EXERCISE 2.4 CLASS 9 MATHS | Algebraic Identities Class 9 Maths скачать в хорошем качестве

EXERCISE 2.4 CLASS 9 MATHS | Algebraic Identities Class 9 Maths

#mathsclass9 #rajansir EXERCISE 2.4 CLASS 9 MATHS | Algebraic Identities Class 9 Maths EXERCISE 2.4 CLASS 9 MATHS | Algebraic Identities Class 9 Maths by ‪@rajansir07‬ Maths Chapter 2 – Polynomials Exercise 2.4 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 (vi) ((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)