# Class 9 RD Sharma Solutions – Chapter 4 Algebraic Identities- Exercise 4.1 | Set 1

**Question 1.** **Evaluate each of the following using identities:**

**(i) (2x – 1/x) ^{2}**

**(ii) (2x + y) (2x – y)**

**(iii) (a**

^{2}b – b^{2}a)^{2}**(iv) (a – 0.1) (a + 0.1)**

**(v) (1.5.x**

^{2}– 0.3y^{2}) (1.5x^{2}+ 0.3y^{2})**Solution:**

i)(2x – 1/x)^{2}We know that, (a -b)

^{2}= a^{2}– 2ab + b^{2}So, (2x – 1/x)

^{2}= (2x)^{2}– (2 × 2x × 1/x) + (1/x)^{2}= 4x

^{2}+ 1/x^{2}– 4

ii)(2x + y) (2x – y)We know that, (a + b)(a – b) = a

^{2}– b^{2}So, (2x + y) (2x – y) = (2x)

^{2}– (y)^{2}= 4x

^{2}– y^{2}

iii)(a^{2}b – b^{2}a)^{2}We know that, (a -b)

^{2}= a^{2}– 2ab + b^{2}So, (a

^{2}b – b^{2}a)^{2}= (a^{2}b)^{2}– (2 × a^{2}b × b^{2}a) + (b^{2}a)^{2}= a

^{4}b^{2}+ b^{4}a^{2}– 2a^{3}b^{3}

iv)(a – 0.1) (a + 0.1)We know that, (a + b)(a – b) = a

^{2}– b^{2}So, (a – 0.1) (a + 0.1) = (a)

^{2}– (0.1)^{2}= a

^{2}– 0.01

v)(1.5.x^{2}– 0.3y^{2}) (1.5x^{2}+ 0.3y^{2})We know that, (a + b)(a – b) = a

^{2}– b^{2}So, (1.5.x

^{2}– 0.3y^{2}) (1.5.x^{2}+ 0.3y^{2}) = (1.5.x^{2})^{2}– (0.3y^{2})^{2}= 2.225x

^{4}– 0.09y^{4}

**Question 2. Evaluate each of the following using identities:**

**(i)(399) ^{2}**

**(ii)(0.98)**

^{2}**(iii)991 × 1009**

**(iv) 117 × 83**

**Solution:**

i)(399)^{2}We can write (399)

^{2}as (400 – 1)^{2}Also, (a -b)

^{2}= a^{2}– 2ab + b^{2}= (400)

^{2}+ (1)^{2}– 2 × 400 ×1= 160000 + 1 – 800 = 159201

Hence, (399)^{2}= 159201

ii)(0.98)^{2}We can write (0.98)

^{2}as (1 – 0.02)^{2}Also, (a -b)

^{2}= a^{2}– 2ab + b^{2}= (1)

^{2}+ (0.02)^{2}– 2 × 0.02 ×1= 1 + 0.0004 – 0.04 = 0.9604

Hence, (0.98)^{2}= 0.9604

iii)991 × 1009We can write 991 × 1009 as (1000 – 9)(1000 + 9)

Also, (a + b)(a – b) = a

^{2}– b^{2}= (1000)

^{2}– (9)^{2}= 1000000 – 81 = 999919

Hence, 991 × 1009 = 999919

iv)117 × 83We can write 117 × 83 as (100 + 17)(100 – 17)

Also, (a + b)(a – b) = a

^{2}– b^{2}= (100)

^{2}– (17)^{2}= 10000 – 289 = 9711

Hence, 117 × 83 = 9711

**Question 3. Simplify each of the following:**

**(i) 175 × 175 +2 × 175 × 25 + 25 × 25****(ii) 322 × 322 – 2 × 322 × 22 + 22 × 22****(iii) 0.76 × 0.76 + 2 × 0.76 × 0.24 + 0.24 × 0.24****(iv) (7.83 × 7.83 – 1.17 × 1.17)/6.66**

**Solution:**

i)175 × 175 +2 × 175 × 25 + 25 × 25It can be written as (175)

^{2}+ 2(175)(25) + (25)^{2}And we also know that, (a + b)

^{2}= a^{2}+ b^{2}+ 2abSo, we can conclude (175)

^{2}+ 2(175)(25) + (25)^{2}as (175 + 25)^{2}= (200)

