# Class 9 RD Sharma Solutions – Chapter 4 Algebraic Identities- Exercise 4.1 | Set 2

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

^{2}+1/x

^{2}= 79, find the value of x +1/x

**Solution:**

Given, x

^{2}+1/x^{2}= 79Attention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the

Demo Class for First Step to Coding Course,specificallydesigned for students of class 8 to 12.The students will get to learn more about the world of programming in these

free classeswhich will definitely help them in making a wise career choice in the future.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}= a^{2}+ b^{2}+ 2abSo,

(x + 1/x)

^{2}= 79 + 2(x + 1/x)

^{2}= 81x + 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

^{2}+ 25y

^{2}= 181 and xy = -6, find the value of 3x + 5y

**Solution:**

Given,

9x

^{2}+ 25y^{2}= 181 and xy = -6Let us take a square of 3x + 5y

(3x + 5y)

^{2}= (3x)^{2}+ (5y)^{2}+ 2 × (3x) × (5y)Since, (a + b)

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

(3x + 5y)

^{2}= 9x^{2}+ 25y^{2}+ 30xy(3x + 5y)

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

^{2}= 181 – 180 = 1So, 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}

^{2}+ 9y

^{2}

**Solution:**

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

(2x + 3y)

^{2}= 4x^{2}+ 9y^{2}+ 12xy(8)

^{2}= 4x^{2}+ 9y^{2}+ 12(2)64 = 4x

^{2}+ 9y^{2}+ 244x

^{2}+ 9y^{2}= 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}

^{2}+ 49y

^{2}

**Solution:**

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}= a^{2}+ b^{2}– 2abSo,

(3x -7y)

^{2}= 9x^{2}+ 49y^{2}– 42xy(10)

^{2}= 9x^{2}+ 49y^{2}– 42(-1)100 = 9x

^{2}+ 49y^{2}+ 429x

^{2}+ 49y^{2}= 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:**

i)(1/2a -3b)(3b +1/2a)(1/4a^{2}+ 9b^{2})The above expression can be written as, (1/2a -3b)(1/2a +3b)(1/4a

^{2}+ 9b^{2})We know that, (a + b)(a – b) = a

^{2}– b^{2}So, [(1/2a)

^{2}-(3b)^{2}](1/4a^{2}+ 9b^{2}) = (1/4a^{2}-9b^{2})(1/4a^{2}+ 9b^{2})We now conclude it as, (1/4a

^{2})^{2}– (9b^{2})^{2}= 1/16a

^{4}-81b^{4}

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) = a

^{2}– b^{2}So, (m +n/7)

^{2}[(m)^{2}-(n/7)^{2}]= (m +n/7)

^{2}(m^{2}-n^{2}/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) -x^{2}+ 2xThe above expression can be written as, [-(x/2 -2/5)(x/2 -2/5)] -x

^{2}+ 2x= [-(x/2 -2/5)

^{2}] -x^{2}+ 2xWe know that, (a -b)

^{2}= a^{2}– 2ab + b^{2}So, -(x

^{2}/4 + 4/25 -2x/5) -x^{2}+ 2x= -x

^{2}/4 -4/25 + 2x/5 -x^{2}+ 2x = -5x^{2}/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)(x^{2}+ x -2)(x^{2}-x + 2)The above expression can be written as, [x

^{2}+ (x -2)][x^{2}-(x -2)]We know that, (a + b)(a – b) = a

^{2}– b^{2}So, (x

^{2})^{2}-(x -2)^{2}= x

^{4}-(x^{2}+ 4 -4x)= x

^{4}-x^{2}-4 + 4x

Hence, (x^{2}+ x -2)(x^{2}-x + 2) = x^{4}-x^{2}-4 + 4x

v)(x^{3}-3x^{2}-x)(x^{2}-3x + 1)The above expression can be written as, x(x

^{2}-3x -1)(x^{2}-3x + 1)= x[(x

^{2}-3x) -1][(x^{2}-3x) + 1]We know that, (a + b)(a – b) = a

^{2}– b^{2}= x [(x

^{2}-3x)^{2}-1^{2}]= x[x

^{4}+ 9x^{2}-6x^{3}-1] = x^{5}+ 9x^{3}-6x^{4}-x

Hence, (x^{3}-3x^{2}-x)(x^{2}-3x + 1) = x^{5}+ 9x^{3}-6x^{4}-x

vi)(2x^{4}-4x^{2}+1)(2x^{4}-4x^{2}-1)The above expression can be written as, [(2x

^{4}-4x^{2})+1][(2x^{4}-4x^{2}) -1]We know that, (a + b)(a – b) = a

^{2}– b^{2}So, (2x

^{4}-4x^{2})^{2}-1^{2}= 4x

^{8}+ 16x^{4}-16x^{6}-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.

^{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 a

^{2}+ b^{2}+ c^{2}– ab – bc – ca by 2= 1/2[2a

^{2}+ 2b^{2}+ 2c^{2}– 2ab – 2bc – 2ca]= 1/2[a

^{2}+ b^{2}+ c^{2}+a^{2}+ b^{2}+ c^{2 }– 2ab – 2bc – 2ca]= 1/2[a

^{2}+ b^{2}-2ab + b^{2}+ c^{2}-2bc + c^{2}+a^{2}-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