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

In the following question two equations I and II are given. You have to solve both the equations and give answer. निम्नलिखित प्रश्न में दो समीकरणों I और II दिए गए हैं।इन दोनों समीकरणों को हल करके निम्नानुसार उत्तर दीजिए। (I) (x + 3)2 - 25 = 0 (II) y2 - 14y + 49 = 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 the relationship cannot be determined यदि x = y या संबंध स्थापित नहीं किया जा सकता
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Solving (x + 3)² - 25 = 0 gives (x + 3)² = 25, so x + 3 = 5 or x + 3 = -5. Therefore x = 2 or x = -8. Solving y² - 14y + 49 = 0 gives (y - 7)² = 0, so y = 7 (repeated root). Comparing the values: both x = 2 and x = -8 are less than y = 7. Therefore x < y is always true.

Multiple choice

Quantity 1 5946x2+991x-11892=0 Quantity 2 5994y2+999y-11988=0 मात्रा 1 5946x2+991x-11892=0 मात्रा 2 5994y2+999y-11988=0

  1. Quantity I > Quantity II मात्रा I > मात्रा II

  2. Quantity I ≥ Quantity II मात्रा I ≥ मात्रा II

  3. Quantity II > Quantity I मात्रा II > मात्रा I

  4. Quantity II ≥ Quantity I मात्रा II ≥ मात्रा I

  5. Quantity I = Quantity II or Relation cannot be established मात्रा I = मात्रा II या सम्बन्ध स्थापित नहीं किया जा सकता है
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Quantity 1: 5946x² + 991x - 11892 = 0. Using quadratic formula: x = [-991 ± sqrt(991² - 4(5946)(-11892))]/(2×5946) = [-991 ± sqrt(982081 + 282783768)]/11892 = [-991 ± sqrt(283765849)]/11892. sqrt(283765849) ≈ 16846. So x ≈ (-991 ± 16846)/11892. x₁ ≈ 15855/11892 ≈ 1.33, x₂ ≈ -17837/11892 ≈ -1.5. Quantity 2: 5994y² + 999y - 11988 = 0. y = [-999 ± sqrt(999² - 4(5994)(-11988))]/(2×5994) = [-999 ± sqrt(998001 + 287481408)]/11988 = [-999 ± sqrt(288479409)]/11988. sqrt(288479409) ≈ 16985. y ≈ (-999 ± 16985)/11988. y₁ ≈ 15986/11988 ≈ 1.33, y₂ ≈ -17984/11988 ≈ -1.5. Both equations have approximately the same roots: x₁ ≈ y₁ and x₂ ≈ y₂. Since both quantities represent the same values, Quantity I = Quantity II. Answer E is correct.

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. $I. (x^2 + 4x - 77 = 0)$ $II. (y^2 + 28y + 195 = 0)$

  1. X > Y

  2. X < Y

  3. X ≤ Y

  4. X ≥ Y

  5. X = Y or the relationship can’t be established
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Solve x²+4x-77=0: (x+11)(x-7)=0, so x=-11 or x=7. Solve y²+28y+195=0: (y+15)(y+13)=0, so y=-13 or y=-15. For x=-11 with y=-13: -11>-13. For x=-11 with y=-15: -11>-15. For x=7 with y=-13: 7>-13. For x=7 with y=-15: 7>-15. In all cases x > y, so option A is correct.

Multiple choice

In each of the following questions, two equations are given. You have to solve them and state the correct relationship- निम्नलिखित प्रश्नों में दो समीकरण (I) और (II) दिये गये है दोनों समीकरण हल कीजिए और सही सम्बन्ध स्थापित कीजिये- I. 3x2 + 17x = - 10 II. 10y2 + 9y + 2 = 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 determined यदि x=y या सम्बन्ध स्थापित नहीं किया जा सकता है।
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Solving equation I: 3x^2 + 17x = -10 gives 3x^2 + 17x + 10 = 0, so x = -2/3 or -5. Solving equation II: 10y^2 + 9y + 2 = 0 gives y = -1/2 or -2/5. For x = -2/3 and y = -1/2: -0.67 < -0.5, so x < y. For x = -5 and y = -2/5: -5 < -0.4, so x < y. In all cases x < y, meaning x ≤ y is correct.

Multiple choice

In following questions two equations (I) and (II) are given you have to solve the questions and find the values of the variables and answer the given questions according to below given pattern.\newline I. (17x^2 + 48x = 9)\newline II. (y^2 - 30y + 225 = 0)
  1. $\(x < y\)$
  2. $\(x \le y\)$
  3. $\(x > y\)$
  4. $\(x \ge y\)$

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

For equation I: 17x² + 48x - 9 = 0. Using quadratic formula: x = (-48 ± √(2304+612))/34 = (-48 ± √2916)/34 = (-48 ± 54)/34. So x = 6/34 ≈ 0.176 or x = -102/34 = -3. For equation II: y² - 30y + 225 = 0. This is (y-15)² = 0 (recognizing 225 = 15² and 30 = 2×15). So y = 15. Both values of x (-3, 0.176) are less than y = 15. Therefore x < y is correct.

Multiple choice

In following questions two equations (I) and (II) are given you have to solve the questions and find the values of the variables and answer the given questions according to below given pattern. I. (4x^2 + 23x + 15 = 0) II. (4y^2 – 14y + 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
A Correct answer
Explanation

For equation I: 4x² + 23x + 15 = 0. Factoring: (4x + 3)(x + 5) = 0. So x = -3/4 = -0.75 or x = -5. For equation II: 4y² - 14y + 10 = 0. Divide by 2: 2y² - 7y + 5 = 0. Factoring: (2y-5)(y-1) = 0. So y = 2.5 or y = 1. Both values of x (-0.75, -5) are less than both values of y (1, 2.5). Therefore x < y is correct.

