### Chapter 6 Algebraic Expressions And Identities – Exercise 6.6 | Set 1

**Question 11. If x – y = 7 and xy = 9, find the value of x**^{2}+y^{2}

^{2}+y

^{2}

**Solution:**

Given in the question x – y = 7 and x y = 9

By squaring on both sides

(x – y)

^{2}= 72x

^{2}+ y^{2}– 2xy = 49x

^{2}+ y^{2}– 2 (9) = 49 (since x y=9)x

^{2}+ y^{2}– 18 = 49x

^{2}+ y^{2}= 49 + 18x

^{2}+ y^{2}=67

**Question 12. If 3x + 5y = 11 and x y = 2, find the value of 9x**^{2} + 25y^{2}

^{2}+ 25y

^{2}

**Solution:**

Given in the question 3x + 5y = 11 and x y = 2

on squaring on both sides

(3x + 5y)

^{2}= 112(3x)

^{2}+ (5y)^{2}+ 2(3x)(5y) = 1219x

^{2}+ 25y^{2}+ 2 (15xy) = 121 (given x y=2)9x

^{2}+ 25y^{2}+ 2(15(2)) = 1219x

^{2}+ 25y^{2}+ 60 = 1219x

^{2}+ 25y^{2}= 121-609x

^{2}+ 25y^{2}= 61

**Question 13. Find the values of the following expressions:**

**(i) 16x**^{2} + 24x + 9 when x = 7/4

^{2}+ 24x + 9 when x = 7/4

We will using the formula (a + b)

^{2}= a^{2}+ b^{2}+ 2ab=(4x)

^{2}+ 2 (4x) (3) + 3^{2}=(4x + 3)

^{2}Putting when x = 7/4

=[4 (7/4) + 3]

^{2}=(7 + 3)

^{2}=100

**(ii) 64x**^{2} + 81y^{2} + 144xy when x = 11 and y = 4/3

^{2}+ 81y

^{2}+ 144xy when x = 11 and y = 4/3

We will using the formula (a + b)

^{2}= a^{2}+ b^{2}+ 2ab=(8x)

^{2}+ 2 (8x) (9y) + (9y)^{2}(8x + 9y)Putting when x = 11 and y = 4/3

=[8 (11) + 9 (4/3)]

^{2}=(88 + 12)

^{2}=(100)

^{2}=10000

**(iii) 81x**^{2} + 16y^{2} – 72xy when x = 2/3 and y = ¾

^{2}+ 16y

^{2}– 72xy when x = 2/3 and y = ¾

We will using the formula (a + b)

^{2}= a^{2}+ b^{2}+ 2ab=(9x)

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

^{2}Putting x = 2/3 and y = 3/4

=[9 (2/3) – 4 (3/4)]

^{2}=(6 – 3)

^{2}=3

^{2}=9

**Question 14. If x + 1/x = 9 find the value of x4 + 1/ x4.**

**Solution:**

Given in the question x + 1/x = 9

squaring both sides

(x + 1/x)

^{2}= (9)^{2}x

^{2}+ 2 × x × 1/x + (1/x)^{2}= 81x

^{2}+ 2 + 1/x^{2}= 81x

^{2}+ 1/x^{2}= 81 – 2x

^{2}+ 1/x^{2}= 79Now again when we square on both sides

(x

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

^{4}+ 2 × x^{2}× 1/x^{2}+ (1/x^{2})^{2}= 6241x

^{4}+ 2 + 1/x^{4}= 6241x

^{4}+ 1/x^{4}= 6241- 2x

^{4}+ 1/x^{4}= 6239

**Question 15. If x + 1/x = 12 find the value of x – 1/x.**

**Solution:**

Given in the question x + 1/x = 12

When squaring both sides

(x + 1/x)

^{2}= (12)^{2}x

^{2}+ 2 × x × 1/x + (1/x)^{2}= 144x

^{2}+ 2 + 1/x^{2}= 144x

^{2}+ 1/x^{2}= 144 – 2x

^{2}+ 1/x^{2}= 142When subtracting 2 from both sides

x

^{2}+ 1/x^{2}– 2 × x × 1/x = 142 – 2(x – 1/x)

