Quantitative Aptitude

Quadratic Equations

367 Questions
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Quadratic Equations Questions

Multiple choice

In these questions, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate option.\newline I. (4x^2 - 17x + 18 = 0)\newline II. (2y^2 - 21y + 40 = 0)

  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
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Solve equation I: 4x² - 17x + 18 = 0 factors to (4x-9)(x-2) = 0, so x = 9/4 = 2.25 or x = 2. Solve equation II: 2y² - 21y + 40 = 0 factors to (2y-5)(y-8) = 0, so y = 5/2 = 2.5 or y = 8. Comparing values: x can be 2 or 2.25, while y can be 2.5 or 8. When x = 2 or 2.25, both values are less than both y values (2.5 and 8). Therefore x < y always.

Multiple choice

In each of the following question, a question is followed by information given in two statement named as Quantity-I(Q1) and Quantity –II (Q2). You have to study the information along with the question and compare the value derived from Quantity-I and Quantity-II. Then answer. Given, (A+B=12) and (A^2 + B^2=74) Quantity I: (A^3 + B^3) Quantity II: (22^2)

  1. Q1 > Q2

  2. Q1 = Q2

  3. Q1 < Q2

  4. Relationship cannot be established

  5. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given A+B=12 and A²+B²=74. First, find AB: (A+B)² = A²+2AB+B², so 144 = 74 + 2AB, giving AB = 35. For A³+B³, use identity: A³+B³ = (A+B)(A²-AB+B²) = 12(74-35) = 12(39) = 468. Quantity I is 468. Quantity II is 22² = 484. Since 468 < 484, Q1 < Q2, making option C correct.

Multiple choice

In these questions, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate option. $I. (2x^{2} – 7x + 3 = 0)$ $II. (2y^{2} – 7y + 6 = 0)$

  1. If x > y

  2. If x < y

  3. If x ≥ y

  4. If x ≤ y

  5. If x = y or relationship between x and y cannot be determined
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Equation I: 2x² - 7x + 3 = 0 factors as (2x-1)(x-3)=0, so x = 1/2 or x = 3. Equation II: 2y² - 7y + 6 = 0 factors as (2y-3)(y-2)=0, so y = 3/2 or y = 2. Comparing values: x values are 0.5 and 3, y values are 1.5 and 2. When x=0.5, xy for y=1.5 and x>y for y=2. Since we can have x>y or x

Multiple choice

In these questions, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate option. $I. (4x^2 + 16x + 15 = 0) \nII. (2y^2 + 3y + 1 = 0)$

  1. If x > y

  2. If x < y

  3. If x ≥ y

  4. If x ≤ y

  5. If x = y or relationship between x and y cannot be determined
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Equation I: 4x² + 16x + 15 = 0 factors as (2x+3)(2x+5)=0, so x = -3/2 or x = -5/2. These are x = -1.5 and x = -2.5. Equation II: 2y² + 3y + 1 = 0 factors as (2y+1)(y+1)=0, so y = -1/2 or y = -1. These are y = -0.5 and y = -1. Comparing: -1.5 < -0.5 (x

Multiple choice

In these questions, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate option. $I. (9x^2 - 45x + 56 = 0) (II. 4y^2 - 17y + 18 = 0)$

  1. If x > y

  2. If x < y

  3. If x ≥ y

  4. If x ≤ y

  5. If x = y or relationship between x and y cannot be determined
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Equation I: 9x² - 45x + 56 = 0. Using quadratic formula: x = (45 ± √(2025-2016))/18 = (45 ± 3)/18, so x = 48/18 = 8/3 ≈ 2.67 or x = 42/18 = 7/2 = 3.5. Equation II: 4y² - 17y + 18 = 0. Using quadratic formula: y = (17 ± √(289-288))/8 = (17 ± 1)/8, so y = 18/8 = 9/4 = 2.25 or y = 16/8 = 2. Comparing: 8/3 > 9/4 (2.67 > 2.25), 8/3 > 2, 7/2 > 9/4 (3.5 > 2.25), 7/2 > 2. In all cases, x > y. Option A is correct.

Multiple choice

In each of the following questions, two equations are given, you have to solve them and-\newline I. (y^2 - 0.7y + 0.12 = 0)\newline II. (x^2 + 0.1 \times x - 0.12 = 0)

  1. x < y

  2. x > y

  3. x ≥ y

  4. x ≤ y

  5. x = y or relationship cannot be established.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Solve each quadratic equation. For I (y²-0.7y+0.12=0): factors as (y-0.4)(y-0.3)=0, so y=0.4 or 0.3. For II (x²+0.1x-0.12=0): factors as (x+0.4)(x-0.3)=0, so x=0.4 or -0.3. Comparing all four combinations, x is always less than or equal to y.

Multiple choice

(\text{In each of the following questions, two equations are given. You have to solve them})(\text{(I)}) (3x^2 - 4x - 4 = 0 )(\text{(II)}) (3y^2 + 14y + 8 = 0)
  1. $\(x > y\)$
  2. $\(x < y\)$
  3. $\(x \ge y\)$
  4. $\(x \le y\)$
  5. $\(x = y \text{ or relation can't be established}\)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Solve (I): 3x² - 4x - 4 = 0 gives x = 2 or x = -2/3. Solve (II): 3y² + 14y + 8 = 0 gives y = -2/3 or y = -4. Comparing: 2 ≥ -2/3, 2 ≥ -4, -2/3 ≥ -2/3, -2/3 ≥ -4. So x ≥ y in all cases.

