Quantitative Aptitude

Quadratic Equations

367 Questions
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Quadratic Equations Questions

Multiple choice

Find the value of variables and state the correct relationship-(2x^2+19x+35=0 )(y^2+3y+2=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

Solving 2x²+19x+35=0: Using quadratic formula or factorization, roots are x = -1.43 and x = -12.5. Solving y²+3y+2=0: Factorizing as (y+2)(y+1)=0, roots are y = -2 and y = -1. All x values (-1.43, -12.5) are less than all y values (-2, -1), therefore x < y. The relationship is consistently x < y across all root combinations.

Multiple choice

$(12x^2 + x - 13 = 0 )(6y^2 + 11y + 3 = 0)$ Find the value of variables and state the correct relationship
  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 12x² + x - 13 = 0 gives x = 1 or x = -13/12. Solving 6y² + 11y + 3 = 0 gives y = -3/2 or y = -1/3. Since x has two possible values and y has two possible values, the relationship depends on which root you take. For x=1, x > both y values. For x=-13/12, x can be greater than y=-3/2 but less than y=-1/3. Thus no single relationship holds in all cases.

Multiple choice

In the following questions, two quantities are given as Quantity I and Quantity II. Find the correct relationship.\nQuantity I: (x^3 - 133 = [56\% \text{ of } 1250] \div 4 + 113 + 124\times 9 \div 2 )\nQuantity II: (y^2 - y - 72 = 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
A Correct answer
Explanation

Quantity I: x^3-133=[56% of 1250]÷4+113+124×9÷2. 56% of 1250=700. 700÷4=175. 124×9÷2=558. Total: x^3-133=175+113+558=846. x^3=979, so x=∛979≈9.93. Quantity II: y^2-y-72=0 factors to (y-9)(y+8)=0, so y=9 or y=-8. Comparing: If y=9, x>y (9.93>9). If y=-8, x>y (9.93>-8). In all cases, Quantity I > Quantity II, so option A is correct.

Multiple choice

$Quantity I: (7x^2 - 34x - 5 = 0) () Quantity II: (2y^2 - 3y - 14 = 0)$

  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

Quantity I: 7x² - 34x - 5 = 0 factors to (7x + 1)(x - 5) = 0, so x = 5 or x = -1/7. Quantity II: 2y² - 3y - 14 = 0 gives y = [3 ± √121]/4 = 3.5 or y = -2. Comparing roots: x can be greater (5 > 3.5, 5 > -2, -0.14 > -2) or less (-0.14 < 3.5) than y depending on which values are chosen. No consistent relationship exists.

Multiple choice

In the given question, two equations numbered l and II are given. Solve both the equations and mark the appropriate answer.\newline I. (x^2 - 17x + 72 = 0)\newline II. (y^2 - y - 72 = 0)

  1. x > y

  2. x < y

  3. x ≥ y

  4. x ≤ y

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

Equation I: x² - 17x + 72 = (x-8)(x-9) = 0, so x = 8 or 9. Equation II: y² - y - 72 = (y-9)(y+8) = 0, so y = 9 or -8. Since x can be >, <, or = y depending on values, no relation can be established.

Multiple choice

In the given question, two equations numbered l and II are given. Solve both the equations and mark the appropriate answer. $I. (x^{2} - 100 = 0) II. (y^{2} - 20y + 100 = 0)$

  1. x > y

  2. x < y

  3. x ≥ y

  4. x ≤ y

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

Equation I: x² - 100 = 0, so x = ±10. Equation II: y² - 20y + 100 = (y-10)² = 0, so y = 10 (repeated root). When x = 10, x = y; when x = -10, x < y. Therefore x ≤ y always. Option D is correct.

Multiple choice

In the given question, two equations numbered l and II are given. Solve both the equations and mark the appropriate answer. $I. (x^2 - 10x + 16 = 0) II. (y^2 - 7y + 10 = 0)$
  1. $\(x > y\)$
  2. $\(x < y\)$
  3. $\(x \ge y\)$
  4. $\(x \le y\)$

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

Equation I: x² - 10x + 16 = (x-2)(x-8) = 0, so x = 2 or 8. Equation II: y² - 7y + 10 = (y-2)(y-5) = 0, so y = 2 or 5. Comparing: x can be > y (8 > 5, 8 > 2), x = y (2 = 2), or x < y (2 < 5). No consistent relation.

Multiple choice

In each of these questions, two equations (a) and (b) are given. You have to solve both the equations and choose the correct option. ( (a) x^{3} = -68921 ) ( (b) y^{2} = 7921 )
  1. $x > y$
  2. $x \le y$
  3. $x \ge y$
  4. $x < y$

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

For x³ = -68921, x = -41 (since 41³ = 68921 and cube root of negative is negative). For y² = 7921, y = ±89 (square root gives both positive and negative). Comparing x = -41 with y = 89 gives x < y, but comparing x = -41 with y = -89 gives x > y. Since the relationship changes, it cannot be established.

