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

**Solution:**

Given,

[?(aâˆ’b)^{2} = a^{2}+b^{2}âˆ’2ab]

Where, a = 2x, b = 1/x

**(ii) (2x+y) (2x-y)**

**Solution:**

Given, (2x+y) (2x-y)

= (2x)^{2}âˆ’(y)^{2}[?(a+b)(aâˆ’b) = a^{2}âˆ’b^{2}]

= 4x^{2}âˆ’y^{2}

?(2x+y)(2xâˆ’y) = 4x^{2}âˆ’y^{2}

**(iii) (a ^{2}bâˆ’ab^{2})^{2}**

Given, (a^{2}bâˆ’ab^{2})^{2}

= (a^{2}b)^{2}+(ab^{2})^{2}âˆ’2âˆ—a^{2}bâˆ—ab^{2 }[?(aâˆ’b)^{2} = a^{2}+b^{2}âˆ’2ab]

Where, a = a^{2}b,b = ab^{2}

= a^{4}b^{2}+b^{4}a^{2}âˆ’2a^{3}b^{3}

?(a^{2}bâˆ’ab^{2})2 = a^{4}b^{2}+b^{4}a^{2}âˆ’2a^{3}b^{3}

*(iv) (a-0.1) (a+0.1)*

**Solution:**

Given, (a-0.1) (a+0.1)

= a^{2}âˆ’(0.1)^{2}[?(a+b)(aâˆ’b) = a^{2}âˆ’b^{2}]

Where, a = a and b = 0.1

= a^{2}âˆ’0.01

?(aâˆ’0.1)(a+0.1) = a^{2}âˆ’0.01

**(v) (1.5x ^{2}âˆ’0.3y^{2})(1.5x^{2}+0.3y^{2})**

**Solution:**

Given, (1.5x^{2}âˆ’0.3y^{2})(1.5x^{2}+0.3y^{2})

= (1.5x^{2})^{2}âˆ’(0.3y^{2})^{2 }[?(a+b)(aâˆ’b) = a^{2}âˆ’b^{2}]

Where, a = 1.5x^{2},b = 0.3y^{2}

= 2.25x^{4}âˆ’0.09y^{4}

?(1.5x^{2}âˆ’0.3y^{2})(1.5x^{2}+0.3y^{2}) = 2.25x^{4}âˆ’0.09y^{4}

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

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

**Solution:**

We have,

399^{2 }= (400-1)^{2}

= (400)^{2}+(1)^{2} â€“ 2x400x1 [ (a-b)^{2} = a^{2}+ b^{2}-2ab ]

Where, a = 400 and b = 1

= 160000 + 1 â€“ 8000

= 159201

Therefore, (399)^{2} = 159201.

**(ii) (0.98) ^{2}**

**Solution:**

We have,

(0.98)^{2} = (1-0.02)^{2}

= (1)^{2 }+ (0.02)^{2} â€“ 2x1x0.02

= 1 + 0.0004 â€“ 0.04 [ Where, a=1 and b=0.02 ]

= 1.0004 â€“ 0.04

= 0.9604

Therefore, (0.98)^{2} = 0.9604

**(iii) 991x1009**

**Solution:**

We have,

991x1009

= (1000-9) (1000+9)

= (1000)^{2} â€“ (9)^{2} [ (a+b) (a-b) = a^{2} â€“ b^{2} ]

= 1000000 â€“ 81 [ Where a=1000 and b=9 ]

= 999919

Therefore, 991x1009 = 999919

**(iv) 117x83**

**Solution:**

We have,

117x83

= (100+17) (100-17)

= (100)^{2} â€“ (17)^{2} [ (a+b) (a-b) = a^{2} â€“ b^{2} ]

= 10000 â€“ 289 [ Where a=100 and b=17 ]

= 9711

Therefore, 117x83 = 9711

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

**(i) 175 x 175 +2 x 175 x 25 + 25 x 25**

**Solution:**

We have,

175 x 175 +2 x 175 x 25 + 25 x 25 = (175)^{2} + 2 (175) (25) + (25)^{2}

= (175+25)^{2} [ a^{2}+ b^{2}+2ab = (a+b)^{2} ]

= (200)^{2} [ Where a=175 and b=25 ]

= 40000

Therefore, 175 x 175 +2 x 175 x 25 + 25 x 25 = 40000.

**(ii) 322 x 322 â€“ 2 x 322 x 22 + 22 x 22**

**Solution:**

We have,

322 x 322 â€“ 2 x 322 x 22 + 22 x 22

= (322-22)^{2} [ a^{2}+ b^{2}-2ab = (a-b)^{2} ]

= (300)^{2} [ Where a=322 and b=22 ]

= 90000

Therefore, 322 x 322 â€“ 2 x 322 x 22 + 22 x 22= 90000.

**(iii) 0.76 x 0.76 + 2 x 0.76 x 0.24 + 0.24 x 0.24**

**Solution:**

We have,

0.76 x 0.76 + 2 x 0.76 x 0.24 + 0.24 x 0.24

= (0.76+0.24)^{2} [ a^{2}+ b^{2}+2ab = (a+b)^{2} ]

= (1.00)^{2} [ Where a=0.76 and b=0.24 ]

= 1

Therefore, 0.76 x 0.76 + 2 x 0.76 x 0.24 + 0.24 x 0.24 = 1.

