Quantitative Aptitude

Quadratic Equations

367 Questions
All Basic factorization 305 Sign scheme based 22 Algebraic substitution 19 Roots from relation 9 Finding extrema 5 Discriminant analysis 2 Finding coefficients 2 Common root 1 Finding quadratic equation 1 Reciprocal equation 1

Quadratic Equations Questions

Multiple choice

(\text{Find the correct relationship between given quantities.}\\text{Quantity I: } 2x^2 + 29x + 50 = 0\\text{Quantity II: } 3y^2 + 18y + 27 = 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

Quantity I: 2x² + 29x + 50 = 0 gives x = -2 or x = -8. Quantity II: 3y² + 18y + 27 = 0 gives y = -3. The relationship between x and y cannot be uniquely determined: if x = -2, then x > y (-2 > -3); if x = -8, then x < y (-8 < -3). Since x can be greater or less than y, the relation cannot be established.

Multiple choice

(\text{Solve both the quantities and state the correct relationship.}\\text{Quantity I:} 3Q_1^2 + 168Q_1 + 2349 = 0\\text{Quantity II:} 2Q_2^2 + 92Q_2 + 1026 = 0)
  1. $\(Q_1 > Q_2\)$
  2. $\(Q_1 \ge Q_2\)$
  3. $\(Q_1 < Q_2\)$
  4. $\(Q_1 \le Q_2\)$
  5. $\(Q_1 = Q_2 \text{ or relationship cannot be established}\)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For Q1: 3x² + 168x + 2349 = 0. Discriminant = 168² - 4(3)(2349) = 28224 - 28188 = 36. Roots = (-168 ± 6)/6 = -27.5, -28.5. For Q2: 2x² + 92x + 1026 = 0. Discriminant = 92² - 4(2)(1026) = 8464 - 8208 = 256. Roots = (-92 ± 16)/4 = -23, -27. Both Q1 roots (-27.5, -28.5) are less than both Q2 roots (-23, -27). Therefore, Q1 ≤ Q2.

Multiple choice

In each of these question, two equations (I) and (II) are given. You have to solve both the equations and give answer.(I. 7x^2 + 42x - 5040 = 0 )(II. 4y^2 + 17y - 30870 = 0)
  1. $\(x > y\)$
  2. $\(x < y\)$
  3. $\(x \le y\)$
  4. $\(x \ge y\)$
  5. $\(x = y\) or relationship cannot be established$
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

For equation I: 7x² + 42x - 5040 = 0 simplifies to x² + 6x - 720 = 0, giving x = 24 or x = -30. For equation II: 4y² + 17y - 30870 = 0 has discriminant = 289 + 493920 = 494209. The roots are approximately y ≈ 87 or y ≈ -88.75. Comparing: x = 24 < y = 87, but x = 24 > y = -88.75. The relationship varies, so it cannot be established.

Multiple choice

Two quantities that is I and II are given in following questions. Students is expected to solve the quantities and answer them according to given options by comparing their numerical values.(3^{x + 5}. 9^{2x – 4} = 9^{5x – 14})and, (2y^{2} – 15y – 28 = 3y^{2} – 23y – 13)Quantity I: Value of ‘x’ Quantity II: Value of ‘y’

  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 No relation
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Equation 1: 3^(x+5) × 9^(2x-4) = 9^(5x-14). Converting 9 to 3²: 3^(x+5) × 3^(4x-8) = 3^(10x-28). So 5x-3 = 10x-28, 5x = 25, x = 5. Equation 2: 2y² - 15y - 28 = 3y² - 23y - 13. Rearranging: y² - 8y + 15 = 0, (y-3)(y-5) = 0, so y = 3 or y = 5. Quantity I = 5, Quantity II = 3 or 5. Since x ≥ y (5 ≥ 3 and 5 ≥ 5), option C is correct.

Multiple choice

In each of the given question below, there are two equations as I and II. You have to solve both equations and give answer- I. 3x2 – 10x + 7 = 0 II. 15y2 – 22y + 8 = 0

  1. If x > y

  2. If x < y

  3. If x ≥ y

  4. If x ≤ y

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

Solving equation I: 3x² - 10x + 7 = 0 factors to (3x-7)(x-1) = 0, giving x = 7/3 or x = 1. Solving equation II: 15y² - 22y + 8 = 0 factors to (5y-4)(3y-2) = 0, giving y = 4/5 or y = 2/3. Since both x values (7/3, 1) are greater than both y values (4/5, 2/3), the relationship is x > y.

Multiple choice

In each of the given question below, there are two equations as I and II. You have to solve both equations and give answer- I. 20x2 – 17x + 3 = 0 II. 20y2 – 9y + 1 = 0

  1. If x > y

  2. If x < y

  3. If x ≥ y

  4. If x ≤ y

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

Solving equation I: 20x² - 17x + 3 = 0 factors to (5x-3)(4x-1) = 0, giving x = 3/5 or x = 1/4. Solving equation II: 20y² - 9y + 1 = 0 factors to (5y-1)(4y-1) = 0, giving y = 1/5 or y = 1/4. Since 3/5 > 1/5 and 3/5 > 1/4 (but 1/4 = 1/4), 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 choose the correct option.\newline I. (x^2 - 40x + 396 = 0)\newline II. (15y^2 + 19y - 8 = 0)

  1. x > y

  2. x < y

  3. x ≥ y

  4. x ≤ y

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

Solving x² - 40x + 396 = 0 gives x = 18 or 22. Solving 15y² + 19y - 8 = 0 gives y = 1/3 or -8/5. Both x values (18, 22) are greater than both y values (0.33, -1.6), so x > y is correct.