^{2}= 140x – 1/x = √140

**Question 16. If 2x + 3y = 14 and 2x – 3y = 2, find value of xy. [Hint: Use (2x+3y)**^{2} – (2x-3y)^{2} = 24xy]

^{2}– (2x-3y)

^{2}= 24xy]

**Solution:**

2x + 3y = 14 equation (1)

2x – 3y = 2 equation (2)

Now square both the equations and subtract equation (2) from equation (1)

(2x + 3y)

^{2}– (2x – 3y)^{2}= (14)^{2}– (2)^{2}4×2 + 9y2 + 12xy – 4×2 – 9y2 + 12xy = 196 – 4

24 xy = 192

xy = 8

**Question 17. If x**^{2} + y^{2} = 29 and xy = 2, find the value of

^{2}+ y

^{2}= 29 and xy = 2, find the value of

**(i) x + y**

**(ii) x – y**

**(iii) x ^{4} + y^{4}**

**Solution:**

(i) x + y

x

^{2}+ y^{2}= 29 (Given in the question)x

^{2}+ y^{2}+ 2xy – 2xy = 29(x + y)

^{2}– 2 (2) = 29(x + y)

^{2}= 29 + 4x + y = ± √33

(ii) x – y

x

^{2}+ y^{2}= 29x

^{2}+ y^{2}+ 2xy – 2xy = 29(x – y)

^{2}+ 2 (2) = 29(x – y)

^{2}+ 4 = 29(x – y)

^{2}= 25(x – y) = ± 5

(iii) x

^{4}+ y^{4}x

^{2}+ y^{2}= 29Squaring both sides

(x

^{2}+ y^{2})^{2}= (29)^{2}x

^{4}+ y^{4}+ 2x^{2}y^{2}= 841x

^{4}+ y^{4}+ 2 (2)^{2}= 841x

^{4}+ y^{4}= 841 – 8= 833

**Question 18. What must be added each of the following expression to make it a whole square?**

**(i) 4x**^{2} – 12x + 7

^{2}– 12x + 7

(2x)

^{2}– 2 (2x) (3) + 32 – 32 + 7(2x – 3)

^{2}– 9 + 7(2x – 3)

^{2}– 2

**(ii) 4x**^{2} – 20x + 20

^{2}– 20x + 20

(2x)

^{2}– 2 (2x) (5) + 52 – 52 + 20(2x – 5)

^{2}– 25 + 20(2x – 5)

^{2}– 5

**Question 19. Simplify:**

**i) (x – y) (x + y) (x**^{2} + y^{2}) (x^{4} + y^{4})

^{2}+ y

^{2}) (x

^{4}+ y

^{4})

(x

^{2}– y^{2}) (x^{2}+ y^{2}) (x^{4}+ y^{4})[(x

^{2})^{2}– (y^{2})^{2}] (x^{4}+ y^{4})(x

^{4}– y^{4}) (x^{4}– y^{4})[(x

^{4})^{2}– (y^{4})^{2}]x

^{8}– y^{8}

**(ii) (2x – 1) (2x + 1) (4x**^{2} + 1) (16x^{4} + 1)

^{2}+ 1) (16x

^{4}+ 1)

[(2x)

^{2}– (1)^{2}] (4x^{2}+ 1) (16x^{4}+ 1)(4x

^{2}– 1) (4x^{2}+ 1) (16x^{4}+ 1) 1[(4x

^{2})^{2}– (1)^{2}] (16x^{4}+ 1) 1(16x

^{4}– 1) (16x^{4}+ 1) 1[(16x

^{4})^{2}– (1)^{2}] 1256x

^{8}– 1

**(iii) (7m – 8n)**^{2} + (7m + 8n)^{2}

^{2}+ (7m + 8n)

^{2}

(7m)

^{2}+ (8n)^{2}– 2(7m)(8n) + (7m)^{2}+ (8n)^{2}+ 2(7m)(8n)(7m)