**(iv) **

**Solution:**

We have,

[?(aâˆ’b)^{2 }= (a+b)(aâˆ’b)]

= 9

**Q.4. If ** **,find the value of**

**Solution:**

**Q.5. If ** **,find the value of**

**Solution: **We have,

Now,

**Q. 6. If ** **find the value** of

**Solution:**

We have,

**Q. 7. If ** **find the value of**

**Solution:**

We have,

**Q. 8. If ** **find the value of **

**Solution:**

We have,

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

**Solution:**

We have,

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

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

= 181 + 30(-6) [Since, 9x^{2 }+ 25y^{2 }= 181 and xy = -6]

â‡’ (3x+5y)^{2} = 1

â‡’ (3x+5y)^{2} = (Â±1)^{2}

â‡’ 3x+5y = Â±1

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

**Solution:**

We have,

(2x + 3y)^{2} = (2x)^{2} + (3y)^{2} + 2*2x*3y

â‡’ (2x + 3y)^{2} = 4x^{2} + 9y^{2} + 12xy [Since, 2x + 3y = 8 and xy = 24 ]

â‡’ (8)^{2} = 4x^{2} + 9y^{2} + 24

â‡’ 64 â€“ 24 = 4x^{2} + 9y^{2}

â‡’ 4x^{2} + 9y^{2} = 40

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

**Solution:**

We have,

(2 - 7y)^{2} = (3x)^{2} + (-7y)^{2} - 2*3x*7y

â‡’ (3x - 7y)^{2} = 9x^{2} + 49y^{2} - 42xy [Since, 3x - 7y = 10 and xy = -1 ]

â‡’ (10)^{2} = 9x^{2} + 49y^{2} + 42

â‡’ 100 â€“ 42 = 9x^{2} + 49y^{2}

â‡’ 9x^{2} + 49y^{2} = 58

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

**(i) **

**Solution:**

**Solution:**

We have,

[?(a+b)(a+b) = (a+b)^{2} and (a+b)(aâˆ’b) = a^{2}âˆ’b^{2}]

**Solution:**

We have,

[?(aâˆ’b)(aâˆ’b) = (aâˆ’b)^{2}]

**(iv) (x ^{2}+xâˆ’2)(x^{2}âˆ’x+2)**

**Solution:**

(x^{2}+xâˆ’2)(x^{2}âˆ’x+2)

[(x)^{2}+(xâˆ’2)][(x^{2}âˆ’(x+2)]

â‡’ (x^{2})^{2}âˆ’(xâˆ’^{2})^{2} [(a â€“ b) (a + b) = a^{2} - b^{2}]

â‡’ x^{4}âˆ’(x^{2}+4âˆ’4x)[?(aâˆ’b)^{2} = a^{2}+b^{2}âˆ’2ab]

â‡’ x^{4}âˆ’x^{2}+4xâˆ’4

?(x^{2}+xâˆ’2)(x^{2}âˆ’x+2) = x^{4}âˆ’x^{2}+4xâˆ’4

**(v) (x ^{3}âˆ’3xâˆ’x)(x^{2}âˆ’3x+1)**

**Solution:**

We have,

(x^{3}âˆ’3xâˆ’x)(x^{2}âˆ’3x+1)

â‡’ x(x^{2}âˆ’3xâˆ’1)(x^{2}âˆ’3x+1)

â‡’ x[(x^{2}âˆ’3x)2âˆ’(1)2][?(a+b)(aâˆ’b) = a^{2}âˆ’b^{2}]

â‡’ x[(x^{2})^{2}+(âˆ’3x)^{2}âˆ’2(3x)(x^{2})âˆ’1]

â‡’ x[x^{4}+9x^{2}âˆ’6x^{3}âˆ’1]

â‡’ x^{5}âˆ’6x^{4}+9x^{3}âˆ’x

?(x^{3}âˆ’3xâˆ’x)(x^{2}âˆ’3x+1) = x^{5}âˆ’6x^{4}+9x^{3}âˆ’x

**(vi) (2x ^{4}âˆ’4x^{2}+1)(2x^{4}âˆ’4x^{2}âˆ’1)**

**Solution:**

We have,

(2x^{4}âˆ’4x^{2}+1)(2x^{4}âˆ’4x^{2}âˆ’1)

â‡’ [(2x^{4}âˆ’4x^{2})^{2}âˆ’(1)^{2}] [?(a+b)(aâˆ’b) = a^{2}âˆ’b^{2}]

â‡’ [(2x^{4})^{2}+(4x^{2})^{2}âˆ’2(2x^{4})(4x^{2})âˆ’1]

â‡’ 4x^{8}âˆ’16x^{6}+16x^{4}âˆ’1 [?(aâˆ’b)^{2 }= a^{2}+b^{2}âˆ’2ab]

?(2x^{4}âˆ’4x^{2}+1)(2x^{4}âˆ’4x^{2}âˆ’1) = 4x^{8}âˆ’16x^{6}+16x^{4}âˆ’1

**Q.13. Prove that a ^{2}+b^{2}+c^{2}âˆ’abâˆ’bcâˆ’ca is always non-negative for all values of a, b and c.**

**Solution: **

We have,

a^{2}+b^{2}+c^{2}âˆ’abâˆ’bcâˆ’ca

Multiply and divide by â€˜2â€™

?a^{2}+b^{2}+c^{2}âˆ’abâˆ’bcâˆ’ca â‰¥ 0

Hence, a^{2}+b^{2}+c^{2}âˆ’abâˆ’bcâˆ’ca â‰¥ 0 is always non-negative for all values of a, b and c.