# NCTB Class 8 Math Chapter Four Exercise 4.1 Solution

NCTB Class 8 Math Chapter Four Exercise 4.1 Solution by Math Expert. Bangladesh Board Class 8 Math Solution Chapter 4 Algebraic formulae and applications Exercise 4.1 Solution.

 Board NCTB Class 8 Subject Mathematics Chapter 4 Chapter Name Algebraic formulae and applications Exercise 4.1 Solution

Exercise 4.1

1>  Find the Square of the following expression with the help of formulae:

(a) (5a+7b)

= (5a)2 + 2.5a.7b + (7b)2

= 25a2 + 70ab + 49b2

(b) (6x + 3 )2

= 36x2  + 2.6x.3 + 9

= 36x2 + 36x +9

(c) (7p – 2q)2

= (7p)2 -2.7p.2q + (2q)2

= 49p2 – 28pq + 4q2

(d) (ax – by )2

= (ax)2  – 2.ax.by + b2y2

= a2 x2 -2axby + b2 y2

(e) (X3+ xy)2

= (x3)2 + 2.x3 .xy + (xy)

= X6 + 2x4y + x2y2

(f) (11a-12b)2

= (11a)2 – 2.11a.12b + (12b)2

= 121a2 – 243ab + 144b2

(g) 6x2 y – 5xy2

= 2xy ( 3x – 5y )2

= 2xy {(3x)2– 2.3x.5y+(5y)2}

= 2xy (9x2 – 30xy + 25y2)

(h) (-X-Y)2

= {-(X+Y)}2

= – (x2 + 2xy + y2)

= -x2 – 2xy – y2

(j) a2x3 – b2y4

= (a2x3 – b2y4)2

= (a2x3)2 – 2.(a2x3).(b2y4) + (b2y4)

= a4x6 – 2a2b2x3y4 + b4y6

(k) (108)2

= (100+8)2

= (100)2 + 2.100.8 + (8)2

= 10000 + 1600 + 64

= 11664

(L) (606)2

= (600+6)2

= (600)2 + 2.600.6 + (6)2

= 360000 + 7200 + 36

= 367236

(m) (597)2

= (600 – 3)2

= (600)2 – 2. 600. 3 + 32

= 360000 – 3600 + 9

= 356409

(n) (a-b+c)2

= (a+ (-b)+ c)2

= a2 + (-b)2 + c2 + 2a (-b) + 2 (b) c + 2ac

= a2 + b2 + c2 – 2ab – 2bc + 2ac

= a2 + b2 + c2 – 2ab – 2bc + 2ac

(o) (ax+b+2)2

= (ax)2 + (b)2 + (2)2 + 2.ax.b + 2b.2 + 2.ax.2

= a2x2 + b2 + 4 + 2abx + 4b + 4ax

(p) (xy + yz – zx)2

= (xy)2 + (yz)2 + (-zx)2 + 2.xy.yz + 2yz.xy + 2 (-zx).xy

= x2y2 + y2z2 + z2x2 + 2xy2z + 2xy2z – 2x2yz

(q) ( 3p+ 2q-5r)2

= (3p)2 + (2q)2 + (-5r)2 + 2.3q.2q + 2. 2q. (-5r) + 2. (-5r). 3p

= 9p2 + 4q2 + 25r2 + 12pq – 20qr – 30pr

(s) 7a2 + 8b2 + 5c2

= (7a2)2 + (8b2) + (-5c2)2 + 2.7a2.8b2 + 2. 8b2. (-5c2) + 2. (-5c2) . (7a)

= 49a4 + 64b4 + 25c4 + 112a2b2 – 80b2c2 – 70a2c2

2> Simplify :

(a) (X+Y)2 + 2(X+Y)(X-Y)+(X+Y)2

= (X+Y+X-Y)2

=(2X)2

=4X2

(b) (2a+3b)2 – 2 (2a + 3b)+ (3b-a)+(3b-a)2

= (2a+3b-3b+a)2

= (3a)2

= 9a2

(c) (3x2 + 7y2)2 + 2 (3x2 + 7y2) (3x2 – 7y2) + (3x2 + 7y2)2

= (3x2+7y2+3x2-7y2)

=(6x2)2

=36x4

(d) (8x + y)2 – (16x+2y) (5x+y) + (5x+y)2

= (8x + y)2 – 2 (8x+y) (5x + y ) + (5x + y)2

= (8x + y – 5x – y )2

= (3x)2

= 9x2

3> Find the product by applying formulae:

(a) ( x+7) (x-7)

= x2 – 72

= x2 – 49

(b) (5x + 13) (5x – 13)

= (5x)2 – (13)2

= 25x2 – 169

(c) (xy + yz) (xy – yz )

= (xy)2 – (yz)2

= x2y2 – y2z2

(d) (ax + b) (ax – b)