^{2}+ (8n)^{2}– 112mn + (7m)^{2}+ (8n)^{2}+ 112mn49m

^{2}+ 64n^{2}+ 49m^{2}+ 64n^{2}grouping the similar expression

98m

^{2}+ 64n^{2}+ 64n^{2}98m

^{2}+ 128n^{2}

**(iv) (2.5p – 1.5q)**^{2} – (1.5p – 2.5q)^{2}

^{2}– (1.5p – 2.5q)

^{2}

on expansion

(2.5p)

^{2}+ (1.5q)^{2}– 2 (2.5p) (1.5q) – (1.5p)^{2}– (2.5q)^{2}+ 2 (1.5p) (2.5q)6.25p

^{2}+ 2.25q^{2}– 2.25p^{2}– 6.25q^{2}grouping the similar expression

4p

^{2}– 6.25q^{2}+ 2.25q^{2}4p

^{2}– 4q^{2}4 (p

^{2}– q^{2})

**(v) (m**^{2} – n^{2}m)^{2} + 2m3n^{2}

^{2}– n

^{2}m)

^{2}+ 2m3n

^{2}

On expansion using (a + b)

^{2}formula(m

^{2})^{2}– 2 (m^{2}) (n^{2}) (m) + (n^{2}m)^{2}+ 2m^{3}n^{2}m

^{4}– 2m^{3}n^{2}+ (n^{2}m)^{2}+ 2m^{3}n^{2}m

^{4}+ n^{4}m^{2}– 2m^{3}n^{2}+ 2m^{3}n^{2}m

^{4}+ m^{2}n^{4}

**Question 20. Show that:**

**(i) (3x + 7)**^{2} – 84x = (3x – 7)^{2}

^{2}– 84x = (3x – 7)

^{2}

LHS => (3x + 7)

^{2}– 84xWe will using the formula (a + b)

^{2}= a^{2}+ b^{2}+ 2ab(3x)

^{2}+ (7)^{2}+ 2 (3x) (7) – 84x(3x)

^{2}+ (7)^{2}+ 42x – 84x(3x)

^{2}+ (7)^{2}– 42x(3x)

^{2}+ (7)^{2}– 2 (3x) (7)(3x – 7)

^{2}= R.H.S

**(ii) (9a – 5b)**^{2} + 180ab = (9a + 5b)^{2}

^{2}+ 180ab = (9a + 5b)

^{2}

LHS => (9a – 5b)

^{2}+ 180abWe will using the formula (a + b)

^{2}= a^{2}+ b^{2}+ 2ab(9a)

^{2}+ (5b)^{2}– 2 (9a) (5b) + 180ab(9a)

^{2}+ (5b)^{2}– 90ab + 180ab(9a)

^{2}+ (5b)^{2}+ 9ab(9a)

^{2}+ (5b)^{2}+ 2 (9a) (5b)(9a + 5b)

^{2}= R.H.S

**(iii) (4m/3 – 3n/4)**^{2} + 2mn = 16m^{2}/9 + 9n^{2}/16

^{2}+ 2mn = 16m

^{2}/9 + 9n

^{2}/16

LHS => (4m/3 – 3n/4)

^{2}+ 2mn(4m/3)

^{2}+ (3n/4)^{2}– 2mn + 2mn(4m/3)

^{2}+ (3n/4)^{2}16/9m

^{2}+ 9/16n^{2}= R.H.S

**(iv) (4pq + 3q)**^{2} – (4pq – 3q)^{2} = 48pq^{2}

^{2}– (4pq – 3q)

^{2}= 48pq

^{2}

LHS => (4pq + 3q)

^{2}– (4pq – 3q)^{2}(4pq)

^{2}+ (3q)^{2}+ 2 (4pq) (3q) – (4pq)^{2}– (3q)^{2}+ 2(4pq)(3q)24pq

^{2}+ 24pq^{2}48pq

^{2}= RHS

**(v) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0**

LHS =>(a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a)

By using the identity (a – b) (a + b) = a

^{2}– b^{2}(a

^{2}– b^{2}) + (b^{2}– c^{2}) + (c^{2}– a^{2})a

^{2}– b^{2}+ b^{2}– c^{2}+ c^{2}– a^{2}0 = R.H.S