NCERT Class 9 Ganita Manjari Chapter 04 Exploring Algebraic Identities PDF Download

Chapter 4: Exploring Algebraic Identities

4.1 Introduction

In earlier chapters, you learnt about linear polynomials and how they can be used to represent and solve real-life problems. You also studied linear equations and discovered how they describe relationships between quantities.

In this chapter, we will take the next step by exploring algebraic identities. These are special mathematical rules that not only make it easier to simplify complicated calculations but also help us work efficiently with algebraic expressions.

Let us begin by exploring a few simple patterns.

Example 1: Consider any three consecutive square numbers. For example, 1, 4, and 9. Add the smallest and the largest squares. Thus, 1 + 9 = 10. Then subtract twice the middle square from this sum. This leads to 10 - (2 × 4) = 10 - 8 = 2.

Now try the same process with another set of three consecutive square numbers. Say 9, 16, 25.

For example, consider the consecutive squares 25, 36, 49. Applying the same rule we get (25 + 49) - (2 × 36) = 74 - 72 = 2.

Repeat this process with other sets of three consecutive square numbers. The result always seems to be 2!

The pattern may look surprising, but soon we will uncover the reason behind it using algebra.

Think and Reflect

Try and find other patterns like this one. For example, you could consider 4 consecutive squares and see if you can find a pattern.

4.2 Visualising Identities

In this section we will revisit some algebraic identities that we have studied in earlier grades and try to visualise them using geometrical models. In particular, we will use squares and rectangles to represent terms.

Consider two line segments of lengths \( a \) and \( b \) units, respectively, and make a longer line segment of length \( (a + b) \) units as shown in Fig. 4.1.

[Figure 4.1: Line segments of lengths a and b units forming a segment of length (a + b) units, See in your textbook]

We can now construct a square of side \( (a + b) \) units and partition it into smaller squares and rectangles as shown in Fig. 4.2.

[Figure 4.2: Square of side (a + b) units, See in your textbook]

Observe that the area of the outer square is \( (a + b)^2 \). The area of the larger square inside the outer square is \( a^2 \) while the area of the smaller square is \( b^2 \). The areas of the two rectangles are \( ab \) each. Together they make the bigger square; hence we can conclude that

\[ (a + b)^2 = a^2 + 2ab + b^2. \]

From Fig. 4.2 it is clear that \( (a + b)^2 = a^2 + 2ab + b^2 \) for all \( a \) and \( b \) when \( a \) and \( b \) are lengths of line segments.

Think of numbers \( a \) and \( b \) where \( a \) and \( b \) do not represent lengths of line segments. What if \( a \) and \( b \) are negative numbers? Let us check for some negative numbers and see if this equation still works.

Teacher's Note

An algebraic identity works for all possible values, not just the ones shown in one picture. When you see the geometric model with positive lengths, remember that the same identity holds when you substitute negative numbers or fractions. Test it with your own numbers to build confidence.

Example 2: Let \( a = -2 \) and \( b = -3 \).

Then \( (a + b) = -5 \) and \( (a + b)^2 = 25 \).

Also \( a^2 = 4 \), \( b^2 = 9 \) and \( 2ab = 12 \).

Thus \( a^2 + 2ab + b^2 = 4 + 12 + 9 = 25 \).

Hence, \( a^2 + 2ab + b^2 = (a + b)^2 \) again!

Now suppose \( a \) and \( b \) are rational numbers, say \( a = -\frac{2}{3} \) and \( b = \frac{3}{4} \)

Then \( (a + b) = \left(-\frac{2}{3} + \frac{3}{4}\right) = \frac{1}{12} \).

\[ (a + b)^2 = \frac{1}{144}. \] \[ a^2 + 2ab + b^2 = \left(-\frac{2}{3}\right)^2 + 2\left(-\frac{2}{3}\right)\left(\frac{3}{4}\right) + \left(\frac{3}{4}\right)^2 \] \[ = \frac{4}{9} - 1 + \frac{9}{16} = \frac{64 - 144 + 81}{144} = \frac{145 - 144}{144} = \frac{1}{144}. \]

So, \( a^2 + 2ab + b^2 = (a + b)^2 \) seems to be true for rational numbers too. But we are still not sure if it is true for all numbers. To verify this, let us investigate further using the distributive property of numbers:

\[ (a + b)^2 = (a + b)(a + b) = a(a + b) + b(a + b) \] \[ = a^2 + ab + ba + b^2 = a^2 + 2ab + b^2. \]

Recall that in Grade 8 you were introduced to \( (a + b)^2 = a^2 + 2ab + b^2 \) as an identity.

Teacher's Note

The difference between an identity and an equation is crucial: an identity is always true no matter what you substitute, while an equation is only true for certain values. When you expand \( (a + b)^2 \) using the distributive property, you are proving the identity works everywhere, not just for the examples you tested.

What is the difference between an equation and an identity?

An algebraic identity is an equation that is true for all values of the variables occurring in it, while an equation need not be true for all values.

Key Points

• An algebraic identity is an equation that is true for all values of the variables in it, unlike a regular equation which may be true only for some values.
• The identity \( (a + b)^2 = a^2 + 2ab + b^2 \) can be visualized as a square divided into smaller regions, and it works for any numbers including negatives and fractions.
• You can verify an identity by testing specific values, but to prove it works everywhere, you must use algebraic properties like the distributive property.

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