# 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}= 79Let 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