Class 9 RD Sharma Solutions Chapter 4 Algebraic Identities Exercise 4.1 | Set 2 (original) (raw)

Last Updated : 23 Jul, 2025

This Exercise question and its concept are repeatedly tested in JEE Mains with slightly difficult twists in the questions and also solid 3-mark questions in the Final Term Examination of Class 9. Check out the sample question paper CBSE and Previous JEE Mains paper for more such questions.

**Read More: **Algebraic Identities

**Question 8. If x 2 +1/x 2 = 79, find the value of x +1/x

**Solution:

We are given the sum of square of x and its reciprocal to be 79. We are asked to just find sum of x and its reciprocal. The first intuition is to square the asked quantity to substitute the value (that is 79) in place of the squared terms. On solving, we could get the value for **x + 1/x

Given, x2 +1/x2 = 79

Let us take the square of x + 1/x

So, (x + 1/x)2 = (x)2 + (1/x)2 + 2 × (x) × (1/x)

Since, (a + b)2 = a2 + b2 + 2ab

So,

(x + 1/x)2 = 79 + 2

(x + 1/x)2 = 81

x + 1/x = ±9

**Hence, the value of x + 1/x is ±9

**Question 9. If 9x 2 + 25y 2 = 181 and xy = -6, find the value of 3x + 5y

**Solution:

**Method 1:

**Assume possible values for xy=-6, (Always start with easier term in this type of question). 2 possibilities,

On further solving, we could arrive at the solution but its time taking approach.

**Method 2:

Start by squaring, the asked quantity. That is, 3x+5y. By this method, we could eliminate squared term with value 181, and hence find the value of (3x+5y)2. Then take root on both sides to get value of 3x+5y. We get 2 values as square root of square results in both + and - term.

Given,

9x2 + 25y2 = 181 and xy = -6

Let us take a square of 3x + 5y

(3x + 5y)2 = (3x)2 + (5y)2 + 2 × (3x) × (5y)

Since, (a + b)2 = a2 + b2 + 2ab

So,

(3x + 5y)2 = 9x2 + 25y2 + 30xy

(3x + 5y)2 = 181 + 30(-6)

(3x + 5y)2 = 181 – 180 = 1

So, 3x + 5y = ±1

**Hence, the value of 3x + 5y is ±1

**Question 10. If 2x + 3y = 8 and xy = 2, find the value of 4x 2 + 9y 2

**Solution:

**Follow the previous question method.

Given,

2x + 3y = 8 and xy = 2

Let us take a square of 2x + 3y

(2x + 3y)2 = (2x)2 + (3y)2 + 2 × (2x) × (3y)

Since, (a + b)2 = a2 + b2 + 2ab

So,

(2x + 3y)2 = 4x2 + 9y2 + 12xy

(8)2 = 4x2 + 9y2 + 12(2)

64 = 4x2 + 9y2 + 24

4x2 + 9y2 = 40

**Hence, the value of 4x 2 + 9y 2 **is 40

**Question 11. If 3x -7y = 10 and xy = -1, find the value of 9x 2 + 49y 2

**Solution:

**Similar method as previous question, again one of the very important 3 mark question in CBSE Class 9th Term Examinations.

Given,

3x -7y = 10 and xy = -1

Let us take a square of 3x -7y

(3x -7y)2 = (3x)2 + (7y)2 - 2 × (3x) × (7y)

Since, (a - b)2 = a2 + b2 - 2ab

So,

(3x -7y)2 = 9x2 + 49y2 - 42xy

(10)2 = 9x2 + 49y2 - 42(-1)

100 = 9x2 + 49y2 + 42

9x2 + 49y2 = 58

**Hence, the value of 9x 2 + 49y 2 is 58

**Question 12. Simplify each of the following products:

****(i) (1/2a - 3b)(3b +1/2a)(1/4a** 2 + 9b 2 )
****(ii) (m +n/7)** 3 (m-n/7)
****(iii) (x/2 -2/5)(2/5 -x/2) -x** 2 + 2x
****(iv) (x** 2 + x -2)(x 2 -x + 2)
****(v) (x** 3 -3x 2 -x)(x 2 -3x + 1)
****(vi) (2x** 4 -4x 2 +1)(2x 4 -4x 2 -1)

