Quantitative Aptitude

Quadratic Equations

367 Questions
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Quadratic Equations Questions

Multiple choice

In the following questions two equations numbered (I) and (II) are given. You have to solve the both equations and give answer— $( (I) \space 18x^2 + 18x + 4 = 0 \space \space \space \space \space (II) \space 12y^2 + 29y + 14 = 0 )$

  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): 18x² + 18x + 4 = 0 gives x = -1/3 or x = -2/3. Solving equation (II): 12y² + 29y + 14 = 0 gives y = 7/12 or y = -11/12. Both values of x are less than both values of y, so x ≥ y holds true.

Multiple choice

In the following question two equations I and II are given. You have to solve both the equations and give answer. $( (I) \space x^2 - 10x + 25 = 0 ) ( (II) \space y^2 + 26y + 169 = 0 )$

  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 determined
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Solving (I): x²-10x+25=0 gives (x-5)²=0, so x = 5. Solving (II): y²+26y+169=0 gives (y+13)²=0, so y = -13. Since 5 > -13, x > y, so answer is B.

Multiple choice

In the following questions two equations I and II are given. You have to solve both the equations and give answer. $( (I) \ (x + 1)^{2} + 1 = 2(x +3) )$ $( (II) \ (2y – 1) (y – 3) = (y – 5) (y – 1) )$

  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 determined.
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Equation I: (x+1)²+1 = 2(x+3) gives x²+2x+1+1 = 2x+6, so x²-4=0, x = ±2. Equation II: (2y-1)(y-3) = (y-5)(y-1) gives 2y²-7y+3 = y²-6y+5, so y²-y-2=0, (y-2)(y+1)=0, y = 2 or y = -1. When x=2, x=y; when x=-2, x < y=-1. The relationship varies, so option E is correct. Option B (x ≥ y) is wrong because x=-2 is less than y=-1.

Multiple choice

In the following questions two equations I and II are given. You have to solve both the equations and give answer.(I) \ 23x^2 – 29x – 88 = 0 (II) \ 29y^2 + 33y – 48 = 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 determined.
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Equation I: 23x²-29x-88=0. Using quadratic formula: x = (29 ± √(841+8084))/46 = (29 ± √8925)/46. √8925 ≈ 94.47, so x ≈ (29 ± 94.47)/46, giving x ≈ 2.68 or x ≈ -1.43. Equation II: 29y²+33y-48=0 gives y = (-33 ± √(1089+5568))/58 = (-33 ± √6657)/58. √6657 ≈ 81.59, so y ≈ 0.84 or y ≈ -1.97. x=2.68 > y=0.84 and > y=-1.97, but x=-1.43 is between y values. Relationship can't be uniquely determined. Option E is correct.

Multiple choice

In the following questions two equations I and II are given. You have to solve both the equations and give answer. ( (I) ) ( 2x^2 + 14x – 120 = 0 ) ( (II) ) ( y^2 + 48y – 324 = 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 determined.
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Equation I: 2x²+14x-120=0 divides by 2: x²+7x-60=0, factors as (x+12)(x-5)=0, so x = -12 or x = 5. Equation II: y²+48y-324=0. Complete square: (y+24)² = 324+576 = 900, so y+24 = ±30, giving y = 6 or y = -54. When x=-12: x > -54 but x < 6. When x=5: x < 6 but x > -54. Relationship depends on which root we pick. Option E is correct as no consistent relationship exists. Option D (x ≤ y) is incorrect because x=-12 > -54.

Multiple choice

In the following questions two equations I and II are given. You have to solve both the equations and give answer. $(\text{(I)}\quad x^2 – 3x – 460 = 0 \qquad \text{(II)}\quad y^2 – 5y – 500 = 0)$

  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 determined.
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Equation I: x² - 3x - 460 = 0. Using quadratic formula or factorization: x = [3 ± √(9+1840)]/2 = [3 ± √1849]/2 = [3 ± 43]/2. So x = 23 or x = -20. Equation II: y² - 5y - 500 = 0. y = [5 ± √(25+2000)]/2 = [5 ± √2025]/2 = [5 ± 45]/2. So y = 25 or y = -20. Comparing: when x = 23 and y = 25: x < y. When x = 23 and y = -20: x > y. Since both x > y and x < y are possible, the relationship cannot be uniquely determined.

Multiple choice

In each of the following questions, two equations I and II are given. You have to solve them and Give answer (I) 7x2 + 26x – 45 = 0 (II) 3y2 – 13y + 14 = 0

  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
C Correct answer
Explanation

Equation I: 7x² + 26x - 45 = 0 factors to (7x-9)(x+5)=0, giving x = 9/7 or x = -5. Equation II: 3y² - 13y + 14 = 0 factors to (3y-7)(y-2)=0, giving y = 7/3 or y = 2. Comparing: 9/7 < 7/3 and 9/7 < 2, so x < y. Option C is correct.

