Math, asked by AmongSUS, 5 months ago

notes of class 9 chapter 2 maths exercise 2.5.​

Answers

Answered by Anonymous
4

\huge\underline\mathcal\purple{BTS \: EXO}

1. Use suitable identities to find the following products:

(i) (x+4)(x +10)

Solution:

Using the identity, (x+a)(x+b) = x 2+(a+b)x+ab

[Here, a = 4 and b = 10]

We get,

(x+4)(x+10) = x2+(4+10)x+(4×10)

= x2+14x+40

(ii) (x+8)(x –10)

Solution:

Using the identity, (x+a)(x+b) = x 2+(a+b)x+ab

[Here, a = 8 and b = −10]

We get,

(x+8)(x−10) = x2+(8+(−10))x+(8×(−10))

= x2+(8−10)x–80

= x2−2x−80

(iii) (3x+4)(3x–5)

Solution:

Using the identity, (x+a)(x+b) = x 2+(a+b)x+ab

[Here, x = 3x, a = 4 and b = −5]

We get,

(3x+4)(3x−5) = (3x)2+[4+(−5)]3x+4×(−5)

= 9x2+3x(4–5)–20

= 9x2–3x–20

(iv) (y2+3/2)(y2-3/2)

Solution:

Using the identity, (x+y)(x–y) = x2–y 2

[Here, x = y2and y = 3/2]

We get,

(y2+3/2)(y2–3/2) = (y2)2–(3/2)2

= y4–9/4

2. Evaluate the following products without multiplying directly:

(i) 103×107

Solution:

103×107= (100+3)×(100+7)

Using identity, [(x+a)(x+b) = x2+(a+b)x+ab

Here, x = 100

a = 3

b = 7

We get, 103×107 = (100+3)×(100+7)

= (100)2+(3+7)100+(3×7))

= 10000+1000+21

= 11021

(ii) 95×96

Solution:

95×96 = (100-5)×(100-4)

Using identity, [(x-a)(x-b) = x2-(a+b)x+ab

Here, x = 100

a = -5

b = -4

We get, 95×96 = (100-5)×(100-4)

= (100)2+100(-5+(-4))+(-5×-4)

= 10000-900+20

= 9120

(iii) 104×96

Solution:

104×96 = (100+4)×(100–4)

Using identity, [(a+b)(a-b)= a2-b2]

Here, a = 100

b = 4

We get, 104×96 = (100+4)×(100–4)

= (100)2–(4)2

= 10000–16

= 9984

3. Factorize the following using appropriate identities:

(i) 9x2+6xy+y2

Solution:

9x2+6xy+y2 = (3x)2+(2×3x×y)+y2

Using identity, x2+2xy+y2 = (x+y)2

Here, x = 3x

y = y

9x2+6xy+y2 = (3x)2+(2×3x×y)+y2

= (3x+y)2

= (3x+y)(3x+y)

(ii) 4y2−4y+1

Solution:

4y2−4y+1 = (2y)2–(2×2y×1)+1

Using identity, x2 – 2xy + y2 = (x – y)2

Here, x = 2y

y = 1

4y2−4y+1 = (2y)2–(2×2y×1)+12

= (2y–1)2

= (2y–1)(2y–1)

(iii) x2–y2/100

Solution:

x2–y2/100 = x2–(y/10)2

Using identity, x2-y2 = (x-y)(x+y)

Here, x = x

y = y/10

x2–y2/100 = x2–(y/10)2

= (x–y/10)(x+y/10)

4. Expand each of the following, using suitable identities:

(i) (x+2y+4z)2

(ii) (2x−y+z)2

(iii) (−2x+3y+2z)2

(iv) (3a –7b–c)2

(v) (–2x+5y–3z)2

((1/4)a-(1/2)b +1)2

Solution:

(i) (x+2y+4z)2

Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx

Here, x = x

y = 2y

z = 4z

(x+2y+4z)2 = x2+(2y)2+(4z)2+(2×x×2y)+(2×2y×4z)+(2×4z×x)