Multiple choice

In the following question two equations numbered I and II are given. You have to solve both the equations and choose the correct option. I. (x^2 – 21x + 98 = 0) II. (y^2 – 35y + 304 = 0)

  1. x > y

  2. x < y

  3. x ≥ y

  4. x ≤ y

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

Solving x² - 21x + 98 = 0 gives x = 7 or 14. Solving y² - 35y + 304 = 0 gives y = 16 or 19. Both x values (7, 14) are less than both y values (16, 19), so 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.\newline\textbf{I.}\space\space4x^{2} - 12x + 5 = 0 \newline\textbf{II.} 8y^{2} + 3y = 38

  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
E Correct answer
Explanation

Solving equation I: 4x² - 12x + 5 = 0 factors to (2x-1)(2x-5)=0, giving x=0.5 or x=2.5. Solving equation II: 8y² + 3y - 38 = 0 using the quadratic formula gives y = (-3 ± √1225)/16 = (-3 ± 35)/16, so y=2 or y=-19/8=-2.375. Comparing values: when x=0.5, x-2.375); when x=2.5, x>y (2.5>2). Since the relationship varies (xy for x=2.5 vs y=2), the relationship cannot be uniquely established.

Multiple choice

Find the values of (x) and (y) and compare them. I. (4x^2 + 3x - 27 = 0) II. (3y^2 - 20y + 32 = 0)

  1. x > y

  2. x < y

  3. x ≥ y

  4. x ≤ y

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

Equation I: 4x² + 3x - 27 = 0. Roots are x = 9/4 and x = -3. Equation II: 3y² - 20y + 32 = 0. Roots are y = 4/3 and y = 8. Comparing all values of x with all values of y: 9/4 = 2.25 < 4/3 ≈ 4.33 and < 8; -3 < 4/3 and -3 < 8. Therefore x < y for all cases.

Multiple choice

In the questions given below you have to find the value of x and y and then compare the value of x and y and give the answer related to this.\newline I. (4x^2 + 19x + 21 = 0), \newline II. (3y^2 – 19y – 14 = 0)
  1. $\(x > y\)$
  2. $\(x < y\)$
  3. $\(x \ge y\)$
  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

For equation I: 4x² + 19x + 21 = 0 factors to (4x+7)(x+3) = 0, giving x = -7/4 or x = -3. For equation II: 3y² - 19y - 14 = 0 factors to (3y+2)(y-7) = 0, giving y = -2/3 or y = 7. Comparing values: when x = -3, x < y for both y values; when x = -7/4, x < y for both y values. Since x is always less than y, option B (x < y) is correct.

Multiple choice

In the questions given below you have to find the value of x and y and then compare the value of x and y and give the answer related to this. $I. (3x^2 - 5x - 12 = 0), II. (3y^2 - 8y - 16 = 0)$

  1. x > y

  2. x < y

  3. x ≥ y

  4. x ≤ y

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

For equation I: 3x² - 5x - 12 = 0 factors to (3x+4)(x-3) = 0, giving x = -4/3 or x = 3. For equation II: 3y² - 8y - 16 = 0 factors to (3y+4)(y-4) = 0, giving y = -4/3 or y = 4. Comparing values: x and y both equal -4/3 in one case (x = y), but x = 3 < y = 4 in the other. Since x can be equal to y or less than y depending on which values we choose, the relation cannot be consistently established.

Multiple choice

In the questions given below you have to find the value of x and y and then compare the value of x and y and give the answer related to this.\newline I. (3x^2 + 2x - 21 = 0), \newline II. (3y^2 - 19y + 28 = 0)

  1. x > y

  2. x < y

  3. x ≥ y

  4. x ≤ y

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

For equation I: 3x² + 2x - 21 = 0 factors to (3x-7)(x+3) = 0, giving x = 7/3 or x = -3. For equation II: 3y² - 19y + 28 = 0 factors to (3y-7)(y-4) = 0, giving y = 7/3 or y = 4. Comparing values: when x = -3, x is less than both y values; when x = 7/3, x equals y in one case and is less than y = 4 in the other. In all cases, x ≤ y, so option D is correct.

Multiple choice

In the following questions two equations numbered I and II are given. You have to solve both the equations and Give answer: $I. (x^{2} – 1 = 0) II. (y^{2} + 4y + 3 = 0)$

  1. If x > y

  2. If x < y

  3. If x ≥ y

  4. If x ≤ y

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

Solve Equation I: x² - 1 = 0 gives x = ±1. Solve Equation II: y² + 4y + 3 = 0 factors to (y+3)(y+1) = 0, so y = -3 or y = -1. When x = 1, x > y (both -3 and -1). When x = -1, x = y (= -1). Therefore x ≥ y always holds. Option C is 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. (x² – 11x + 30 = 0) II. (2y² – 9y + 10 = 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
A Correct answer
Explanation

Solve equation I: x² - 11x + 30 = 0 factors to (x-5)(x-6) = 0, so x = 5 or 6. Solve equation II: 2y² - 9y + 10 = 0 factors to (2y-5)(y-2) = 0, so y = 5/2 or 2. Comparing values: x can be 5 or 6, while y can be 2 or 2.5. Both possible values of x (5 and 6) are greater than both possible values of y (2 and 2.5). Therefore x > y always.