Multiple choice

In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer. इनमें से प्रत्येक प्रश्न में दो समीकरण (I) और (II) दिए गए हैं। आपको दोनों समीकरणों को हल करना है और उत्तर देना है। I. 2x2 + 3x – 35 = 0 II. 4y2 + 10y – 104 = 0

  1. x > y

  2. x ≤ y

  3. x ≥ y

  4. x < y

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

Solve equation I: 2x² + 3x - 35 = 0 factors to (2x - 7)(x + 5) = 0, so x = 3.5 or x = -5. Solve equation II: 4y² + 10y - 104 = 0. Divide by 2: 2y² + 5y - 52 = 0. Using quadratic formula: y = (-5 ± √(25 + 416))/4 = (-5 ± √441)/4 = (-5 ± 21)/4, so y = 4 or y = -6.5. When x = -5, both x < y (y=4) and x > y (y=-6.5) are possible. The relationship cannot be uniquely established.

Multiple choice

In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer. इनमें से प्रत्येक प्रश्न में दो समीकरण (I) और (II) दिए गए हैं। आपको दोनों समीकरणों को हल करना है और उत्तर देना है। I. x2 -7x +12 = 0 II. y2 -5y + 6= 0

  1. x > y

  2. x ≤ y

  3. x ≥ y

  4. x < y

  5. x = y या x और y के बीच संबंध स्थापित नहीं किया जा सकता
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Solve equation I: x² - 7x + 12 = 0 factors to (x - 3)(x - 4) = 0, so x = 3 or x = 4. Solve equation II: y² - 5y + 6 = 0 factors to (y - 2)(y - 3) = 0, so y = 2 or y = 3. Comparing values: x = 3 equals y = 3, and x = 4 is greater than both y = 2 and y = 3. Therefore, x ≥ y is always true (x is always equal to or greater than y).

Multiple choice

In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer. इनमें से प्रत्येक प्रश्न में दो समीकरण (I) और (II) दिए गए हैं। आपको दोनों समीकरणों को हल करना है और उत्तर देना है। I. x 3 – 125 = 0 II. y 4 – 625 = 0

  1. x > y

  2. x ≤ y

  3. x ≥ y

  4. x < y

  5. if x = y or relationship between x and y can't be established x = y या x और y के बीच संबंध स्थापित नहीं किया जा सकता
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Solve equation I: x³ - 125 = 0, so x³ = 125, giving x = 5 (real root). Solve equation II: y⁴ - 625 = 0, so y⁴ = 625, giving y = ±5 (since 5⁴ = 625 and (-5)⁴ = 625). The real roots are y = 5 and y = -5. When x = 5, x equals y = 5 but x > y = -5. Therefore, x ≥ y is always true (x is either equal to or greater than y).

Multiple choice

In each of the following questions two equations are given. Solve these equations and give answer: निम्नलिखित प्रश्नों में दो समीकरण (I) और (II) दिए गए है आपको दोनों समीकरणों को हल करना है और उत्तर देना है: (I) 5x² -20x + 15 = 0 (II) 3y² -20y + 17 = 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 no relationship can be established between x and y. यदि x = y अथवा x और y के बीच कोई सम्बन्ध स्थापित नहीं किया जा सकता।
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Equation (I): 5x² - 20x + 15 = 0 gives x = 1 or x = 3. Equation (II): 3y² - 20y + 17 = 0 gives y ≈ 1.55 or y ≈ 3.47. When x = 1, both y values are greater than x. When x = 3, one y value (3.47) is greater, the other (1.55) is smaller. No consistent relationship exists between x and y.

Multiple choice

In each of these questions, two equations (a) and (b) are given. You have to solve both the equations and choose the correct option. a) (x^2 – 1089 = 0) b) (3y^2 – 363 = 0)

  1. x > y

  2. x ≤ y

  3. x ≥ y

  4. x < y

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

x² - 1089 = 0 gives x = ±33. 3y² - 363 = 0 gives y² = 121, so y = ±11. Since x can be +33 or -33 and y can be +11 or -11, we cannot establish a definite relationship. When x = 33 and y = 11, x > y; when x = -33 and y = 11, x < y.

Multiple choice

In each of these questions, two equations (a) and (b) are given. You have to solve both the equations and choose the correct option. (a) x^2 – 12x - 108 =0 (b) y^2 – 24y + 95 = 0
  1. $x > y$
  2. $x ≤ y$
  3. $x ≥ y$
  4. $x < y$

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

Solving equation (a): x^2-12x-108=0 gives x=(12±√(144+432))/2=(12±√576)/2=(12±24)/2, so x=18 or x=-6. Solving equation (b): y^2-24y+95=0 gives y=(24±√(576-380))/2=(24±√196)/2=(24±14)/2, so y=19 or y=5. Since x can be -6 to 18 and y can be 5 to 19, we cannot establish a definite relationship between x and y. Option E is correct.

Multiple choice

In each of these questions, two equations (a) and (b) are given. You have to solve both the equations and choose the correct option. a) (2x^2 – x - 78 = 0) b) (y^2 + 26y + 168 = 0)

  1. x > y

  2. x ≤ y

  3. x ≥ y

  4. x < y

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

Solving equation (a): 2x^2-x-78=0 gives x=(1±√(1+624))/4=(1±25)/4, so x=6.5 or x=-6. Solving equation (b): y^2+26y+168=0 gives y=(-26±√(676-672))/2=(-26±2)/2, so y=-12 or y=-14. Since x values (6.5, -6) are always greater than y values (-12, -14), x > y is always true. Option A is correct.