Quadratic Equation
Direction: In the following questions two equations numbered I and II are given. You have to solve both the equations and give answer
( 1 ) if x > y
( 2 ) if x ≥ y
( 3 ) if x < y
( 4 ) if x ≤ y
( 5 ) if x = y or the relationship cannot be established.
 Ⅰ. 5x + 2y = 31
Ⅱ. 3x + 7y = 36

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Given that : Ⅰ. 5x + 2y = 31 .....( 1 )
Ⅱ. 3x + 7y = 36 .....( 2 )Correct Option: A
Given that : Given that : Ⅰ. 5x + 2y = 31 .....( 1 )
Ⅱ. 3x + 7y = 36 .....( 2 )
Solving these two linear equations, we get x = 5, y = 3 .
∴ x > y is correct answer .
 Ⅰ. 2x^{2} + 11x + 14 = 0
Ⅱ. 4y^{2} + 12y + 9 = 0

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From the given equations , we have
Ⅰ. 2x^{2} + 11x + 14 = 0
⇒ 2x^{2} + 7x + 4x + 14 = 0
⇒ x( 2x + 7 ) + 2( 2x + 7 ) = 0
Ⅱ. 4y^{2} + 12y + 9 = 0
⇒ ( 2y + 3 )^{2} = 0Correct Option: C
From the given equations , we have
Ⅰ. 2x^{2} + 11x + 14 = 0
⇒ 2x^{2} + 7x + 4x + 14 = 0
⇒ x( 2x + 7 ) + 2( 2x + 7 ) = 0
⇒ ( x + 2 )( 2x + 7 ) = 0⇒ x = 7 or 2 2
Ⅱ. 4y^{2} + 12y + 9 = 0
⇒ ( 2y + 3 )^{2} = 0
⇒ ( 2y + 3 )( 2y + 3 ) = 0
From above both equations it is clear that x < y is correct answer .⇒ y = 3 , 3 2 2
 Ⅰ. x^{2}  7x + 12 = 0
Ⅱ. y^{2}  12y + 32 = 0

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As per the given above equations , we have
From equation Ⅰ. x^{2}  7x + 12 = 0
⇒ x^{2}  4x  3x + 12 = 0
Ⅱ. y^{2}  12y + 32 = 0
⇒ y^{2}  8y  4y + 32 = 0Correct Option: D
As per the given above equations , we have
From equation Ⅰ. x^{2}  7x + 12 = 0
⇒ x^{2}  4x  3x + 12 = 0
⇒ x( x  4)  3( x  4) = 0
⇒ ( x  4) ( x  3) = 0
⇒ x = 4 or, 3
Ⅱ. y^{2}  12y + 32 = 0
⇒ y^{2}  8y  4y + 32 = 0
⇒ y( y  8)  4( y  8) = 0
⇒ ( y  8) ( y  4) = 0
⇒ y = 4 or, 8
Thus , x ≤ y is required answer .
 Ⅰ. x^{2} = 729
Ⅱ. y = √729

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According to question ,we have
From equation Ⅰ.
x^{2} = 729
⇒ x = ± √729 = ± 27
From equation Ⅱ.
y = √729Correct Option: D
According to question ,we have
From equation Ⅰ. x^{2} = 729
⇒ x = ± √729 = ± 27
From equation Ⅱ. y = √729
⇒ y = 27
Hence , x ≤ y is required answer .
 The equation x^{2} − px + q = 0, p, q ∈ R has on real root if :

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According to question , we can say that
The equation x^{2} − px + q = 0, p, q ∈ R has no real root if B^{2} < 4ACCorrect Option: B
According to question , we can say that
The equation x^{2} − px + q = 0, p, q ∈ R has no real root
On comparing with quadratic eq. Ax^{2} + Bx + C = 0 , we get
∴ A =1, B=  p, C = q
Now ,we have B^{2} < 4AC
⇒ (  P )^{2} < 4 x 1 x q
⇒ p^{2} < 4q.
Hence , option B is correct answer .