Class 9 Maths Chapter 4 Solutions

🟧 Square Tile

Represents a square with side x.

🟩 Rectangle Tile

Represents one variable tile.

🟨 Unit Tile

Represents one unit square.

Every algebraic identity, multiplication, factorisation and quadratic expression in Class 9 Maths Chapter 4 is created using different combinations of these three simple tiles.

Build an Expression

Collect the Algebra Tiles for the expression before arranging them into a rectangle.

Arrange the Algebra Tiles

Don’t guess the factors. Arrange the tiles into one complete rectangle and let the rectangle reveal the answer.

The Rectangle Reveals the Factors

Once every tile fits perfectly, you don’t have to guess the answer. Simply read the two side lengths of the rectangle.

Ask This First Question

Before solving any factorisation question, don’t start multiplying or guessing. Simply follow this decision guide to choose the correct method.

Always Check for a Common Factor First

This is the quickest method of factorisation. If every term has a common number or variable, take it outside first.

Factorise by Grouping

If there is no common factor and the expression has four or more terms, group the terms in pairs and look for a common factor again.

Split the Middle Term

Use this method when the expression has three terms, no common factor, and begins with a squared term.

Factorise Using Algebraic Identities

Some expressions already match a well-known identity. Instead of splitting the middle term, apply the identity directly.

Now You Know Which Method to Choose

Every factorisation question becomes easier when you ask the right question first instead of guessing the method.

⭐ Golden Order to Remember

What is a Rational Expression?

A Rational Expression is an algebraic fraction in which both the numerator and the denominator are algebraic expressions.

How to Simplify a Rational Expression

Always factorise first, then cancel the common factors. Never cancel individual terms.

Worked Example

Chapter 4 Revision Hub

Before solving NCERT questions, spend one minute revising the five most important ideas from this chapter.

Can You Answer These Questions?

If your answer is “Yes” to every question below, you’re ready to solve the NCERT exercises confidently.

🔍 First Recall the Previous Pattern

On the previous page, we observed a surprising pattern for three consecutive square numbers.

In both examples, we added the smallest and largest square numbers and then subtracted twice the middle square(s). The result was always the same.

Now NCERT asks us to investigate whether a similar pattern exists for four consecutive square numbers.

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

✍️ Solution

Instead of guessing the answer directly, let us try one example and carefully observe what happens.

Take four consecutive square numbers:

Now add the first and last square numbers.

Also add the two middle square numbers.

Now compare these two sums.

Try another set of four consecutive squares:

Again, we get the same answer.

🎯 Observation

For these examples, we observe the pattern:

So, for four consecutive square numbers, the difference between the sum of the first and last squares and the sum of the two middle squares is always 4.

1. What can you say about a and b if
   (a + b)² < a² + b²
2. What can you say about a and b if
   (a + b)² > a² + b²
3. When will
   (a + b)² = a² + b²
4. Did you observe that a + b and a² + b² are both positive? What term will decide which is larger? Use the expansion of
   (a + b)²
   to decide.

✍️ Solution

To answer all four questions, first recall the algebraic identity

Now compare this with

Notice that the only extra term is 2ab. Therefore, the comparison depends only on whether 2ab is positive, negative or zero.

① When will (a + b)² be less than a² + b²?

This is possible only when a and b have opposite signs.

② When will (a + b)² be greater than a² + b²?

This happens when a and b have the same sign (both positive or both negative).

③ When will they be equal?

This is possible only when a = 0 or b = 0.

④ Which term decides the comparison?

The deciding term is

If 2ab is positive, then (a + b)² is larger. If 2ab is negative, then a² + b² is larger. If 2ab = 0, both expressions are equal.

Solution

We already know the identity

Now replace b by −b everywhere.

Label the squares and rectangles in Fig. 4.4 so that it represents the identity

Solution

The large square has side (a + b + c), so its total area is (a + b + c)². Divide this square into smaller squares and rectangles as shown below.

Observe the Figure

• There are three squares: a², b² and c².
• There are two rectangles of area ab, so together they give 2ab.
• There are two rectangles of area bc, so together they give 2bc.
• There are two rectangles of area ac, so together they give 2ac.

Suppose 7x is split as 2x + 5x. Can a similar rectangular arrangement be formed? Consider other possibilities and check.

📚 Recall the Previous Figure

In the previous example, the algebra tiles were arranged by splitting 7x = 3x + 4x, which formed a rectangle of dimensions (x + 3) and (x + 4). Now, we shall investigate whether a similar rectangle can also be formed by splitting 7x = 2x + 5x.

🧩 Step 1: Split 7x as 2x + 5x

In the previous arrangement, the 7x tiles were divided into 3x and 4x. Now, split them differently as

Arrange the tiles again so that they form a complete rectangle without leaving any gaps.

📏 Step 2: Find the Dimensions of the Rectangle

From the new arrangement of algebra tiles, observe the sides of the rectangle carefully.

Since the figure is a rectangle, its area is equal to

🧮 Step 3: Find the Polynomial Represented by the Rectangle

The dimensions of the rectangle are (x + 2) and (x + 5). Therefore,

Now compare this result with the previous arrangement.

✅ Conclusion

Yes, a similar rectangular arrangement can be formed by splitting 7x = 2x + 5x. However, the rectangle now represents the polynomial x² + 7x + 10, which is different from the previous polynomial x² + 7x + 12. Hence, different ways of splitting the middle term can produce different rectangular arrangements and different polynomials.