Test: Algebraic Identities Quadratic

By subtracting 1 from 2,you will get 1 that makes the whole square of (x+1)=1

The correct option is Option C.

(a + b)² = a² + b² + 2ab

(12)² = 60 + 2ab

ab = (144 - 60)/2

ab = 84/2

ab = 42

Now, 

(a + b)³ = a³ + b³ + 3ab(a+b)

(12)³ = a³ + b³ + (3 × 42 × 12)

a³ + b³ = 1728 - 1512

Therefore, a³ + b³ = 216

(100-1)²= (100)²+(1)²+2×100×1 10000+1+200 =10201

= ( a² + b² + c² - ab - bc - ca )

By multiplying it by 2 and dividing it by 2.

= 2 ( a² + b² + c² - ab - bc - ca ) ÷ 2

= ( 2a² + 2b² + 2c² - 2ab - 2bc - 2ca ) ÷ 2

= ( a² + a² + b² + b² + c² + c² - 2ab - 2bc - 2ca ) ÷ 2

= ( a² + b² - 2ab + b² + c² - 2bc + a² + c² - 2ca ) ÷ 2

= [ ( a - b )² + ( b - c )² + ( a - c )² ] ÷ 2

Now , whatever is the value of  ( a - b ) , ( b - c ) and ( a - c ) but its square will be always positive, and 2 is also a positive number.

So, the sum of ( a - b )², ( b - c )² and ( a - c )² will be a positive number and if a positive number is divided by a positive number then the result is also a positive number.

So, it is a positive number,hence it can't be a negative number.

(100+2)×(100-2). [(a+b)(a-b)=a^2-b^2
100^2 - 2^2
10000 - 4
9996

2ab = (a² + b²) - (a - b)²
= 29 - 9 = 20
⇒ ab = 10.

Let p(x) = x² + 2x + 0.5

To find what must be added to p(x) to make it perfect square.

if we add 0.5 to p(x) we will get
p(x) = x2 + 2x + 0.5 + 0.5
p(x) = x2 + 2x + 1
p(x) = x2 + 1x + 1x + 1
p(x) = x(x + 1) + 1(x + 1)
p(x) = (x+1)(x+1)
p(x) = (x+1)²
which is a perfect square.

Therefore 0.5 must be added to the given expression to make it perfect square.

(100+1)×(100 - 1) it is the formula of a²-b² so, (100)² -(1)² = 10000-1 = 9999.

The expansion of (2a - 3b + 5c)² -: (2a)² + (- 3b)² + (5c)² + 2× 2a × (-3b) + 2× (-3b)× 5c + 2× 5c + 2a. = 4a² + 9b² + 25c² - 12ab - 30bc + 20ac