^{2}= 40000

Hence, 175 × 175 +2 × 175 × 25 + 25 × 25 = 40000

ii)322 × 322 – 2 × 322 × 22 + 22 × 22It can be written as (322)

^{2}– 2(322)(22) + (22)^{2}And we also know that, (a – b)

^{2}= a^{2}+ b^{2}– 2abSo, we can conclude (322)

^{2}– 2(322)(22) + (22)^{2}as (322 – 22)^{2}= (300)

^{2}= 90000

Hence, 322 × 322 – 2 × 322 × 22 + 22 × 22 = 90000

iii)0.76 × 0.76 + 2 × 0.76 × 0.24 + 0.24 × 0.24It can be written as (0.76)

^{2}+ 2(0.76)(0.24) + (0.24)^{2}And we also know that, (a + b)

^{2}= a^{2}+ b^{2}+ 2abSo, we can conclude (0.76)

^{2}+ 2(0.76)(0.24) + (0.24)^{2}as (0.76 + 0.24)^{2}= (1.0)

^{2}= 1

Hence, 0.76 × 0.76 + 2 × 0.76 × 0.24 + 0.24 × 0.24 = 1

iv)(7.83 × 7.83 – 1.17 × 1.17)/6.66It can be written as (7.83

^{2}– 1.17^{2})/6.66And we also know that, (a + b)(a – b) = a

^{2}– b^{2}So, we can conclude (7.83

^{2}– 1.17^{2})/6.66 as [(7.83 + 1.17)(7.83 – 1.17)]/6.66= (9 × 6.66)/6.66 = 9

Hence, (7.83 × 7.83 – 1.17 × 1.17)/6.66 = 9

**Question 4.** **If x + 1/x = 11, find the value of x**^{2} +1/x^{2}

^{2}+1/x

^{2}

**Solution:**

Given, x + 1/x = 11

So, (x + 1/x)

^{2}= (x)^{2}+ (1/x)^{2}+ 2 × (x) × (1/x)We also know that, (a + b)

^{2}= a^{2}+ b^{2}+ 2abSo,

(11)

^{2}= x^{2}+ 1/x^{2}+ 2121 – 2 = x

^{2}+ 1/x^{2}x

^{2}+ 1/x^{2}= 119

Hence, value of x^{2}+ 1/x^{2}is 119

**Question 5. If x – 1/x = -1, find the value of x**^{2} +1/x^{2}

^{2}+1/x

^{2}

**Solution:**

Given, x – 1/x = -1

So, (x – 1/x)

^{2}= (x)^{2}+ (1/x)^{2}– 2 × (x) × (1/x)We also know that, (a – b)

^{2}= a^{2}+ b^{2}– 2abSo,

(-1)

^{2}= x^{2}+ 1/x^{2}– 21 + 2 = x

^{2}+ 1/x^{2}x

^{2}+ 1/x^{2}= 3

Hence, value of x^{2}– 1/x^{2}is 3

**Question 6. If x + 1/x = √5, find the value of x**^{2} +1/x^{2} and x^{4} +1/x^{4}

^{2}+1/x

^{2}and x

^{4}+1/x

^{4}

**Solution:**

Given, x + 1/x = √5

So, (x + 1/x)

^{2}= (x)^{2}+ (1/x)^{2}+ 2 × (x) × (1/x)We also know that, (a + b)

^{2}= a^{2}+ b^{2}+ 2abSo,

(√5)

^{2}= x^{2}+ 1/x^{2}+ 25 – 2 = x

^{2}+ 1/x^{2}x

^{2}+ 1/x^{2}= 3Now, taking the square of x

^{2}+ 1/x^{2}(x

^{2}+ 1/x^{2})^{2}= x^{4}+ 1/x^{4}+ 2 × (x)^{2}× (1/x^{2})(3)

^{2}= x^{4}+ 1/x^{4}+ 2x

^{4}+ 1/x^{4}= 7

Hence, value of x^{2}+ 1/x^{2}is 3 and that of x^{4}+1/x^{4}is 7

**Question 7. If x**^{2} +1/x^{2} = 66, find the value of x – 1/x

^{2}+1/x

^{2}= 66, find the value of x – 1/x

**Solution:**

Given, x

^{2}+1/x^{2}= 66Let 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}= a^{2}+ b^{2}– 2abSo,

(x – 1/x)

^{2}= 66 – 2(x – 1/x)

^{2}= 64x – 1/x = ±8

Hence, the value of x – 1/x is 8