- Factorize the following:
(i) 9x2 – 16y2
Solution:
= (3x)2 – (4y)2
Using a2 – b2 = (a + b) (a – b)
= (3x + 4y) (3x – 4y)
(ii) 5x2 – 7y2 [ Hint : ( 𝟓 )2 = 5 ]
Solution:
( 5 x)2 – ( 7 y)2
Using a2 – b2 = (a + b) (a – b)
= ( 5 x + 7 y) ( 5 x – 7 y)
(iii) 100 – 9x2
Solution:
= 102 – (3x)2
= (10 + 3x) (10 – 3x) [∵a2 – b2 = (a + b) (a – b)]
(iv) (x + 4y)2 – 4z2
Solution:
= (x + 4y)2 – (2z)2 [∵a2 – b2 = (a + b) (a – b)]
= (x + 4y + 2z) (x + 4y – 2z)
(v) x – 64xy4
Solution:
= x (1 – 64y4)
= x (1 + (8y2)2) [∵a2 – b2 = (a + b) (a – b)]
= x (1 + 8y2) (1 – 8y2)
(vi) a2 + 2ab + b2 – 9c2
Solution:
= (a + b)2 – (3c)2 [∵(a + b)2 = a2 + 2ab + b2 ;a2 – b2 = (a + b) (a – b)]
= (a + b + 3c) (a + b – 3c)
(vii) 4x2 – 9y2 – 2x – 3y
Solution:
= (2x)2 – (3y)2 – (2x + 3y)
= [(2x + 3y) (2x – 3y)] – (2x – 3y) [∵a2 – b2 = (a + b) (a – b)]
= (2x + 3y) (2x – 3y – 1)
(viii) a2 + b2 – a + b
Solution:
= a2 + b2 – (a – b) [∵a2 – b2 = (a + b) (a – b)]
= (a + b) (a – b) – (a – b)
= (a – b) (a + b – 1)
(ix) x4 – 625
Solution:
= (x2)2 – 252 [∵a2 – b2 = (a + b) (a – b)]
= (x + 25) (x2 – 25)
= (x2 + 25) (x2 – 52)
= (x2 + 25) (x + 5) (x – 5)
(x) 36 (5x + y)2 – 25 (4x – y)2
Solution:
= [6 (5x + y)]2 – [5 (4x – y)]2 [∵a2 – b2 = (a + b) (a – b)]
= (30x + 6y + 20x – 5y) (30x + 6y – 20x + 5y)
= (50x + y) (10x + 11y)
(xi) (a + 𝟏/𝐚 )2 – 4 (x – 𝟏/𝐱 )2
Solution:
= (a + 𝟏/𝐚 )2 – [2 (x – 𝟏/𝐱 )]2
= [a+ 𝟏/𝐚 + 2( x− 𝟏/𝐱)][ a+ 𝟏/𝐚 – 2(x− 𝟏/𝐱 )]
(xii) 12m2 – 75n2
Solution:
= ( √12 m2)2 – (√75 n2)2 [∵a2 – b2 = (a + b) (a – b)]
= (2 √3m2)2 – (5√3 n2)2 [ 12 =2√3 and 75 = 5√3
= (2√3m2 + 5√3 n2) (2√3 m2 – 5√3 n2)
= √3 (2m2 + 5n2) (2m – 5n) (2m + 5n)
- Factorize by adding and subtracting appropriate quantity.