= x2+4y2+16z2+4xy+16yz+8xz

(ii) (2x−y+z)2

Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx

Here, x = 2x

y = −y

z = z

(2x−y+z)2 = (2x)2+(−y)2+z2+(2×2x×−y)+(2×−y×z)+(2×z×2x)

= 4x2+y2+z2–4xy–2yz+4xz

(iii) (−2x+3y+2z)2

Solution:

Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx

Here, x = −2x

y = 3y

z = 2z

(−2x+3y+2z)2 = (−2x)2+(3y)2+(2z)2+(2×−2x×3y)+(2×3y×2z)+(2×2z×−2x)

= 4x2+9y2+4z2–12xy+12yz–8xz

(iv) (3a –7b–c)2

Solution:

Using identity (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx

Here, x = 3a

y = – 7b

z = – c

(3a –7b– c)2 = (3a)2+(– 7b)2+(– c)2+(2×3a ×– 7b)+(2×– 7b ×– c)+(2×– c ×3a)

= 9a2 + 49b2 + c2– 42ab+14bc–6ca

(v) (–2x+5y–3z)2

Solution:

Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx

Here, x = –2x

y = 5y

z = – 3z

(–2x+5y–3z)2 = (–2x)2+(5y)2+(–3z)2+(2×–2x × 5y)+(2× 5y×– 3z)+(2×–3z ×–2x)

= 4x2+25y2 +9z2– 20xy–30yz+12zx

(vi) ((1/4)a-(1/2)b+1)2

Solution:

Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx

Here, x = (1/4)a

y = (-1/2)b

z = 1

Ncert solutions class 9 chapter 2-5

5. Factorize:

(i) 4x2+9y2+16z2+12xy–24yz–16xz

(ii ) 2x2+y2+8z2–2√2xy+4√2yz–8xz

Solution:

(i) 4x2+9y2+16z2+12xy–24yz–16xz

Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx

We can say that, x2+y2+z2+2xy+2yz+2zx = (x+y+z)2

4x2+9y2+16z2+12xy–24yz–16xz = (2x)2+(3y)2+(−4z)2+(2×2x×3y)+(2×3y×−4z)+(2×−4z×2x)

= (2x+3y–4z)2

= (2x+3y–4z)(2x+3y–4z)

(ii) 2x2+y2+8z2–2√2xy+4√2yz–8xz

Using identity, (x +y+z)2 = x2+y2+z2+2xy+2yz+2zx

We can say that, x2+y2+z2+2xy+2yz+2zx = (x+y+z)2

2x2+y2+8z2–2√2xy+4√2yz–8xz

= (-√2x)2+(y)2+(2√2z)2+(2×-√2x×y)+(2×y×2√2z)+(2×2√2×−√2x)

= (−√2x+y+2√2z)2

= (−√2x+y+2√2z)(−√2x+y+2√2z)

6. Write the following cubes in expanded form:

(i) (2x+1)3

(ii) (2a−3b)3

(iii) ((3/2)x+1)3

(iv) (x−(2/3)y)3

Solution:

(i) (2x+1)3

Using identity,(x+y)3 = x3+y3+3xy(x+y)

(2x+1)3= (2x)3+13+(3×2x×1)(2x+1)

= 8x3+1+6x(2x+1)

= 8x3+12x2+6x+1

(ii) (2a−3b)3

Using identity,(x–y)3 = x3–y3–3xy(x–y)

(2a−3b)3 = (2a)3−(3b)3–(3×2a×3b)(2a–3b)

= 8a3–27b3–18ab(2a–3b)

= 8a3–27b3–36a2b+54ab2

(iii) ((3/2)x+1)3

Using identity,(x+y)3 = x3+y3+3xy(x+y)

((3/2)x+1)3=((3/2)x)3+13+(3×(3/2)x×1)((3/2)x +1)

Ncert solutions class 9 chapter 2-6

(iv) (x−(2/3)y)3

Using identity, (x –y)3 = x3–y3–3xy(x–y)

Ncert solutions class 9 chapter 2-7

\huge\underline\mathcal\purple{BTS \: EXO}

Similar questions