Multiple choice

In each question two equations numbered I and II are given, you have to solve both the equation and state the correct relationship. $I. (5x^2 + 28x = -15)$ $II. (3y^2 + 11y + 6 = 0)$
  1. $\(x > y\)$
  2. $\(x < y\)$
  3. $\(x \le y\)$
  4. $\(x \ge y\)$

  5. X = Y or the relationship cannot be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Solve equation I: 5x² + 28x + 15 = 0 gives x = -0.6, -5. Solve equation II: 3y² + 11y + 6 = 0 gives y = -2/3, -3. Since x has values both less than and greater than y (e.g., -5 < -3 and -0.6 > -2/3), no consistent inequality relationship exists.

Multiple choice

In each question two equations numbered I and II are given, you have to solve both the equation and state the correct relationship. I. (x^2 + 30x + 81 = 0) II. (y^2 - 9y - 162 = 0)

  1. X > Y

  2. X < Y

  3. X ≤ Y

  4. X ≥ Y

  5. X = Y or the relationship cannot be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Solve equation I: x² + 30x + 81 = 0 gives x = -3, -27. Solve equation II: y² - 9y - 162 = 0 gives y = -9, 18. The value sets overlap and interleave (x has both negative values, y has one negative and one positive), so no single relationship like x > y or x < y holds for all pairs.

Multiple choice

In each question two equations numbered I and II are given, you have to solve both the equation and state the correct relationship. $I. (x^2 = 4)$ $II. (y^2 = y)$
  1. $\(X > Y\)$
  2. $\(X < Y\)$
  3. $\(X \le Y\)$
  4. $\(X \ge Y\)$

  5. X = Y or the relationship cannot be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Equation I: x² = 4 gives x = 2, -2. Equation II: y² = y means y² - y = 0, so y(y-1) = 0, giving y = 0, 1. The values are {2, -2} and {0, 1}. Since x has values both above and below y (2 > 1 and -2 < 0), no consistent relationship exists.

Multiple choice

In each question two equations numbered I and II are given. You have to solve both the equations and mark the appropriate answer. (I) (x^{2}-49x+594=0) (II) (y^{2}-58y+837=0)

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

  5. If x = y or no relation can be established between x and y.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Solving (I): x² - 49x + 594 = 0 gives x = 27, 22 (factors 27 and 22). Solving (II): y² - 58y + 837 = 0 gives y = 27, 31 (factors 27 and 31). Maximum x value is 27, minimum y value is 27. Since all x values (27, 22) are less than or equal to all y values (27, 31), x ≤ y is correct.

Multiple choice

In each of the following question, two equations (I) and (II) are given. You have to solve them and answer the question. I. 4x2 = 49 II. 9y2 – 66y + 121 = 0

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

  5. If x = y or relationship cannot be established
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Solve 4x² = 49 to get x = ±3.5. For 9y² - 66y + 121 = 0, note this is (3y - 11)² = 0, giving y = 11/3 ≈ 3.67. Since x = ±3.5 and y ≈ 3.67, we have x < y (as 3.5 < 3.67 and -3.5 < 3.67). The relationship is consistently x < y.

Multiple choice

In each question two equations numbered I and II are given, you have to solve both the equation and choose the correct answer. $I. (x^2 - 40x + 391 = 0)$ $II. (y^2 - 8y - 345 = 0)$
  1. $\(x > y\)$
  2. $\(x < y\)$
  3. $\(x \le y\)$
  4. $\(x \ge y\)$

  5. x = y or the relationship can’t be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Solving equation I: x² - 40x + 391 = 0 factors to (x-17)(x-23) = 0, so x = 17 or 23. Solving equation II: y² - 8y - 345 = 0 gives y = (8 ± √1444)/2 = (8 ± 38)/2, so y = 23 or -15. When x = 17, we have 17 > -15 but 17 < 23. When x = 23, we have 23 > -15 and 23 = 23. Since x can be less than, greater than, or equal to y depending on which values are chosen, the relationship cannot be established.

Multiple choice

In each question two equations numbered I and II are given, you have to solve both the equation and choose the correct answer. I. (x^2 + 42x + 405 = 0) II. (y^2 + 45y + 506 = 0)
  1. $\(x > y\)$
  2. $\(x < y\)$
  3. $\(x \le y\)$
  4. $\(x \ge y\)$
  5. $\(x = y\) or the relationship can’t be established$
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Solving equation I: x² + 42x + 405 = 0. Using quadratic formula: x = (-42 ± √(1764-1620))/2 = (-42 ± √144)/2 = (-42 ± 12)/2. So x = -15 or x = -27. Solving equation II: y² + 45y + 506 = 0 gives y = (-45 ± √(2025-2024))/2 = (-45 ± 1)/2. So y = -22 or y = -23. Comparing: -15 > -22 (x > y), -15 > -23 (x > y), -27 < -22 (x < y), -27 < -23 (x < y). Since x can be greater or less than y, the relationship cannot be established.

Multiple choice

In each of the following question, two equations are given. You have to solve them and give answer- निम्नलिखित प्रश्न में दो समीकरण (I) और (II) दिये गये है दोनों समीकरण हल कीजिए और उत्तर दीजिये - I. 6x2 - 7x + 1 = 0 II. y2 -2y +1 = 0

  1. If x>y यदि x>y

  2. If x≥y यदि x≥y

  3. If x<y यदि x<y

  4. If x≤y यदि x≤y

  5. If x=y or relationship can not be established यदि x=y या सम्बन्ध स्थापित नहीं किया जा सकता है।
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Solve equation I: 6x² - 7x + 1 = 0 factors as (2x-1)(3x-1)=0, giving x = 1/2 or x = 1/3. Solve equation II: y² - 2y + 1 = 0 is (y-1)² = 0, giving y = 1. For both values of x (0.5 and 0.33), we have x < y. Therefore x ≤ y is true, making option D correct.