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 each of the following questions, a question is followed by information given in two statements 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 - $(Q_1^2 - 7Q_1 + 10.56 = 0 \\ Q_2^2 + 3Q_2 - 5.04 = 0)$

  1. Q1 > Q2

  2. Q1 ≥ Q2

  3. Q1 < Q2

  4. Q1 ≤ Q2

  5. Q1 = Q2 or relationship cannot be determined
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Solve both quadratic equations to compare Q1 and Q2. For Q1^2 - 7Q1 + 10.56 = 0, the roots are Q1 = 2.4 and Q1 = 4.4 (using quadratic formula or factoring as (Q1-2.4)(Q1-4.4)=0). For Q2^2 + 3Q2 - 5.04 = 0, roots are Q2 = 1.2 and Q2 = -4.2. Comparing positive roots: Q1 (2.4, 4.4) > Q2 (1.2). Therefore Q1 > Q2, making option A correct.

Multiple choice

Find the value of variables and state the correct relation.(1.5 x^{2} + 8x - 49.5 = 0)(4.5 y^{2} - 2.8y + 0.4 = 0)
  1. $x > y$
  2. $x \ge y$
  3. $x < y$
  4. $x \le y$

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

For x: 1.5x² + 8x - 49.5 = 0. Using quadratic formula: x = [-8 ± √(64 + 297)]/3 = [-8 ± √361]/3 = [-8 ± 19]/3. So x = 11/3 ≈ 3.67 or x = -27/3 = -9. For y: 4.5y² - 2.8y + 0.4 = 0 → y = [2.8 ± √(7.84 - 7.2)]/9 = [2.8 ± √0.64]/9 = [2.8 ± 0.8]/9. So y = 3.6/9 = 0.4 or y = 2/9 ≈ 0.22. Since x can be 3.67 or -9 and y can be 0.4 or 0.22, we cannot establish a unique relationship - option E is correct.

Multiple choice

Find the value of variables and state the correct relation. (6x^2 + 41x + 13 = 0) (2y^2 + 27y + 81 = 0)
  1. $x > y$
  2. $x \ge y$
  3. $x < y$
  4. $x \le y$
  5. $x = y or relationship cannot be established$
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

For x: 6x² + 41x + 13 = 0. Discriminant = 1681 - 312 = 1369 = 37². Roots: x = [-41 ± 37]/12 → x = -4/12 = -1/3 or x = -78/12 = -6.5. For y: 2y² + 27y + 81 = 0. Discriminant = 729 - 648 = 81 = 9². Roots: y = [-27 ± 9]/4 → y = -18/4 = -4.5 or y = -36/4 = -9. Comparing: When x = -1/3, x > both y values. When x = -6.5, x > -9 but x < -4.5. Since the relationship varies, E (cannot be established) is correct.

Multiple choice

Find the value of variables and state the correct relation. (x^2 - 89x + 1278 = 0) (y^2 -35y + 306 = 0)
  1. $\(x > y\)$
  2. $\(x \ge y\)$
  3. $\(x < y\)$
  4. $\(x \le y\)$

  5. x = y or relationship cannot be stablished
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For x: x² - 89x + 1278 = 0. Factoring: (x - 53)(x - 36) = 0 → x = 53 or x = 36. For y: y² - 35y + 306 = 0. Factoring: (y - 18)(y - 17) = 0 → y = 18 or y = 17. In all cases (x=53 or 36, y=18 or 17), we have x > y. This means x ≥ y is true, so option B is correct.

Multiple choice

Find the value of variables and state the correct relation. (x^2 - 3158 = 0) (y^2 = 86159)
  1. $x > y$
  2. $x \ge y$
  3. $x < y$
  4. $x \le y$

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

For x: x² = 3158 → x = ±√3158 ≈ ±56.2. For y: y² = 86159 → y = ±√86159 ≈ ±293.5. We have multiple possible sign combinations. If both are positive: 56.2 < 293.5 → x < y. If x negative, y positive: x < y. If both negative: -56.2 > -293.5 → x > y. Since the relationship depends on sign choices and cannot be uniquely determined, option E is correct.

Multiple choice

For each of the following pairs of equations find the relationship between (x) & (y). ( )(I) (2x^2 - 10x + 12 = 0) ( )(II) (4y^2 - 16 = 0)
  1. $If \(x > y\)$
  2. $If \(x < y\)$
  3. $If \(x \geq y\)$
  4. $If \(x \leq y\)$
  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

Equation I: 2x² - 10x + 12 = 0 factors to (x-2)(x-3)=0, so x = 2 or 3. Equation II: 4y² - 16 = 0 gives y² = 4, so y = 2 or -2. Comparing: when x=2, x=y=2; when x=3, x>y (3 > 2, 3 > -2). Thus x is always greater than or equal to y, making option C correct.

