A polynomial equation with degree equal to two is known as **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 an equation 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 method.

**Completing the Square: fundamentals**

We have our quadratic equation as

\( ax^2~+~bx~+~c\) =0

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 fig.1, we have the geometrical equivalent of the expression \(x^2\), bx and c.

Figure 1: 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 area of each equal to \( \frac b2 \) x square units. The figures are arranged accordingly in the fig. 2. Now, we have

\(x^2~+~bx~ and~ c \)

Figure 2: Rearranging the figures

But our square is not complete yet. To complete the** square**, one square of side \( \frac b2 \) 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 place, it results in fig. 3.

Figure 3: Completing the Square

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

\( \left( x~+~\frac b2 \right)^2 \) square units

The remaining area is equal to

\( \left( c~-~\frac {b^2}{4} \right) \) square units

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

\( x^2~+~bx~+~c\) = \( \left( x~+~\frac b2 \right)^2~+~\left( c~-~\frac{b^2}{4}\right)\)

This method is known as** completing the square method**. We have achieved it geometrically. We know that \(x^2~+~bx~+~c\) = 0. So,

\( \left( x~+~\frac b2 \right)^2~+~\left(c~-~\frac{b^2}{4}\right) \) = 0

⇒ \( \left( x~+~\frac b2 \right)^2 \) = \( – \left( c~-~ \frac {b^2}{4} \right) \)

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.

**Example 1:**** \( x^2~+~4x~-~5\)**

b = 4; c = -5

So, \( \left( x~+~\frac 42 \right)^2 \) = \( – \left( -5~-~\frac{4^2}{4}\right) \)

⇒ \((x~ +~ 2)^2\) = 9

⇒ \( (x~+~2)\) = ±\( \sqrt{9}\) , 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.

\( x^2~-~\frac 53 x~+~\frac 23\) = 0

b = \( – \frac 53\); \( c~=~\frac 23 \)

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

⇒ \( \left( x~-~\frac56 \right)^2 \) = \( \frac {1}{36} \)

⇒ \( \left( x~-~\frac56 \right) \) = ± \( \sqrt{\frac{1}{36}}\)

⇒ \( \left( x~-~\frac 56 \right) \) = ± \( \frac 16 \)

⇒ \( x \) = \( 1, -\frac 23 \)

The below video will help you visualize the concepts of quadratic equations

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