Theory:

When it comes to solving the algebraic equation, identities play a vital role to solve them. In this lesson, we are going to explore the geometrical proof of identities.
 
Let us first see the varies identities which we will discuss here:
 
1. (x+a)(x+b)=x2+x(a+b)+ab
 
2. \((a + b^2) = a^2 + 2ab + b^2\)
 
3. \((a - b^2) = a^2 - 2ab + b^2\)
 
4. (a + b)(a  b)= a2b2
 
Let us take each identity one by one and discuss the proof of that identity.
 
Identity \(1\): (x+a)(x+b)=x2+x(a+b)+ab.
 
Let us construct a rectangular diagram of four regions. One region is square-shaped with a dimension of \(4 × 4\) (Orange). Also, the other three regions are rectangle in shape with dimensions \(4×2\) (Blue), \(3 × 4\) (Yellow) and \(3 × 2\) (Green).
 
pic 6.png
 
By observing the above rectangle, we can notice that:
 
\(\text{Area of the bigger rectangle}\) \(=\) \(\text{Area of a square (Orange)}\) \(+\) \(\text{Area of three rectangles}\)
 
\((4 + 3)\) \((4 + 2)\) \(=\) \((4 × 4) + (4 × 2) + (3 × 4) + (3 × 2)\)
 
Now, we simplify the LHS and RHS of the above expression.
 
LHS \(=\) \((4 + 3)\) \((4 + 2)\) \(=\) \(7×6 = 42\)
 
RHS \(=\) \((4 × 4) + (4 × 2) + (3 × 4) + (3 × 2)\)
 
RHS \(=\) \(16+8+12+6 = 42\)
 
Therefore, LHS \(=\) RHS
 
Similarly, if we use the variables in this case instead of number we get:
pic 7.png
 
Let one side of a rectangle be \((x +a)\), and the other side be \((x + b)\) units.
 
Then, \(\text{the total area of the rectangle }AEGI\) \(=\) \(\text{length } \times \text{breadth}\) \(=\) \((x+a)(x+b)\)………….\((1)\)
 
\(\text{The area of the rectangle }AEGI\) \(=\) \(\text{The area of the square }ABCD\) \(+\) \(\text{The area of the rectangle }BEFD\) \(+\) \(\text{The area of the rectangle }DFGH\) \(+\) \(\text{The area of the rectangle }CDHI\).
 
\(\text{The area of the rectangle AEGI}\) \(=\) x2+ax+ab+bx
 
=x2+x(a+b)+ab……………..\((2)\)
 
From the equation, \((3)\) and \((4)\) we get (x+a)(x+b)=x2+x(a+b)+ab.
 
Therefore, (x+a)(x+b)=x2+x(a+b)+ab is a identity.
Example:
Simplify the expression (x+2)(x+9) using the identity (x+a)(x+b)=x2+x(a+b)+ab.

The expression is (x+2)(x+9).
 
Now write the given expression (x+2)(x+9) with respect to the given identity \((x+a)(x+b)\) \(=\) \(x^2+(a+b)x+ab\).
 
(x+a)(x+b)=(x+2)(x+9)=(x2+x(2+9)+2×9)
 
Now simplify the expression.
 
(x2+11x+18).
 
Therefore, (x+2)(x+9) \(=\) (x2+11x+18).