Quadratic Equation
Direction: In each of these questions, two equations are given. You have to solve these equations and find out the values of x and y and give answer
( 1 ) x < y
( 2 ) x > y
( 3 ) x ≤ y
( 4 ) x ≥ y
( 5 ) x = y
- Ⅰ. 16x2 + 20x + 6 = 0
Ⅱ. 10y2 + 38y + 24 = 0
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As per the given above equations , we can say that
Ⅰ. 16x2 + 20x + 6 = 0
⇒ 8x2 + 10x + 3 = 0
Ⅱ. 5y2 + 38y + 24 = 0
⇒ 5y2 + 19y + 12 = 0Correct Option: B
As per the given above equations , we can say that
Ⅰ. 16x2 + 20x + 6 = 0
⇒ 8x2 + 10x + 3 = 0
⇒ (4x + 3)(2x + 1) = 0
Ⅱ. 5y2 + 38y + 24 = 0∴ x = - 3 or, - 1 4 2
⇒ 5y2 + 19y + 12 = 0
∴ (y + 3)(5y + 4) = 0
Hence, required answer will be x > y .∴ y = - 3 or, - 4 5
- Ⅰ. 18x2 + 18x + 4 = 0
Ⅱ. 12y2 + 29y + 14 =0
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From the given above question , we have
From equation Ⅰ. 18x2 + 18x + 4 = 0
⇒ 9x2 + 9x + 2 = 0
From equation Ⅱ. 12y2 + 29y + 14 =0
⇒ (3y + 2)(4y + 7) = 0Correct Option: D
From the given above question , we have
From equation Ⅰ. 18x2 + 18x + 4 = 0
⇒ 9x2 + 9x + 2 = 0
⇒ (3x + 2)(3x + 1) = 0
From equation Ⅱ. 12y2 + 29y + 14 =0∴ x = - 2 or, - 1 3 3
⇒ (3y + 2)(4y + 7) = 0
Hence, x ≥ y is correct answer .∴ y = - 2 or, - 7 3 4
- Ⅰ. 4x + 7y = 209
Ⅱ. 12x – 14y = –38
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As per the given above equations , we can say that
Ⅰ. 4x + 7y = 209 ....................( 1 )
Ⅱ. 12x – 14y = –38 ...................( 2 )
Now, (1) × 2 + (2), we have
12x – 14y = – 38Correct Option: E
As per the given above equations , we can say that
Ⅰ. 4x + 7y = 209 ....................( 1 )
Ⅱ. 12x – 14y = –38 ...................( 2 )
Now, (1) × 2 + (2), we have
12x – 14y = –38
8x + 14y = 418
⇒ 20x = 380
Now, putting the value of x = 19 in equation (1), We have,∴ x = 380 = 19 20
4 × 19 + 7y = 209
or, 7y = 209 – 76 = 133
Hence, x = y is correct answer .∴ y = 133 = 19 7
- Ⅰ. 17x2 + 48x = 9
Ⅱ. 13y2 = 32y – 12
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According to question ,we have
From equation Ⅰ. 17x2 + 48x - 9 = 0
⇒ (x + 3)(17x – 3) = 0
From equation Ⅱ. 13y2 - 32y + 12 = 0
⇒ (y – 2)(13y – 6) = 0Correct Option: A
According to question ,we have
From equation Ⅰ.
17x2 + 48x - 9 = 0
⇒ (x + 3)(17x – 3) = 0∴ x = - 3 or, 3 17
From equation Ⅱ.
13y2 - 32y + 12 = 0
⇒ (y – 2)(13y – 6) = 0
Hence, required answer will be x < y .∴ y = 2 or, 6 13
- Ⅰ. 8x2 + 6x = 5
Ⅱ. 12y2 – 22y + 8 = 0
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According to question ,we can say that
Ⅰ. 8x2 + 6x - 5 = 0
⇒ (4x + 5) (2x – 1) = 0
Ⅱ. 12y2 – 22y + 8 = 0
⇒ 6y2 – 11y + 4 = 0
⇒ 2y( 3y – 4 ) - 1( 3y - 4 ) = 0Correct Option: C
According to question ,we can say that
Ⅰ. 8x2 + 6x - 5 = 0
⇒ (4x + 5) (2x – 1) = 0
Ⅱ. 12y2 – 22y + 8 = 0∴ x = - 5 or, 1 4 2
⇒ 6y2 – 11y + 4 = 0
⇒ 2y( 3y – 4 ) - 1( 3y - 4 ) = 0
⇒ (2y – 1) (3y – 4) = 0
From above both equations it is clear that x ≤ y is required answer .∴ y = 1 or, 4 2 3
