According to question ,we have
From equation Ⅰ.
x² = 729
⇒ x = ± √729 = ± 27

From equation Ⅱ.
y = √729

Correct Option: D

According to question ,we have
From equation Ⅰ. x² = 729
⇒ x = ± √729 = ± 27
From equation Ⅱ. y = √729
⇒ y = 27
Hence , x ≤ y is required answer .

According to question ,we have
From equation Ⅰ. x² – 4 = 0

From equation Ⅱ. y² + 6y + 9 = 0
⇒ ( y + 3 )² = 0

Correct Option: A

According to question ,we have
From equation Ⅰ. x² – 4 = 0
⇒ x = ± 2,
From equation Ⅱ. y² + 6y + 9 = 0
⇒ ( y + 3 )² = 0
⇒ ( y + 3 )( y + 3 ) = 0
⇒ y + 3 = 0 ⇒ y = -3 , -3
Clearly , x > y is required answer .

As we know that ,
For reciprocal roots, product of roots must be 1

Correct Option: C

As we know that ,
For reciprocal roots, product of roots must be 1

Let, the two natural consecutive odd numbers be n and (n + 2)
Now, according to the question,
⇒ n² + ( n + 2 )² = 394
⇒ n² + n² + 4 + 4n = 394

Correct Option: D

Let, the two natural consecutive odd numbers be n and (n + 2)
Now, according to the question,
⇒ n² + ( n + 2 )² = 394
⇒ n² + n² + 4 + 4n = 394
⇒ 2n² + 4n - 390= 0
⇒ n² + 2n - 195 = 0
⇒ n² + 15n - 13n - 195 = 0
⇒ n( n + 15 ) - 13( n + 15 ) = 0
⇒ ( n + 15 ) ( n - 13 ) = 0
⇒ n = 13 and n ≠ - 15
∴ The two natural consecutive odd numbers be 13 and (13 + 2) = 15 .
∴ the sum of the numbers = 13 + 15 = 28
Quicker Approach:
By mental operation, 13² + 15² = 169 + 225 = 394
∴ Required sum = 13 + 15 = 28

According to question ,we can say that
Let , x = √30 + √30 + √30 +........
On squaring both sides, we have
x² = 30 + √30 + √30 + √30 +........
⇒ x² = 30 + x ⇔ x² - x - 30 = 0
⇒ x² - 6x + 5x - 30 = 0

Correct Option: C

According to question ,we can say that
Let , x = √30 + √30 + √30 +........
On squaring both sides, we have
x² = 30 + √30 + √30 + √30 +........
⇒ x² = 30 + x ⇔ x² - x - 30 = 0
⇒ x² - 6x + 5x - 30 = 0
⇒ x( x - 6 ) + 5( x - 6 ) = 0
⇒ ( x - 6 ) ( x + 5 ) = 0
⇒ x = 6 because x ≠ - 5
Hence , required answer is 6 .