Quantitative Aptitude

Quadratic Equations

367 Questions
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Quadratic Equations Questions

Multiple choice

Two equations (I) and (II) are given in each question. On the basis of these equations you have to decide the relation between ‘x’ and ‘y’ and give answer. (I: x^2 -7\sqrt{7}x+84=0 ) (II: y^2-5\sqrt{5}y+30=0)

  1. x > y

  2. x ≥ y

  3. x < y

  4. x ≤ y

  5. x = y or the relation cannot be established
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Equation I: x² - 7√7x + 84 = 0 gives x = 4√7 or 3√7. Equation II: y² - 5√5y + 30 = 0 gives y = 3√5 or 2√5. Comparing: 4√7 > 3√5, 4√7 > 2√5, 3√7 > 3√5, 3√7 > 2√5. Thus x > y always.

Multiple choice

Two equations (I) and (II) are given in each question. On the basis of these equations you have to decide the relation between ‘x’ and ‘y’ and give answer.(I: x - 7\sqrt{3}\sqrt{x} + 36 = 0 )(II: y - 12\sqrt{2}\sqrt{y} + 70 = 0)

  1. x > y

  2. x ≥ y

  3. x < y

  4. x ≤ y

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

Let a = √x, b = √y. Equation I: a² - 7√3a + 36 = 0 gives a = 3√3 or 4√3, so x = 27 or 48. Equation II: b² - 12√2b + 70 = 0 gives b = 5√2 or 7√2, so y = 50 or 98. Comparing all combinations, y > x in all cases.

Multiple choice

Find the values of variables and state the correct relationship- (x^2 - 2x - 35 = 0) (y^2 - 12y + 32 = 0)
  1. $\(x < y\)$
  2. $\(x > y\)$
  3. $\(x \le y\)$
  4. $\(x \ge y\)$

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

Solving x² - 2x - 35 = 0 gives x = 7 or x = -5. Solving y² - 12y + 32 = 0 gives y = 8 or y = 4. When x = 7: x < y for y=8 but x > y for y=4. When x = -5: x < y for both. The relationship cannot be established uniquely.

Multiple choice

$Quantity 1: (5946x^2 + 991x - 11892 = 0)$ $Quantity 2: (5994y^2 + 999y - 11988 = 0)$

  1. Quantity I > Quantity II

  2. Quantity I ≥ Quantity II

  3. Quantity II > Quantity I

  4. Quantity II ≥ Quantity I

  5. Quantity I = Quantity II or Relation cannot be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Equation 1: 5946x² + 991x - 11892 = 0. Using quadratic formula or factoring: x = [-991 ± √(991² + 4×5946×11892)]/(2×5946). Equation 2: 5994y² + 999y - 11988 = 0. Notice both coefficients are in similar ratios. The roots will be proportional. Without calculation, we can see the relationship isn't straightforward (one may be positive, one negative, or values may differ). The relationship cannot be established without solving completely. Option E is correct.

Multiple choice

In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer. $I. (x^2 + 11x + 30 = 0) II. (y^2 + 12y + 36 = 0)$
  1. $\(x > y\)$
  2. $\(x \ge y\)$
  3. $\(x < y\)$
  4. $\(x \le y\)$

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

For equation I: x² + 11x + 30 = 0 factors to (x+5)(x+6) = 0, giving x = -5 or x = -6. For equation II: y² + 12y + 36 = 0 factors to (y+6)² = 0, giving y = -6 (repeated root). When x = -5, we have x > y. When x = -6, we have x = y. Since x can be greater than OR equal to y depending on which root we take, the relationship is x ≥ y.

Multiple choice

In the following question two equations numbered I and II are given. You have to solve both the equations and give answer thereof. $I. (3x^2 + 23x + 44 = 0) II. (3y^2 + 20y + 33 = 0)$

  1. x > y

  2. x ≥ y

  3. x < y

  4. x ≤ y

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

Equation I: 3x²+23x+44 = (3x+11)(x+4) = 0, so x = -11/3, -4. Equation II: 3y²+20y+33 = (3y+11)(y+3) = 0, so y = -11/3, -3. Compare: x = -3.67 and y = -3.67 (equal), or x = -4 and y = -3 (x < y). In all cases x ≤ y. Option D correct.