Multiple choice

In following questions two equations (I) and (II) are given you have to solve the questions and find the values of the variables and answer the given questions according to below given pattern. I. 2x2 – 13x + 20 = 0 II. 3y2 – 19y + 28 = 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

For equation I: 2x² - 13x + 20 = 0. Using quadratic formula: x = (13 ± √(169-160))/4 = (13 ± 3)/4. So x = 4 or x = 2.5. For equation II: 3y² - 19y + 28 = 0. Using quadratic formula: y = (19 ± √(361-336))/6 = (19 ± √25)/6 = (19 ± 5)/6. So y = 4 or y ≈ 2.33. When x = 4, x = y. When x = 2.5, x > y (2.33). When y = 4, x = y. When y = 2.33, x > y. So x is either equal to or greater than y, meaning x ≥ y 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 4x2 – 7x + 3 = 0 II y2 – 2y + 1 = 0

  1. If x>y

  2. If x≥y

  3. If x<y

  4. If x≤y

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

Solving equation I: 4x² - 7x + 3 = 0 gives x = 1 or x = 0.75. Solving equation II: y² - 2y + 1 = 0 gives y = 1 (repeated root). Comparing the values: when x = 1, x = y; when x = 0.75, x < y. Therefore, x ≤ y is always true.

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. $I. (3x^2 - 15x + 12 = 0) II. (3y^2 - 19y + 20 = 0)$

  1. X > Y

  2. X < Y

  3. X ≤ Y

  4. X ≥ Y

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

Equation I: 3x² - 15x + 12 = 0. Simplifying: x² - 5x + 4 = 0, so (x-1)(x-4) = 0. Values: x = 1 or x = 4. Equation II: 3y² - 19y + 20 = 0. Solving: y = [19 ± √(361 - 240)]/6 = [19 ± √121]/6 = [19 ± 11]/6. So y = 30/6 = 5 or y = 8/6 = 4/3. Comparing: When x=1, y=5 gives xy. The relationship varies, so it cannot be established.

Multiple choice

There are two quantities provided in the following questions. You have to find out the values of the quantities and compare them according to the given codes as follows:-\nQuantity I: (7x^2 - 34x - 5 = 0) \nQuantity II: (2y^2 - 3y - 14 = 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 the relation cannot be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Solve both quadratic equations using the quadratic formula. For Quantity I: 7x² - 34x - 5 = 0 gives x = 5 or x = -1/7. For Quantity II: 2y² - 3y - 14 = 0 gives y = 3.5 or y = -2. When x = 5, x > y (both values). When x = -1/7, x < y (both values). Since the relationship varies, no consistent comparison exists.

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. $I. (x^3 = -70496)$ $II. (y^2 = 7921)$

  1. X > Y

  2. X < Y

  3. X ≤ Y

  4. X ≥ Y

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

For equation I: x³ = -70496, so x ≈ -41.3 (negative because cube of negative is negative). For equation II: y² = 7921, so y = ±89 (positive or negative square root). Since y can be either +89 or -89, we cannot establish a definite relationship between x and y - sometimes x < y, sometimes x > y.

Multiple choice

In the given question, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate answer- I. x = √529 II. y2 -21y + 104 = 0

  1. x > y

  2. x < y

  3. x ≥ y

  4. x ≤ y

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

Equation I: x = √529 = 23. Equation II: y² - 21y + 104 = 0. Using quadratic formula: y = (21 ± √(441 - 416))/2 = (21 ± √25)/2 = (21 ± 5)/2. Thus y = 13 or y = 8. Since x = 23, and both y values (13, 8) are less than 23, we have x > y. Option A is correctly marked. Note: Option C incorrectly shows the same text as A but is marked wrong.

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.\newline I. (x^2 - 17x - 84 = 0)\newline II. (y^2 + 4y - 117 = 0)

  1. X > Y

  2. X < Y

  3. X ≤ Y

  4. X ≥ Y

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

Equation I: x² - 17x - 84 = 0 gives x = (17 ± 25)/2 = 21 or -4. Equation II: y² + 4y - 117 = 0 gives y = (-4 ± 22)/2 = 9 or -13. Comparing: When x = 21, y could be 9 (x > y) or -13 (x > y). When x = -4, y could be 9 (x < y) or -13 (x > y). The relationship varies, so it 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.\newline I. (x^2 = 81)\newline II. ((y - 9)^2 = 0)

  1. X > Y

  2. X < Y

  3. X ≤ Y

  4. X ≥ Y

  5. X = Y or the relationship can’t be established
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Equation I: x² = 81 gives x = 9 or -9. Equation II: (y - 9)² = 0 gives y = 9. Comparing: When x = 9, y = 9 (x = y). When x = -9, y = 9 (x < y). So x is either equal to or less than y, meaning x ≤ y.

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) 17x2+48x=9 (II) 13y2=32y-12

  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
C Correct answer
Explanation

Equation I: 17x^2 + 48x - 9 = 0 gives x = -3 or x = 9/17. Equation II: 13y^2 - 32y + 12 = 0 gives y = 2 or y = 6/13. Comparing: -3 < 2 and -3 < 6/13; 9/17 < 2 but 9/17 > 6/13. For all valid comparisons, x < y.