Multiple choice

In the following question, two statements, numbered as I and II are given. On solving them, we get quantities I and II respectively. Solve both the quantities and choose the correct relationship.\n\n\textbf{Quantity I:} $Q_1^2 + 56Q_1 + 783 = 0$\n\textbf{Quantity II:} $Q_2^2 + 46Q_2 + 513 = 0$
  1. $Q_1 > Q_2$
  2. $Q_1 \ge Q_2$
  3. $Q_1 < Q_2$
  4. $Q_1 \le Q_2$
  5. $Q_1 = Q_2$ or relationship cannot be established
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Solving Q1^2+56Q1+783=0 gives roots Q1=-27 and Q1=-29. Solving Q2^2+46Q2+513=0 gives roots Q2=-27 and Q2=-19. Since both Q1 values (-27, -29) are less than or equal to only one Q2 value (-27), we get Q1 <= Q2.

Multiple choice

Find the value of variables and state the correct relationship.(x^2 + 2x - 15 = 0)(y^2 - y - 20 = 0)
  1. $\(x > y\)$
  2. $\(x \ge y\)$
  3. $\(x < y\)$
  4. $\(x \le y\)$

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

x² + 2x - 15 = 0 factors to (x+5)(x-3) = 0, so x = -5 or x = 3. y² - y - 20 = 0 factors to (y-5)(y+4) = 0, so y = -4 or y = 5. Comparing all pairs: -5 < -4, -5 < 5, 3 > -4, 3 < 5. Since x can be less than, greater than, or equal to y depending on which roots we compare, no unique relationship exists.

Multiple choice

Find the value of variables and state the correct relationship.(x^2 - 29x + 204 = 0)(y^2 - 38y + 357 = 0)
  1. $x > y$
  2. $x \ge y$
  3. $x < y$
  4. $x \le y$

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

x² - 29x + 204 = 0 has discriminant 25, so x = (29 ± 5)/2 = 17 or 12. y² - 38y + 357 = 0 has discriminant 16, so y = (38 ± 4)/2 = 21 or 17. Comparing all pairs: x=17 equals y=17; otherwise x=12 or 17 is always ≤ y=17 or 21. Therefore x ≤ y is always true.

Multiple choice

(\text{Find the value of variables and state the correct relationship.}\\text{I. } (x - 18)^2 = 0\\text{II. } y^2 = 324)
  1. $\(x < y\)$
  2. $\(x > y\)$
  3. $\(x \ge y\)$
  4. $\(x \le y\)$
  5. $\(x = y \text{ or relation cannot be established}\)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Solving equation I: (x - 18)^2 = 0 gives x = 18 (only one value). Solving equation II: y^2 = 324 gives y = 18 or y = -18. Comparing: if y = 18, then x = y; if y = -18, then x > y. In all cases, x is greater than or equal to y, so x ≥ y is correct.

Multiple choice

Find the value of variables and state the correct relationship.\newline I. (x^2 - 1 = 0)\newline II. (y^2 + 4y + 3 = 0)
  1. $\(x > y\)$
  2. $\(x < y\)$
  3. $\(x \ge y\)$
  4. $\(x \le y\)$

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

Solving equation I: x^2 - 1 = 0 gives x^2 = 1, so x = 1 or x = -1. Solving equation II: y^2 + 4y + 3 = 0. Factoring: (y + 3)(y + 1) = 0, so y = -3 or y = -1. Comparing all values: when x = 1, x > y for both y values; when x = -1, x > y when y = -3, but x = y when y = -1. Therefore, x is always greater than or equal to y.

Multiple choice

Solve the following equations and determine the relationship between (x) and (y): (I) \ 3x^2 + 29x + 70 = 0 \ (II) \ 2y^2 + 27y + 91 = 0

  1. x > y

  2. x ≥ y

  3. x < y

  4. x ≤ y

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

For equation (I): 3x² + 29x + 70 = 0. Factor: (3x + 14)(x + 5) = 0, so x = -14/3 or x = -5. For equation (II): 2y² + 27y + 91 = 0. Factor: (2y + 13)(y + 7) = 0, so y = -13/2 = -6.5 or y = -7. Comparing values: when x = -14/3 ≈ -4.67 and y = -7, then x > y. When x = -5 and y = -7, then x > y. Both cases give x > y, so option A is correct.

Multiple choice

In the following question two equation numbered I and II are given you have to solve both the equations and give answer. $I. (2x^2 + 11x + 14 = 0) II. (4y^2 + 12y + 9 = 0)$

  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
C Correct answer
Explanation

Solving equation I: 2x^2 + 11x + 14 = 0 gives x = -2 or x = -3.5. Solving equation II: 4y^2 + 12y + 9 = 0 gives y = -1.5 only. When x = -2, we have x < y (-2 < -1.5). When x = -3.5, we also have x < y (-3.5 < -1.5). Therefore, x is always less than y.

Multiple choice

In each of these questions, two equations (I) and (II) are given. You have to solve both equations to give the answer: ( (I) x^2 – 4x - 165 = 0 ) ( (II) y^2 + 28y + 192 = 0 )

  1. If x < y

  2. If x > y

  3. If x ≥ y

  4. If x ≤ y

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

Equation (I): x² - 4x - 165 = 0 factors to (x-15)(x+11)=0, so x = 15 or -11. Equation (II): y² + 28y + 192 = 0 factors to (y+16)(y+12)=0, so y = -16 or -12. Comparing: 15 > -16, 15 > -12, -11 > -16, -11 > -12. In all cases, x > y.