**Solution:

**Direct formula question,

**Apply formulas like

  1. a2 - b2 = (a+b)(a-b)
  2. (a-b)2 = a2 + b2 -2ab

**i) (1/2a -3b)(3b +1/2a)(1/4a2 + 9b2)

The above expression can be written as, (1/2a -3b)(1/2a +3b)(1/4a2 + 9b2)

We know that, (a + b)(a – b) = a2 – b2

So, [(1/2a)2 -(3b)2](1/4a2 + 9b2) = (1/4a2 -9b2)(1/4a2 + 9b2)

We now conclude it as, (1/4a2)2 - (9b2)2

= 1/16a4 -81b4

**Hence, (1/2a -3b)(3b +1/2a)(1/4a 2 + 9b 2 ) = 1/16a 4 -81b 4

**ii) (m +n/7)3 (m-n/7)

The above expression can be written as, (m +n/7)2 (m +n/7)(m-n/7)

We know that, (a + b)(a – b) = a2 – b2

So, (m +n/7)2 [(m)2 -(n/7)2]

= (m +n/7)2 (m2 -n2/49)

**Hence, (m +n/7) 3 (m-n/7) = (m +n/7) 2 (m 2 -n 2 /49)

**iii) (x/2 -2/5)(2/5 -x/2) -x2 + 2x

The above expression can be written as, [-(x/2 -2/5)(x/2 -2/5)] -x2 + 2x

= [-(x/2 -2/5)2] -x2 + 2x

We know that, (a -b)2 = a2 – 2ab + b2

So, -(x2/4 + 4/25 -2x/5) -x2 + 2x

= -x2/4 -4/25 + 2x/5 -x2 + 2x = -5x2/4 + 12x/5 -4/25

**Hence, (x/2 -2/5)(2/5 -x/2) -x 2 + 2x = -5x 2 /4 + 12x/5 -4/25

**iv) (x2 + x -2)(x2 -x + 2)

The above expression can be written as, [x2 + (x -2)][x2 -(x -2)]

We know that, (a + b)(a – b) = a2 – b2

So, (x2)2 -(x -2)2

= x4 -(x2 + 4 -4x)

= x4 -x2 -4 + 4x

**Hence, (x 2 + x -2)(x 2 -x + 2) = x 4 -x 2 -4 + 4x

**v) (x3 -3x2 -x)(x2 -3x + 1)

The above expression can be written as, x(x2 -3x -1)(x2 -3x + 1)

= x[(x2 -3x) -1][(x2 -3x) + 1]

We know that, (a + b)(a – b) = a2 – b2

= x [(x2 -3x)2 -12]

= x[x4 + 9x2 -6x3 -1] = x5 + 9x3 -6x4 -x

**Hence, (x 3 -3x 2 -x)(x 2 -3x + 1) = x 5 + 9x 3 -6x 4 -x

**vi) (2x4 -4x2 +1)(2x4 -4x2 -1)

The above expression can be written as, [(2x4 -4x2)+1][(2x4 -4x2) -1]

We know that, (a + b)(a – b) = a2 – b2

So, (2x4 -4x2)2 -12

= 4x8 + 16x4 -16x6 -1

**Hence, (2x 4 -4x 2 +1)(2x 4 -4x 2 -1) = 4x 8 + 16x 4 -16x 6 -1

**Question 13. Prove that a 2 + b 2 + c 2 – ab – bc – ca is always non-negative for all values of a, b and c.

**Solution:

In order to prove the given expression,

First Multiply and Divide a2 + b2 + c2 – ab – bc – ca by 2

= 1/2[2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca]

= 1/2[a2 + b2 + c2 +a2 + b2 + c2 – 2ab – 2bc – 2ca]

= 1/2[a2 + b2 -2ab + b2 + c2 -2bc + c2 +a2 -2ac]

= 1/2[(a -b)2 + (b -c)2 + (c -a)2]

As we can clearly see the above expression is sum of square terms which will always be positive.

**Hence, **a 2 + b 2 + c 2 – ab – bc – ca is always non-negative for all values of a, b and c is proved