Multiple choice

In each question below two equations have been given on the basis of these you have to find out the relationship between 'x' and 'y'. $( (I) \space 3x + 12 = 4x + 6 )$ $( (II) \space y^2 + 17y + 72 = 0 )$

  1. If x = y or relation can't 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
B Correct answer
Explanation

From equation (I): 3x + 12 = 4x + 6, solving gives x = 6. From equation (II): y² + 17y + 72 = 0 factors to (y+8)(y+9) = 0, so y = -8 or y = -9. Since x = 6 and y is either -8 or -9, we have x > y in both cases. Therefore the relationship is x > y.

Multiple choice

In each of the following questions, two equations (\textbf{(I)}) and (\textbf{(II)}) are given. You have to solve them and Give answer:(\textbf{(I)}) (2x^2 - 11x + 15 = 0)(\textbf{(II)}) (2y^2 - 15y + 28 = 0)

  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
C Correct answer
Explanation

Solve (I): 2x²-11x+15=0 factors to (2x-5)(x-3)=0, so x=2.5 or x=3. Solve (II): 2y²-15y+28=0 factors to (2y-7)(y-4)=0, so y=3.5 or y=4. All x values (2.5, 3) are less than all y values (3.5, 4), so x

Multiple choice

In the following questions two equations (I) and (II) are given. You have to solve both the equations and give answer. ( (I) \space 5x^2 + 8x + 3 = 0 \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space (II) \space 16y^2 - 40y + 25 = 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
C Correct answer
Explanation

Solving equation (I): 5x² + 8x + 3 = 0 gives x = -0.6 or x = -1. Solving equation (II): 16y² - 40y + 25 = 0 factors as (4y-5)² = 0, giving y = 1.25 (repeated root). Both x values are less than y, so x < y.

Multiple choice

In the following question two equation numbered I and II are given .You have to solve both the equations and give answer if - I. (x^2 - 22x + 120 = 0) II. (y^2 - 26y + 168 = 0)
  1. $\(x > y\)$
  2. $\(x \ge y\)$
  3. $\(x < y\)$
  4. $\(x \le y\)$

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

Equation I: x²-22x+120=(x-10)(x-12) gives x=10,12. Equation II: y²-26y+168=(y-12)(y-14) gives y=12,14. Since x can be 12 (equal to y's 12) but never exceeds 14, and x=10<12,14, we have x≤y.

Multiple choice

Two equations I & II are given in each question. On the basis of these equations, you have to decide the relation between x & y and give the answer. $I. (x^2 + 11x + 30 = 0) II. (y^2 + 7y + 12 = 0)$

  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

Solving x² + 11x + 30 = 0 gives (x+5)(x+6) = 0, so x = -5 or -6. Solving y² + 7y + 12 = 0 gives (y+3)(y+4) = 0, so y = -3 or -4. For all combinations: x = -5 < y(-3 or -4) and x = -6 < y(-3 or -4). Therefore x < y always. Option C is correct.

Multiple choice

$I. (20x^2 - x - 12 = 0)$ $II. (20y^2 + 27y + 9 = 0)$ Solve the given equations and determine the relationship between x and y.

  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

For I: 20x²-x-12=0, using quadratic formula: x=(1±√(1+960))/40=(1±31)/40, so x=0.8 or x=-0.75. For II: 20y²+27y+9=0, y=(-27±√(729-720))/40=(-27±3)/40, so y=-0.6 or y=-0.75. Comparing: x=0.8 > y=-0.6, x=-0.75 = y=-0.75, x=-0.75 < y=-0.6. The relationship varies between values, so it cannot be established.

Multiple choice

In each of the following question, a question is followed by information given in two statement named as Quantity-I(Q1) and Quantity –II (Q2). You have to study the information along with the question and compare the value derived from Quantity-I and Quantity-II. Then answer - (Q_1^2 - 5.4 Q_1 + 5.04 = 0 ) (Q_2^2 - 9.2 Q_2 + 21 = 0)

  1. Q1 > Q2

  2. Q1 ≥ Q2

  3. Q1 < Q2

  4. Q1 ≤ Q2

  5. Q1 = Q2 or relationship can not be determined
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For Q1: Solving Q1² - 5.4Q1 + 5.04 = 0 using quadratic formula: Q1 = (5.4 ± √(29.16 - 20.16))/2 = (5.4 ± 3)/2, giving Q1 = 4.2 or Q1 = 1.2. For Q2: Solving Q2² - 9.2Q2 + 21 = 0: Q2 = (9.2 ± √(84.64 - 84))/2 = (9.2 ± 0.8)/2, giving Q2 = 5 or Q2 = 4.2. Since Q1 values (1.2, 4.2) are both ≤ Q2 values (4.2, 5), Q1 ≤ Q2.

Multiple choice

In the following two equations numbered I and II are given. you have to solve both the equations and choose the correct option.\newline I. (x^{\frac{3}{2}} = 25^2 + 104) \newline II. (y^{\frac{1}{4}} = 4^2 - 13)
  1. $\(p < q\)$
  2. $\(p \ge q\)$
  3. $\(p > q\)$
  4. $\(p \le q\)$
  5. $\(p = q\) or the relationship between \(p\) and \(q\) cannot be determined$
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

y^(1/4) = 3, so y = 81. x^(3/2) = 729 = 3^6, so x = (3^6)^(2/3) = 3^4 = 81. Thus x = y = 81, making option E correct.