Quantitative Aptitude

Quadratic Equations

367 Questions
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Quadratic Equations Questions

Multiple choice

Directions: In the following question, two equations (I) and (II) are given. You have to solve both the equations and give answer. (I) 12x2 = -17x - 5 (II) 5y = y2 + 6

  1. x > y

  2. x < y

  3. x ≥ y

  4. x ≤ y

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

(I) 12x^2 + 17x + 5 = 0. Factors: 12x^2 + 12x + 5x + 5 = 0 -> 12x(x+1) + 5(x+1) = 0. Roots: x = -1, -5/12 (-0.41). (II) y^2 - 5y + 6 = 0. Factors: (y-2)(y-3) = 0. Roots: y = 2, 3. Comparing roots: -1 < 2, -1 < 3, -0.41 < 2, -0.41 < 3. Thus x < y.

Multiple choice

Directions: In the following question, two equations numbered I and II are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Mark your answer accordingly. I. 3x2 + 18x + 15 = 0 II. 14y2 + 3y = 2

  1. x > y

  2. x < y

  3. x ≥ y

  4. x ≤ y

  5. x = y, or no relationship can be established between 'x' and 'y'.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

I: 3x^2 + 18x + 15 = 0 -> x^2 + 6x + 5 = 0 -> (x+5)(x+1) = 0. x = -5, -1. II: 14y^2 + 3y - 2 = 0 -> (7y-2)(2y+1) = 0. y = 2/7, -1/2. Comparing values: -5 < -1 < -0.5 < 0.28. Thus, x < y.

Multiple choice

Directions: In the following question, you have to compare Quantity I and Quantity II and then determine which of the following options describes the comparison. 1. 18x2 + 18x + 4 = 0 2. 12y2 + 29y + 14 = 0 Quantity I = x Quantity II = 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

Solving 18x^2 + 18x + 4 = 0 gives 9x^2 + 9x + 2 = 0, so (3x+1)(3x+2) = 0, x = -1/3, -2/3. Solving 12y^2 + 29y + 14 = 0 gives (3y+2)(4y+7) = 0, y = -2/3, -7/4. Comparing values, x >= y.

Multiple choice

In the following questions one or two equations (I) and (II) are given. On the basis of these you have to find out the relation between x & y.((I)) (2x + 3y = 5)((II)) ((1 - y)^2 = (x - 1)^2)

  1. If x = y

  2. If x > y

  3. If y > x

  4. If x ≥ y

  5. If y ≥ x
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

From (I): 2x + 3y = 5. From (II): (1-y)² = (x-1)² means 1-y = ±(x-1). Testing: if x=y, then 2x+3x=5 gives 5x=5, x=1. So x=y=1. Check (II): (1-1)² = (1-1)² gives 0=0, which holds true. Therefore x=y is correct.

Multiple choice

In the following questions one or two equations (I) and (II) are given. On the basis of these you have to find out the relation between x & y. $( (I) \quad x^2 + 9x + 20 = 0 )$ $( (II) \quad y^2 + 13y + 42 = 0 )$

  1. If x = y

  2. If x > y

  3. If y > x

  4. If x ≥ y

  5. If y ≥ x
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Solving (I): x² + 9x + 20 = 0 gives (x+4)(x+5) = 0, so x = -4 or -5. Solving (II): y² + 13y + 42 = 0 gives (y+6)(y+7) = 0, so y = -6 or -7. Comparing: when x = -4, y = -6 or -7, so x > y. When x = -5, y = -6 or -7, so x > y. In all cases x > y.

Multiple choice

In the following questions one or two equations (I) and (II) are given. On the basis of these you have to find out the relation between x & y.( (I) ) (x^2 + 28 = 11x)( (II) ) (y^2 + 56 = 15y)

  1. If x = y

  2. If x > y

  3. If y > x

  4. If x ≥ y

  5. If y ≥ x
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Solving (I): x² - 11x + 28 = 0 gives (x-4)(x-7) = 0, so x = 4 or 7. Solving (II): y² - 15y + 56 = 0 gives (y-7)(y-8) = 0, so y = 7 or 8. Comparing: when x=4, y can be 7 or 8, so y > x. When x=7, y can be 7 or 8. If y=7, x=y. If y=8, y > x. So y ≥ x in all cases.

Multiple choice

$In each of the following questions, two equations (I) and (II) are given. You have to solve them and(I) \space x^2 - 7x = -12\(II) \space y^2 - 9y = -20$ Give answer

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

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

Solving equation (I): x² - 7x = -12 becomes x² - 7x + 12 = 0, which factors as (x-3)(x-4) = 0, giving x = 3 or x = 4. Solving equation (II): y² - 9y = -20 becomes y² - 9y + 20 = 0, which factors as (y-4)(y-5) = 0, giving y = 4 or y = 5. When x = 3, x < y (since y is 4 or 5). When x = 4, we have x = y for y = 4, but x < y for y = 5. The relationship cannot be established for all cases, but the minimum value of x is 3 and the minimum value of y is 4, so x ≤ y is generally true.