(i) x2 + 6x + 3
Solution:
= x2 + 2. x. 3 + 32 – 32 + 3
= (x + 3)2 – 9 + 3
= (x + 3)2 – 6 [∵a2 – b2 = (a + b) (a – b)]
= (x + 3)2 – 62
= (x + 3 + 6 ) (x + 3 – 6 )
(ii) x2 + 10x + 8
Solution:
= x2 + 2. x. 5 + 52 – 52 + 8
= (x + 5)2 – 25 + 8
= (x + 5)2 – 17 [∵a2 + 2ab + b2 = (a + b)2]
= (x + 5)2 – 172
= (x + 5 + 17 ) (x + 5 – 17 )
(iii) x2 + 10x + 20
Solution:
= x2 + 2. x. 5 + 52 – 52 + 20 [∵a2 + 2ab + b2 = (a + b)2]
= (x + 5)2 – 25 + 20
= (x + 5)2 – 5
= (x + 5)2 – 52 [∵a2 – b2 = (a + b) (a – b)]
= (x + 5 + 5 ) (x + 5 – 5 )
(iv) x2 + 2x – 1
Solution:
= x2 + 2. x. 1 + 12 – 12 + 1 [∵a2 + 2ab + b2 = (a + b)2]
= (x + 1)2 – 1 + 1
= (x + 1)2 – 2
= (x + 1)2 – 22 [∵a2 – b2 = (a + b) (a – b)]
= (x + 1 + 2 ) (x + 1 – 2 )
- prove that (1 + 𝐱/𝐲 ) (1 – 𝐱/𝐲 ) (1 + 𝐱𝟐/𝐲𝟐 ) (1 + 𝐱𝟒/𝐲𝟒 ) + 𝐱𝟖/𝐲𝟖 = 1
Solution:
L.H.S = (1 + 𝐱/𝐲 ) (1 – 𝐱/𝐲 ) (1 + 𝐱𝟐/𝐲𝟐) (1 + 𝐱𝟒/𝐲𝟒 ) + 𝐱𝟖/𝐲𝟖
[∵a2 – b2 = (a + b) (a – b)]
= (1 – 𝐱𝟐/𝐲𝟐 ) (1 + 𝐱𝟐/𝐲𝟐 ) (1 + 𝐱𝟒/𝐲𝟒) + 𝐱𝟖/𝐲𝟖
= (1 – 𝐱𝟒/𝐲𝟒) (1 + 𝐱𝟒/𝐲𝟒 ) + 𝐱𝟖/𝐲𝟖
= (1)2 – (𝐱𝟒/𝐲𝟒)2 + 𝐱𝟖/𝐲𝟖
= 1 – 𝐱𝟖/𝐲𝟖 + 𝐱𝟖/𝐲𝟖
= 1 R.H.S
- If x = 𝐚+𝐛/𝐚−𝐛 and y = 𝐚−𝐛/𝐚+𝐛 find x2 – y2.
Solution:
x2 – y2 = (x + y) (x – y)
= (a+b/a−b+a−b/a+b)(a+b/a−b – a−b/a+b)
= [[(a+b)2+ (a−b)2 ]/(a2− b2)][[(a+b)2− (a−b)2]/(a2− b2)]
= {(a2+ 2ab + b2+ a2+ b2− 2ab)/[(a2− b2)2]} x (a2 + 2ab + b2 – a2 – b2 + 2ab)
= [(2a2+ 2b2)4ab]/[(a2− b2)2]
= [2(a2+ b2)4ab]/(a2− b2)2
= [8ab(a2+ b2)]/(a2− b2)2
- If p = x – y, q = y – z and r = z – x, simplify r2 – p + 2pq – q2
Solution:
(z – x)2 – (x – y)2 + 2 (x – y) (y – z) – (y – z)2
= z2 – 2xy + x2 – (x2 – 2xy + y2) + 2 (xy – y2 – xz + yz) – (y2 + z2 – 2yz)
= z2 + x2 – 2xz – x2 + 2xy – y2 + 2xy – 2y2 – 2xz + 2yz – y2 – z2 + 2yz
= 4xy – 4xz + 4yz – 4y2
= 4[xy – xz + yz – y2]
= 4[x (y – z) – y (y – z)]
= 4 (x – y) (y – z)
= 4pq
- The radius of a circle is 13cm in which a chord of 10cm is drawn. Find the distance of the chord from the centre of the circle.
[Hint: use Pythagoras rule and the fact that the perpendicular line drawn from the centre of the circle to any o the chords bisects the chord]
Solution:
Let AC = 10cm is the chord of the circle with radius 13cm then using Pythagoras theory
OA2 = OB2 + AB2
132 = x2 + 52
x2 = 132 – 52
= (13 + 5) (13 – 5)
= 18 (8)
x2 = 144 (∵ the perpendicular line drawn from the centre of the circle to a chord bisects the chord)
x = 144
x = 12 cm
- Is it possible to factorize x2 + 4x + 20? Give reason
Solution:
x2 + 4x + 20
= x2 + 2. x. 2 + 22 – 22 + 20
= x2 + 4x + 4 – 4 + 20
= (x + 4)2 + 16
= (x + 4)2 + 42
We cannot factorize their further since this is nit of the form a2 – b2
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