From the given equations , we have
Ⅰ. 2x² + 11x + 14 = 0
⇒ 2x² + 7x + 4x + 14 = 0
⇒ x( 2x + 7 ) + 2( 2x + 7 ) = 0

Ⅱ. 4y² + 12y + 9 = 0
⇒ ( 2y + 3 )² = 0

Correct Option: C

From the given equations , we have
Ⅰ. 2x² + 11x + 14 = 0
⇒ 2x² + 7x + 4x + 14 = 0
⇒ x( 2x + 7 ) + 2( 2x + 7 ) = 0
⇒ ( x + 2 )( 2x + 7 ) = 0

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

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 .

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

The given quadratic equation is 3x² - 10x + 3 = 0
Comparing with ax² + bx + c = 0, we get
a = 3, b = −10, c = 3

Correct Option: A

The given quadratic equation is 3x² - 10x + 3 = 0
Comparing with ax² + bx + c = 0, we get
a = 3, b = −10, c = 3