Multiple choice

In each question below two equations are given. On the basis of these you have to find out the relationship between x and y. $I. (x^2 + 23x + 132 = 0) II. (y^2 + 21y + 110 = 0)$

  1. If x = y or relation cannot be established

  2. If x > y

  3. If x < y

  4. If x ≤ y

  5. If x ≥ y
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

x² + 23x + 132 = 0 gives x = -11, -12. y² + 21y + 110 = 0 gives y = -10, -11. x values: -11, -12. y values: -10, -11. Maximum x = -11, maximum y = -10. Minimum x = -12, minimum y = -11. All x values ≤ corresponding y values, so x ≤ y is always true.

Multiple choice

In each question below two equations are given. On the basis of these you have to find out the relationship between x and y. $I. (x(x+25) + 156 = 0) II. (y^2 - 144 = 0)$

  1. If x = y or relation cannot be established

  2. If x > y

  3. If x < y

  4. If x ≤ y

  5. If x ≥ y
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

x(x+25) + 156 = 0 gives x² + 25x + 156 = 0, x = -12, -13. y² - 144 = 0 gives y = 12, -12. x values: -12, -13. y values: 12, -12. When y = 12, x < y. When y = -12, x = -12 gives x = y, and x = -13 gives x < y. So x ≤ y always holds.

Multiple choice

$(\text{In each of the following questions, two equations are given. You have to solve them and state the correct relationship-})$ $(\text{I. } x^3 - 468 = 1729 \qquad \text{II. } y^2 - 1733 + 1564 = 0 )$
  1. $\( x > y \)$
  2. $\( x \ge y \)$
  3. $\( x < y \)$
  4. $\( x \le y \)$
  5. $\( x = y \text{ or relationship can not be established} \)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Equation I: x³ - 468 = 1729, so x³ = 1729 + 468 = 2197, therefore x = ∛2197 = 13. Equation II: y² - 1733 + 1564 = 0, so y² = 1733 - 1564 = 169, therefore y = ±13. Since x = 13 and y can be +13 or -13, we have x ≥ y (x equals or is greater than y). Option B is correct.

Multiple choice

In each of the following questions, two equations (I) and (II) are given. You have to solve them and state the correct relationship-(\text{I.}) (8x^2 + 12x - 8 = 0)(\text{II.}) (9y^2 + 18y + 9 = 0)
  1. $\(x > y\)$
  2. $\(x \ge y\)$
  3. $\(x \le y\)$
  4. $\(x < y\)$

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

Solving equation I: 8x² + 12x - 8 = 0. Divide by 4: 2x² + 3x - 2 = 0 gives (2x-1)(x+2) = 0, so x = 1/2 or x = -2. Solving equation II: 9y² + 18y + 9 = 0. Divide by 9: y² + 2y + 1 = 0 gives (y+1)² = 0, so y = -1. Comparing: When x = 1/2, x > y. When x = -2, x < y. Since x can be greater or less than y depending on which root we take, the relationship cannot be established uniquely.

Multiple choice

In each of the following questions, two equations are given. You have to solve them and determine the relationship between x and y. $(I. x(x+2) = 0 )(II. y(y-3) = 0)$

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

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

Equation I: x(x+2) = 0 gives x = 0 or x = -2. Equation II: y(y-3) = 0 gives y = 0 or y = 3. Comparing all values: x can be 0 or -2, y can be 0 or 3. When x = -2, x < y. When x = 0 and y = 3, x < y. When x = 0 and y = 0, x = y. In all cases, x ≤ y holds.

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. (12x^2 - 2x - 4 = 0) II. (10y^2 + 2 = 9y)
  1. $\(x > y\)$
  2. $\(x < y\)$
  3. $\(x \ge y\)$
  4. $\(x \le y\)$

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

Equation I: 12x² - 2x - 4 = 0 → 6x² - x - 2 = 0. Solving: x = (1 ± √49)/12 = (1 ± 7)/12, so x = 2/3 or x = -1/2. Equation II: 10y² - 9y + 2 = 0. Solving: y = (9 ± √81 - 80)/20 = (9 ± 1)/20, so y = 1/2 or y = 2/5. Since x can be > y or < y depending on which root you pick, no unique relation exists.

Multiple choice

In the following questions two equations numbered I and II are given. You have to solve both the equations and ————— (I. \space 20x^{2} - x - 12 = 0 ) (II.\space 20y^{2} + 27y + 9 = 0)

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

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

Equation I: 20x² - x - 12 = 0. Using quadratic formula: x = [1 ± √(1 + 960)] ÷ 40 = [1 ± √961] ÷ 40 = [1 ± 31] ÷ 40. So x = 32/40 = 4/5 = 0.8, or x = -30/40 = -3/4 = -0.75. Equation II: 20y² + 27y + 9 = 0. Using quadratic formula: y = [-27 ± √(729 - 720)] ÷ 40 = [-27 ± √9] ÷ 40 = [-27 ± 3] ÷ 40. So y = -24/40 = -3/5 = -0.6, or y = -30/40 = -3/4 = -0.75. Comparing: x values are 0.8 and -0.75. y values are -0.6 and -0.75. Since x can be 0.8 > -0.6 or x can be -0.75 = -0.75, the relationship cannot be uniquely established. Option E is correct.