= (ax)2 – (b)2

= a2x2 – b2

(e) (a + 3) (a- 4)

= a2 – 4a + 3a – 12

= a2 – a – 12

(f) (ax + 3 ) (ax + 4)

= a2x+ 4ax + 3ax + 12

= a2x2 + 7ax + 12

(g) (6x + 17 ) (6x – 13)

= 36x2 – 78x + 102x – 221

= 36x2 + 24x -221

(h) (a2 + b2) (a2 – b2) (a4 + b4)

= (a2)2 – (b2)2 (a4+ b4)

= (a4 – b4) (a4– b4)

= (a4)2 – (b4)2

= a8 – b8

(i) (ax + by + cz) (ax + by – cz)

= (ax +by)2 – (cz)2

= a2x2 + 2.ax.by + b2y2 – c2z2

= a2x2 + 2abxy + b2y2 – c2z2

(j) (3a – 10) (3a – 5)

= 9a2 – 15 – 30a + 50

= 9a2 – 30a + 35

(k) (5a + 2b – 3c) (5a + 2b + 3c)

= (5a+ 2b)2 – (3c)2

= (5a)2 + 2.5a.2b + (2b)2 – 9c2

= 25a2 +20ab + 4b2 – 9c2

(l) (ax + by + 5) (ax + by + 3)

= a2x2 +axby+ 3ax+ abxy + b2y2 + 3by+ 5ax+5by + 15

= a2x2 + 2axby+ 8ax+b2y2 +8by+ 15

4> If a=4, b=6 and c=3, find the 4a2b2 – 16ab2c + 16b2c2

= (2ab)2 – 2.2ab.4bc + (4bc)2

= (2ab + 4bc)2

= (2x4x6+4x6x3)2

= (48 + 72)2

= (120)2

= 14400

5> x- 1/x = 3, find the value of x2+1/x2

= (x-1/x)2 + 2.x.1/x

= (3)+ 2

= 9+2

=11

6> If a+1/a=4, what is the value of a4 + 1/a4 ?

Or, (a+1/a)2 = 42

Or, a2 + 2.a.1/a + (1/a)2 = 16

Or, a2 + 1/a2 = 16-2

= 14

Therefore, a4 + 1/a4

= (a2)2 + (1/a2)2

= (a2 + 1/a2) – 2.a2.1/a2

= (14)2 – 2

= 196-2

= 194

8> If a-1/a =m, show that a4+ 1/a4 =m4+4m2 +2

Or (a-1/a)2 = m2

Or, a2-2.a.1/a2 = m2

Or, a2 + 1/a2 = m2+2

Or, (a2+1/a2)2 = (m2 + 2)2

Or, (a2)2 + 2.a2.1/a2 + 1/a4 = m4 + 2.m2.2 + 4

Or, a4 + 2+ 1/a4 = m4 + 4m2 + 4

Or, a4 + 2 + 1/a4 = m4 + 4m2+4

Or, a4+ 1/a4 = m4 + 4m2 + 4-2

= m4+4m2 +2 (Proved)

9> If x-1/x=4, Prove that x2+(1/x)2 =18

x- 1/x =4 (given)

or, (x – 1/x)2 = (4)2

or, x2 – 2.x.1/x + 1/x2 =16

or, x2 + 1/x2 = 16+2

= 18 (proved)

10> If m+1/m = 2, prove that m4 + 1/m4 =2

M+1/m=2 (Given)

Or, (m+1/m)2 =4

Or, m2+2.m.1/m+1/m2 =4

Or, m2 + 1/m2 = 4-2

=2

Or, ( m2+1/m2)2 =4

Or, m4 + 2.m2.1/m2+1/m4 =4

Or, m4 + 1/m4 = 4-2

=2 (proved)

11> If x+y=12 and xy=27, find the value of (x-y)2 and x2 + y2

=  x2+y2 – 2.x.y

= 90 – 2.27

= 90 – 54

= 36

X2 + y2

= (x+y)2 + 2xy

= (12)2 – 2.27

= 144- 54

=90

12> If a+b=13 and a-b=3, find the Value of 2a2 +2b2 and ab.

2a2 + 2b2

= 2(a2 + b2)

= 2{(a+b)2 – 2.ab}

= 2 {(13)2 – 2.ab

a-b=3

or, (a-b)2 =9

or, a2 – 2ab + b2 =9

or, (a+b)2 + 2ab =9

13> Express the difference between of the square of two expression

(a) (5p – 3q)2 – (p+7q)2

= (5q-3q+p+7q) (5p-3q-p-7q)

= (6p+4q) (4p-10q)

= 24p2+60pq + 16pq – 40q2

= 24p2 – 44pq – 40q2

Updated: March 22, 2021 — 2:49 pm

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