1. Find the square with the help of the formulae:

1) 2a + 3b

Solution: (2a + 3b)²
=(2a)² + 2.2a.3b + (3b)²
=4a² + 12ab + 9b²

খ) x² + 2/y²

Solution: ( x² + 2/y²)²
= (x²)² + 2.x².2/y² + (2/y²)²
= x⁴ + 4x²/y² + 4/y²

3) 4y – 5x

Solution: (4y – 5x)²
= (4y)²– 2.4y.5x + (5x)²
= 16y² – 40xy + 25x²

4) 5x² – y

Solution: (5x² – y)²
= (5x²)² – 2.5x².y + y²
= 25x⁴–  10x²y + y²

5) 3b –  5c – 2a

Solution: (3b – 5c- 2a)²
= {(3b – 5c) – 2a}²
= (3b – 5c)²-2.(3b – 5c).2a + (2a)²
= (3b)² – 2.3b.5c +( 5c)²-(6b + 10c).2a + 4a²
=9b²  – 30bc + 25c² – 12ab – 20ac + 4a²

6) ax – by – cz

Solution: (ax – by – cz)²
= {(ax – by) – cz}²
= (ax – by)²-2.(ax – by).cz + (cz)²
= (ax)² – 2.ax.by + (by)² – 2axcz + 2bycz + c²z²
= a²x²– 2axby + b²y² – 2axcz + 2bycz + c²z²

7) 2a + 3x – 2y – 5z

Solution: (2a + 3x – 2y – 5z)²
= {(2a + 3x) – (2y + 5z)}²
= (2a + 3x)²-2.(2a + 3x).(2y + 5z) + (2y + 5z)²
= (2a)² + 2.2a.3x + (3x)² -2.(4ay + 6xy + 10az + 15xz + (2y)² + 2.2y.5z + (5z)²
= 4a² + 12ax + 9x² – 8ay – 12xy – 20az – 30xz + 4y² + 20yz + 25z²

8) 1007

Solution: (1007)²
= (1000 + 7)²
= (1000)² + 2.1000.7 + (7)²
= 1000000 + 14000 + 49
= 1014049

2. Simplify:

1) (7p + 3q – 5r)² – 2(7p + 3q – 5r)(8p – 4q – 5r) + (8p – 4q – 5r)²

Solution: Let, 7p + 3q – 5r = a and 8p – 4q – 5r=b
∴  Given Expression = a² – 2.a.b + b²
= (a – b)²
= {(7p + 3q – 5r) – (8p – 4q – 5r)}²  [Substituting the values of a and b]
= (7p + 3q – 5r – 8p + 4q + 5r)²
= (7q – p)²
= (7q)² – 2.7q.p + (p)²
= 49q² – 14pq + p²

2) (2m + 3n – p)² + (2m-  3n + p)² – 2(2m + 3n – p)(2m – 3n + p)

Solution: Let, a = 2m + 3n – p and b = 2m – 3n + p
∴ Given Expression = a² + b² – 2.a.b = (a – b)²
= {(2m + 3n – p) – (2m – 3n + p)}²  [Substituting the values of a and b]
= (2m + 3n – p – 2m + 3n – p)²
= (6n – 2p)²
= (6n)² – 2.6n.2p + (2p)²
= 36n² – 24np + 4p²

3) 6.35 × 6.35 + 2 × 6.35 × 3.65 + 3.65 × 3.65

Solution: Let, a = 6.35 and b = 3.65
∴ Given Expression = a × a+ 2 × a × b + b × b = a² + b² + 2.a.b = (a + b)²
= (6.35 + 3.65)²   [Substituting the values of a and b] = (10)²
= 100

3. If a – b = 4 and ab = 60 , what is the value of a + b?

Solution: Given, a – b = 4 and ab = 60
We know, (a + b)² = (a – b)² + 4ab
Or, (a + b)² = 4² + 4 × 60
Or, (a + b)² = 16 +  240 [substituting the value]
Or, (a + b)² = 256
Or, a + b = ± √256
Or, a + b = ± 16

4. If a + b = 9m and ab = 18m² , what is the value of a – b ?

Solution: Given, a + b = 9m and ab = 18m²
We know, (a -b)² = (a + b)² – 4ab
= (9m)² – 4 × 18m² [substituting the value]
= 81m² – 72m²
= 9m²
∴ (a-b) = ± √9m² = ±3m

5. If x – 1/x = 4 ,  prove that, x⁴ + 1/x⁴ =322.

Solution: Given, x – 1/x = 4
Or, (x – 1/x)² = 4²  (square)

Or, x² – 2.x.1/x + (1/x)² = 16
Or, x² – 2 + 1/x² = 16
Or, x² + 1/x² = 16 + 2
Or, x² + 1/x² = 18
Or, (x² + 1/x²)² = (18)²  (Square)

Or, (x²)² + 2.x².1/x² + (1/x²)² = 324
Or, x⁴ + 2 + 1/x⁴ = 324
Or, x⁴ + 1/x⁴ = 324 – 2
Or, x⁴ + 1/x⁴ = 322 (Proved)