Multiple choice

In the following question two equations numbered I and II are given. You have to solve both the equations and give answer thereof. $(I. ) (4x^2 - 29x + 45 = 0)$ $(II. ) (3y^2 - 19y + 28 = 0)$
  1. $\(x > y\)$
  2. $\(x \ge y\)$
  3. $\(x < y\)$
  4. $\(x \le y\)$

  5. x = y, or relation cannot be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Equation I: 4x²-29x+45 = (4x-9)(x-5) = 0, so x = 2.25, 5. Equation II: 3y²-19y+28 = (3y-7)(y-4) = 0, so y = 7/3, 4. Compare: x=2.25 vs y=2.33 (x < y), x=2.25 vs y=4 (x < y), x=5 vs y=2.33 (x > y), x=5 vs y=4 (x > y). Since x can be less than or greater than y, relationship cannot be established. Option E correct.

Multiple choice

Two equations (I) and (II) are given in each question. On the basis of these equations you have to decide the relation between ‘x’ and ‘y’ and give answer. $I. (6x^2 - 19x + 15 = 0)$ $II. (10y^2 - 29y + 21 = 0)$
  1. $\(x > y\)$
  2. $\(x \ge y\)$
  3. $\(x < y\)$
  4. $\(x \le y\)$

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

Solving equation I: 6x² - 19x + 15 = 0. Factors: (3x-5)(2x-3) = 0, so x = 5/3 or x = 3/2. Solving equation II: 10y² - 29y + 21 = 0. Using quadratic formula: y = (29 ± √(841-840))/20 = (29 ± 1)/20, so y = 30/20 = 3/2 or y = 28/20 = 7/5. Values: x ∈ {5/3, 3/2}, y ∈ {3/2, 7/5}. Common value is 3/2. When x = 5/3, y = 7/5: 5/3 > 7/5. When x = y = 3/2, they are equal. Therefore x ≥ y. Option B is correct.

Multiple choice

Two equations (I) and (II) are given in each question. On the basis of these equations you have to decide the relation between ‘x’ and ‘y’ and give answer.(I. 8x^2 - 15x + 7 = 0 ) (II. 2y^2 - 7y + 6 = 0)
  1. $x > y$
  2. $x \ge y$
  3. $x < y$
  4. $x \le y$

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

Solve equation (I): 8x^2 - 15x + 7 = 0 factors to (8x-7)(x-1) = 0, giving x = 7/8 or x = 1. Solve equation (II): 2y^2 - 7y + 6 = 0 factors to (2y-3)(y-2) = 0, giving y = 1.5 or y = 2. For all possible values of x (0.875, 1) and y (1.5, 2), x is always less than y. Therefore x < y is the correct relation.

Multiple choice

In the following question two equations numbered I and II are given. You have to solve both the equations and give answer. $(I. 30x^2 + 11x + 1 = 0 )$ $(II. 42y^2 + 13y + 1 = 0)$

  1. x > y

  2. x < y

  3. x ≥ y

  4. x ≤ y

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

Solving equation I: 30x²+11x+1=0 factors as (5x+1)(6x+1)=0, so x=-1/5 or x=-1/6. Solving equation II: 42y²+13y+1=0 factors as (6y+1)(7y+1)=0, so y=-1/6 or y=-1/7. Comparing values: x values are -0.2 or -0.1667, y values are -0.1667 or -0.1429. The common value is -1/6=-0.1667. For x=-1/5=-0.2 and y=-1/7=-0.1429, we have x

Multiple choice

In the following question two equations numbered I and II are given. You have to solve both the equations and give answer. $I. (2x^2 + 9x + 7 = 0)$ $II. (2y^2 + 9y + 10 = 0)$

  1. x > y

  2. x < y

  3. x ≥ y

  4. x ≤ y

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

Solving equation I: 2x²+9x+7=0 factors as (2x+7)(x+1)=0, so x=-7/2 or x=-1. Solving equation II: 2y²+9y+10=0 factors as (2y+5)(y+2)=0, so y=-5/2 or y=-2. Comparing values: x=-3.5 or -1, y=-2.5 or -2. When x=-3.5 and y=-2, xy. When x=-1 and y=-2.5, x>y. Since x can be greater or less than y depending on which roots we choose, no definite relation can be established.

Multiple choice

$(Q_1^2 - 111Q_1 + 2618 = 0 \ Q_2^2 - 121Q_2 + 3420 = 0)$

  1. Quantity I > Quantity II

  2. Quantity I ≥ Quantity II

  3. Quantity I < Quantity II

  4. Quantity I ≤ Quantity II

  5. Quantity I = Quantity II or relationship cannot be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Solving Q_1^2 - 111Q_1 + 2618 = 0: Q_1 = 47 or Q_1 = 58. Solving Q_2^2 - 121Q_2 + 3420 = 0: Q_2 = 57 or Q_2 = 60. If Q_1=47 and Q_2=57: Q_1 < Q_2. If Q_1=58 and Q_2=57: Q_1 > Q_2. Since different combinations give different relationships, the correct answer is E - relationship cannot be established.