equation with 3.

x^{2 }– 5/3 x + 2/3 = 0

b = -5/3; c = 2/3

So, \( \left( x~+~\frac{\left( – \frac 53 \right)}{2} \right)^2 \) = –\( \left( \frac 23~-~\frac{\left( – \frac 53 \right)^2}{4} \right) \)

⇒ (x-5/6)^{2} = 1/36

⇒ (x-5/6)= ± √(1/36)

⇒ x-5/6 = ±1/6

⇒ x = 1, -2/3