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. x2 – 7x + 12 = 0
Ⅱ. y2 + y – 12 = 0
-
View Hint View Answer Discuss in Forum
From the given equations , we can say that
From equation Ⅰ.
x2 - 7x + 12 = 0
⇒ x2 - 4x - 3x + 12 = 0
From equation Ⅱ.
y2 + y - 12 = 0
⇒ y2 + 4y - 3y - 12 = 0Correct Option: B
From the given equations , we can say that
From equation Ⅰ.
x2 - 7x + 12 = 0
⇒ x2 - 4x - 3x + 12 = 0
⇒ x( x - 4 ) - 3( x - 4 ) = 0
⇒ ( x - 3 )( x - 4 ) = 0
⇒ x = 3 or x = 4
From equation Ⅱ.
y2 + y - 12 = 0
⇒ y2 + 4y - 3y - 12 = 0
⇒ y( y + 4 ) - 3( y + 4 ) = 0
⇒ ( y - 3 )( y + 4 ) = 0
⇒ y = 3 or y = - 4
From above both equations it is clear that x ≥ y is correct answer .
Direction: In the following questions, two equations numbered I and II are given. You have to solve both questions 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. x3 – 468 = 1729
Ⅱ. y2 – 1733 + 1564 = 0
-
View Hint View Answer Discuss in Forum
According to question,
From equation Ⅰ. x3 – 468 = 1729
⇒ x3 = 2197 = 133
From equation Ⅱ. y2 – 1733 + 1564 = 0
⇒ y2 = 169 = 132Correct Option: B
According to question,
From equation Ⅰ. x3 – 468 = 1729
⇒ x3 = 2197 = 133
⇒ x = 13
From equation Ⅱ. y2 – 1733 + 1564 = 0
⇒ y2 = 169 = 132
⇒ y = ±13
Thus , x ≥ y is required answer .
- Ⅰ. √784x + 1234 = 1486
Ⅱ. √1089y + 2081 = 2345
-
View Hint View Answer Discuss in Forum
According to question ,we can say that
From equation Ⅰ. √784x + 1234 = 1486
⇒ √784x = 252
From equation Ⅱ. √1089y + 2081 = 2345
⇒ √1089y = 264Correct Option: A
According to question ,we can say that
From equation Ⅰ. √784x + 1234 = 1486
⇒ √784x = 252
⇒ 28x = 252
⇒ x = 9
From equation Ⅱ. √1089y + 2081 = 2345
⇒ √1089y = 264
⇒ 33y = 264
⇒ y = 8
∴ x > y
Thus , required answer will be x > y .
-
Ⅰ. 9 + 19 = √x √x √x Ⅱ. y5 - ( 2 × 14 )11/2 = 0 √y
-
View Hint View Answer Discuss in Forum
As per the given above question , we have
Ⅰ. 9 + 19 = √x √x √x
⇒ 9 + 19 = √x × √x ⇒ x = 28Ⅱ. y5 - ( 2 × 14 )11/2 = 0 √y
⇒ y5 √y - ( 2 × 14 )11/2 = 0
Correct Option: E
As per the given above question , we have
Ⅰ. 9 + 19 = √x √x √x
⇒ 9 + 19 = √x × √x ⇒ x = 28Ⅱ. y5 - ( 2 × 14 )11/2 = 0 √y
⇒ y5 √y - ( 2 × 14 )11/2 = 0
⇒ y5 + ( 1/2 ) = ( 2 × 14 )11/2
⇒ y11/2 = ( 2 × 14 )11/2
⇒ y = 2 × 14 = 28
∴ x = y
From above it is clear that x = y is required answer .
-
Ⅰ. 12 - 23 = 5√x √x √x Ⅱ. √y - 5 √y = 1 12 12 √y
-
View Hint View Answer Discuss in Forum
From the given above equations , we have
Ⅰ. 12 - 23 = 5√x √x √x
⇒ 12 - 23 = 5√x × √x⇒ √y 
1 - 5 
= 1 12 12 √y
Correct Option: A
From the given above equations , we have
Ⅰ. 12 - 23 = 5√x √x √x
⇒ 12 - 23 = 5√x × √x∴ x = - 11 = - 2.2 5 Ⅰ. √y - 5 √y = 1 12 12 √y ⇒ √y 1 - 5 = 1 12 12 √y ⇒ y - 4 = 1 12
∴ x > y⇒ y = - 12 = - 3 4
From above it is clear that required answer is x > y .
