Elementary Algebra
 The line passing through the points (– 2, 8) and (5, 7)

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As indicated in the graph, the line passing through the points cuts Yaxis only.
Correct Option: C
As indicated in the graph, the line passing through the points cuts Yaxis only.
 Let a = √6  √5, b = √5  2, c = 2  √3 Then point out the correct alternative among the four alternatives given below.

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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 = 0.23Correct 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
 If the difference of two numbers is 3 and the difference of their squares is 39; then the larger number is:

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Let the numbers are p and q.
p – q = 3 … (1)
p^{2} – q^{2} = 39
(p – q) (p + q) = 39 ⇒ x + y = 13 … (2)
Adding eq^{n} (1) and (2)
p + q + p – q = 16 ⇒ p = 8Correct Option: D
Let the numbers are p and q.
p – q = 3 … (1)
p^{2} – q^{2} = 39
(p – q) (p + q) = 39 ⇒ x + y = 13 … (2)
Adding eq^{n} (1) and (2)
p + q + p – q = 16 ⇒ p = 8
∴ q = 3
Hence, 8 is the larger number.
 If a^{2} + b^{2} + c^{2} = 2 (a – b – c) – 3, then the value of 2a – 3b + 4c is

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Given that : a^{2} + b^{2} + c^{2} = 2 (a – b – c) – 3
⇒ a^{2} + b^{2} + c^{2}  2 (a – b – c) + 3 = 0
⇒ a^{2} + b^{2} + c^{2}  2a + 2b + 2c + 3 = 0
⇒ ( a^{2} + 1  2a ) + ( b^{2} + 1 + 2b ) + ( c^{2} + 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 = –1Correct Option: A
Given that : a^{2} + b^{2} + c^{2} = 2 (a – b – c) – 3
⇒ a^{2} + b^{2} + c^{2}  2 (a – b – c) + 3 = 0
⇒ a^{2} + b^{2} + c^{2}  2a + 2b + 2c + 3 = 0
⇒ ( a^{2} + 1  2a ) + ( b^{2} + 1 + 2b ) + ( c^{2} + 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.
 If a, b, c are real and a^{3} + b^{3} + c^{3} = 3abc and a + b + c ≠ 0 , then the relation between a, b, c will be

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Given that :
Now ,a^{3} + b^{3} + c^{3} = 3abc
( a+b+c )( a^{2} + b^{2} + c^{2}  ab  bc  ca ) + 3abc = 3abc
( a+b+c )( a^{2} + b^{2} + c^{2}  ab  bc  ca ) = 0
∴ a² + b² + c²  ab  bc  ca = 0 or a + b + c ≠ 0Correct Option: C
Given that :
Now ,a^{3} + b^{3} + c^{3} = 3abc
( a+b+c )( a^{2} + b^{2} + c^{2}  ab  bc  ca ) + 3abc = 3abc
( a+b+c )( a^{2} + b^{2} + c^{2}  ab  bc  ca ) = 0
∴ a² + b² + c²  ab  bc  ca = 0 or a + b + c ≠ 0
Where ,
a = b = c