Quadratic Equation
Direction: In each of these questions, two equations 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 no relationship can be established between x and y.
- Ⅰ. x2 -24x + 144 = 0
Ⅱ. y2 − 26y + 169 = 0
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As per the given above question , we have
From equation Ⅰ. x2 - 24x + 144 = 0
⇒ x2 − 12x - 12x + 144 = 0
From equation Ⅱ. y2 − 26y + 169 = 0
⇒ y2 − 13y − 13y + 169 = 0Correct Option: A
As per the given above question , we have
From equation Ⅰ. x2 - 24x + 144 = 0
⇒ x2 − 12x - 12x + 144 = 0
⇒ x(x − 12) − 12(x − 12) = 0
⇒ (x − 12)2 = 0
∴ x = 12
From equation Ⅱ. y2 − 26y + 169 = 0
⇒ y2 − 13y − 13y + 169 = 0
⇒ y(y − 13) − 13(y − 13) = 0
⇒ (y − 13)2 = 0
∴ y = 13
Hence, required answer will be x < y .
- Ⅰ. 2x2 + 3x – 20 = 0
Ⅱ. 2y2 + 19y + 44 = 0
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According to question ,
From equation Ⅰ. 2x2 + 3x– 20 = 0
⇒ 2x2 + 8x − 5x − 20 = 0
⇒ 2x(x + 4) − 5(x + 4) = 0
From equation Ⅱ. 2y2 + 19y + 44 = 0
⇒ 2y2 + 11y + 8y + 44 = 0
⇒ 2y(2y + 11) + 4(2y + 11) = 0Correct Option: D
According to question ,
From equation Ⅰ. 2x2 + 3x– 20 = 0
⇒ 2x2 + 8x − 5x − 20 = 0
⇒ 2x(x + 4) − 5(x + 4) = 0
⇒ (2x − 5)(x + 4) = 0∴ x = 5 , - 4 2
From equation Ⅱ. 2y2 + 19y + 44 = 0
⇒ 2y2 + 11y + 8y + 44 = 0
⇒ 2y(2y + 11) + 4(2y + 11) = 0
⇒ (y + 4)(2y + 11) = 0
Hence, x ≥ y is correct answer .∴ y = - 4, - 5 2
- Ⅰ. 10x2 − 7x + 1 = 0
Ⅱ. 35y2 -12y + 1 = 0
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From the above given equations , we have
Ⅰ. 10x2 − 7x + 1 = 0
⇒ 10x2 − 5x − 2x + 1 = 0
⇒ 5x(2x − 1) − 1(2x − 1) = 0
Ⅱ. 35y2 - 12y + 1 = 0
⇒ 35y2 − 7y + 5y + 1 = 0
⇒ 7y(5y − 1) − 1(5y − 1) = 0Correct Option: D
From the above given equations , we have
Ⅰ. 10x2 − 7x + 1 = 0
⇒ 10x2 − 5x − 2x + 1 = 0
⇒ 5x(2x − 1) − 1(2x − 1) = 0
⇒ (5x − 1)(2x − 1) = 0∴ x = 1 , 1 5 2
Ⅱ. 35y2 - 12y + 1 = 0
⇒ 35y2 − 7y + 5y + 1 = 0
⇒ 7y(5y − 1) − 1(5y − 1) = 0
⇒ (7y − 1)(5y − 1) = 0
Hence, required answer is x ≥ y .∴ y = 1 , 1 7 5
- Ⅰ. 6x2 + 77x + 121 = 0
Ⅱ. y2 + 9y − 22 = 0
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From the given above equations , we have
Ⅰ. 6x2 + 77x + 121 = 0
⇒ 6x2 + 66x + 11x + 121 = 0
Ⅱ. y2 + 9y − 22 = 0
⇒ y2 + 11y − 2y − 22 = 0Correct Option: E
From the given above equations , we have
Ⅰ. 6x2 + 77x + 121 = 0
⇒ 6x2 + 66x + 11x + 121 = 0
⇒ 6x(x + 11) + 11(x + 11) = 0
⇒ (6x + 11)(x + 11) = 0∴ x = - 11 , - 11 6
Ⅱ. y2 + 9y − 22 = 0
⇒ y2 + 11y − 2y − 22 = 0
⇒ y(y + 11) - 2(y + 11)
⇒ (y − 2)(y + 11) = 0
⇒ y = 2, –11
Hence, no relationship can be established between x and y.
- Ⅰ. x2 − 6x = 7
Ⅱ. 2y2 + 13y + 15 = 0
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As per the given above question , we have
From equation Ⅰ.
x2 − 6x = 7
⇒ x2 − 6x - 7 = 0
⇒ x2 − 7x + x − 7 = 0
From equation Ⅱ.
2y2 + 13y + 15 = 0
2y2 + 10y + 3y + 15 = 0
⇒ 2y(y + 5) + 3(y + 5) = 0Correct Option: B
As per the given above question , we have
From equation Ⅰ. x2 − 6x = 7
⇒ x2 − 6x - 7 = 0
⇒ x2 − 7x + x − 7 = 0
⇒ x(x − 7) + 1(x − 7) = 0
⇒ (x + 1)(x − 7) = 0
⇒ x = -1, 7
From equation Ⅱ. 2y2 + 13y + 15 = 0
2y2 + 10y + 3y + 15 = 0
⇒ 2y(y + 5) + 3(y + 5) = 0
⇒ (2y + 3)(y + 5) = 0∴ y = - 3 , - 5 2
Hence, x > y is correct answer .
