## Elementary Algebra

#### Elementary Algebra

1. The line passing through the points (– 2, 8) and (5, 7)

1. As indicated in the graph, the line passing through the points cuts Y-axis only.

##### Correct Option: C

As indicated in the graph, the line passing through the points cuts Y-axis only.

1. Let a = √6 - √5, b = √5 - 2, c = 2 - √3 Then point out the correct alternative among the four alternatives given below.

1. We know that , √6 = 2.44, √5 = 2.23, √3 = 1.73
a = √6 - √5 = 2.44 - 2.23 = 0.21
b = √5 - 2 = 2.23 - 2 = 0.23

##### Correct Option: A

We know that , √6 = 2.44, √5 = 2.23, √3 = 1.73
a = √6 - √5 = 2.44 - 2.23 = 0.21
b = √5 - 2 = 2.23 - 2 = 0.23
c = 2 - √3 = 2 - 1.73 = 0.27

1. If the difference of two numbers is 3 and the difference of their squares is 39; then the larger number is:

1. Let the numbers are p and q.
p – q = 3 … (1)
p2 – q2 = 39
(p – q) (p + q) = 39 ⇒ x + y = 13 … (2)
p + q + p – q = 16 ⇒ p = 8

##### Correct Option: D

Let the numbers are p and q.
p – q = 3 … (1)
p2 – q2 = 39
(p – q) (p + q) = 39 ⇒ x + y = 13 … (2)
p + q + p – q = 16 ⇒ p = 8
∴ q = 3
Hence, 8 is the larger number.

1. If a2 + b2 + c2 = 2 (a – b – c) – 3, then the value of 2a – 3b + 4c is

1. Given that :- a2 + b2 + c2 = 2 (a – b – c) – 3
⇒ a2 + b2 + c2 - 2 (a – b – c) + 3 = 0
⇒ a2 + b2 + c2 - 2a + 2b + 2c + 3 = 0
⇒ ( a2 + 1 - 2a ) + ( b2 + 1 + 2b ) + ( c2 + 1 + 2c ) = 0
⇒ ( a - 1 )2 + ( b + 1 )2 + ( c + 1 )2 = 0
This is possible when ( a - 1 )2 = 0, ( b + 1 )2 = 0, and ( c + 1 )2 = 0
⇒ a = 1, b = –1, c = –1

##### Correct Option: A

Given that :- a2 + b2 + c2 = 2 (a – b – c) – 3
⇒ a2 + b2 + c2 - 2 (a – b – c) + 3 = 0
⇒ a2 + b2 + c2 - 2a + 2b + 2c + 3 = 0
⇒ ( a2 + 1 - 2a ) + ( b2 + 1 + 2b ) + ( c2 + 1 + 2c ) = 0
⇒ ( a - 1 )2 + ( b + 1 )2 + ( c + 1 )2 = 0
This is possible when ( a - 1 )2 = 0, ( b + 1 )2 = 0, and ( c + 1 )2 = 0
⇒ a = 1, b = –1, c = –1
Thus, 2a – 3b + 4c = 2 (1) – 3 (–1) + 4 (–1)
2a – 3b + 4c = 2 + 3 – 4 = 1.

1. If a, b, c are real and a3 + b3 + c3 = 3abc and a + b + c ≠ 0 , then the relation between a, b, c will be

1. Given that :-
Now ,a3 + b3 + c3 = 3abc
( a+b+c )( a2 + b2 + c2 - ab - bc - ca ) + 3abc = 3abc
( a+b+c )( a2 + b2 + c2 - ab - bc - ca ) = 0
∴ a² + b² + c² - ab - bc - ca = 0 or a + b + c ≠ 0

##### Correct Option: C

Given that :-
Now ,a3 + b3 + c3 = 3abc
( a+b+c )( a2 + b2 + c2 - ab - bc - ca ) + 3abc = 3abc
( a+b+c )( a2 + b2 + c2 - ab - bc - ca ) = 0
∴ a² + b² + c² - ab - bc - ca = 0 or a + b + c ≠ 0
Where ,
a = b = c