FORMULA FOR A MINUS B THE WHOLE SQUARE

Expand and simplify :

Problem 1 :

(x - 3)²

Solution :

By comparing (a - b)² and (x - 3)² , we get

a = x, b = 3

(a - b)² = a² - 2ab + b²

Substitute x for a and 3 for b

(x - 3)² = x² - 2(x)(3) + 3²

(x - 3)² = x² - 6x + 9

Problem 2 :

(2 - x)²

Solution :

By comparing (a - b)² and (2 - x)² , we get

a = 2, b = x

(2 - x)² = 2² - 2(2)(x) + x²

(2 - x)² = 4 - 4x + x²

Problem 3 :

(3x - 1)²

Solution :

By comparing (a - b)² and (3x - 1)² , we get

a = 3x, b = 1

(3x - 1)² = (3x)² - 2(3x)(1) + (1)²

(3x - 1)² = 9x² - 6x + 1

Problem 4 :

(x - y)²

Solution :

By comparing (a - b)² and (x - y)² , we get

a = x, b = y

(x - y)² = x² - 2(x)(y) + y²

(x - y)² = x² - 2xy + y²

So, the expansion of (x - y)² is

x² - 2xy + y²

Problem 5 :

(2x – 5y)²

Solution :

Here a = 2x, b = 5y

(2x - 5y)² = (2x)² - 2(2x)(5y) + (5y)²

(2x - 5y)² = 4x² - 20xy + 25y²

So, the expansion of (2x - 5y)² is

4x² - 20xy + 25y²

Problem 6 :

(ab - 2)²

Solution:

Here a = ab, b = 2

(ab - 2)² = (ab)² - 2(ab)(2) + (2)²

(ab - 2)² = ab² - 4ab + 4

Problem 7 :

Evaluate (48)²

Solution :

(48)² = (50 - 2)²

= (50)² - 2(50)(2) + (2)²

= 2500 – 200 + 4

     = 2304

Problem 8 :

(2x/3 – 3y/2)²

Solution :

Here a = 2x/3, b = 3y/2

(2/3 x – 3/2 y)² = (2x/3)² - 2(2x/3)(3y/2) + (3y/2)²

= 4/9 x² - 2xy + 9/4 y²

Problem 9 :

By using the suitable identity, evaluate x² + 1/x², if x - 1/x = 5

Solution :

Problem 10 :

Find the length of the side of the given square if area of the square is 625 square units and then find the value of x.

Solution :

Area of the square = 625

(4x+5)² = 625

Using the algebraic identity (a + b)2 we get 

(4x)² + 2(4x)(5) + 5² = 625

16x² + 40x + 25 = 625

Subtracting 625 on both sides, we get

16x² + 40x -600 = 0

2x² + 5x - 75 = 0

(x - 5) (2x + 15) = 0

Equating each factor to zero, we get

x = 5 and x = -15/2

By applying -15/2, we get the side length with negative sign, so the answer is 5.