6. If 2x + 2/x = 3 , what is the value of x² + 1/x² ?

Solution: Given, 2x + 2/x = 3
Or, 2(x + 1/x) = 3
Or, x + 1/x = 3/2
Or, (x + 1/x)² = (3/2)²
Or, x² + 2.x.1/x + (1/x)² = 9/4 (square)
Or, x²+  1/x² + 2 = 9/4
Or, x²  + 1/x² = 9/4 – 2
Or, x² + 1/x² = (9 – 8)/4
Or, x² + 1/x² = 1/4

7. If a + 1/a = 2 , show that, a² + 1/a² = a⁴ + 1/a⁴

Solution: Given, a +  1/a= 2
Or, (a² +  1)/a =  2
Or, a² + 1 = 2a
Or, a² – 2a + 1 =0
Or, (a – 1)² = 0
Or, a – 1 = 0
Or, a = 1
Now,
L.H.S = a² + 1/a² = 1² + 1/1² = 2
R.h.s = a⁴ + 1/a⁴ =1⁴  + 1/1⁴  = 2
∴ L.H.S = R.H.S (Showed)

8. If a + b = √7 and a – b =  √5 , prove that, 8ab(a² + b²) = 24

Solution: Given, a + b  =√7 and a – b = √5
L.H.S = 8ab(a² + b²)
= 4ab × 2(a² + b²)
= {(a + b)² – (a – b)²}{(a + b)² + (a – b)²} [Apply Corollary 5 and 6]
= {(√7)² – (√5)²}{(√7)² + (√5)²}
= (7 – 5)(7 + 5)
= 2 × 12
= 24
= R.H.S (Proved)

9. If a + b + c = 9 and ab + bc + ca = 31 , what is the value of a² + b² + c² ?

Solution:  Given, a + b + c = 9 and ab + bc + ca = 31
We know, a² + b² + c² = (a + b + c)² – 2(ab +bc + ca)
                = 9² – 2 × 31 [substituting the value]
                = 81 – 62
                = 19

10. If a² + b² + c² = 9 and ab + bc + ca = 8 , what is the value of (a + b + c)² ?

Solution: Given, a² + b² + c² = 9 এবং ab + bc + ca = 8
We know, (a  + b + c)² = a² + b² +  c² +2(ab + bc + ca)
                = 9 + 2 × 8 [substituting the value]
                = 9 + 16
                = 25

11. If a + b + c = 6 and a² + b² + c² = 14 , what is the value of (a – b)² + (b – c)² + (c – a)² ?

Solution: Given, a + b + c = 6 and a² + b² + c² = 14
Now, (a – b)² + (b –  c)² + (c – a)²
= a² – 2ab + b² + b² – 2bc + c² + c² – 2ca + c²
= 2a² +2b²  + 2c² – 2ab -2bc  – 2ca
= 2(a² + b² + c²) – 2(ab + bc + ca)
= 2(a² + b² +  c²) + ( a² + b²  + c²) – (a + b + c)²     [a² + b² +c² = (a + b  + c)²-2(ab + bc + ca)]
= 2×14 + 14 – (6)²
= 28 + 14 – 36
= 42 – 36
= 6

12.If x = 3, y = 4 and z = 5, what is the value of 9x² + 16y²  + 4z² – 24xy -16yz + 12zx ?

Solution: Given, x = 3, y = 4 and z = 5
Now,
9x² + 16y² + 4z² – 24xy – 16yz + 12zx
= (3x)² + (-4y)² + (2z)² + 2 × 3x × (- 4y) + 2 × (- 4y) × 2z + 2 × 2z × 3x
= {3x + (- 4y) + 2z}²
= (3x –  4y + 2z)²
= (3 ×  3 – 4× 4 + 2 × 5)² [substituting the value]

= (9 – 16 + 10)²
= 3²
= 9

14. Express x² + 10x + 24 as the difference of two squares.

Solution: x² + 10x + 24
= x² +  2.x.5 + 5² –  1
=(x  + 5)² – 1²

15. If a⁴ + a²b² + b⁴ = 8 and a² + ab + b² =  4 , find the value of 1) a² + b², 2) ab

Solution:

(1) Given,
a⁴ + a²b² + b4 = 8 … … … (i)
a² + ab+  b² = 4 … … … (ii)
From (1) we have, a⁴ + a²b² + b⁴ = 8
Or, (a²)² + 2.a².b²  + (b²)²– a²b²  = 8
Or, (a²  +b²)²  – (ab)² =  8
Or, (a² + b² + ab)(a² + b² – ab) = 8
Or, 4(a² + b² –  ab) =  8
Or, a² + b² – ab =  8/4 = 2 … … … (iii)
Adding equation (ii) and (iii) we have,
2a² + 2b² = 4 + 2
Or, a² + b² = 6/2
Or, a² + b² = 3

2) Subtract equation from (ii) to (iii) we have,
2ab = 4 -2
Or, ab = 2/2
Or, ab = 1