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

Solve the following system of equations: y = x^2 - 1 y = 2x - 3

  1. (x, y) = (2, 1)

  2. (x, y) = (1, 2)

  3. (x, y) = (3, -1)

  4. (x, y) = (-1, 3)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To solve this system of equations, we can substitute the first equation into the second equation. Substituting y = x^2 - 1 into y = 2x - 3, we get x^2 - 1 = 2x - 3. Rearranging the equation, we get x^2 - 2x + 2 = 0. Factoring the quadratic equation, we get (x - 2)(x - 1) = 0. Setting each factor equal to zero, we get x = 2 and x = 1. Substituting x = 2 back into the first equation, we get y = 2^2 - 1 = 3. Therefore, the solution is (x, y) = (2, 1).

Multiple choice

Solve the Diophantine equation: x^2 + y^2 = 17.

  1. (1, 4)

  2. (2, 3)

  3. (3, 2)

  4. (4, 1)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We can try different values of x and y to find solutions. Substituting x = 1, we get y^2 = 16, which gives y = 4. Therefore, one solution is (1, 4).

Multiple choice

Directions: In the following question, two equations numbered I and II are given. Solve both the equations and determine the relationship between x and y. Choose the correct option accordingly. I. x2 + 12x + 32 = 0 II. y2 + 17y + 72 = 0

  1. x < y

  2. x ≤ y

  3. x > y

  4. x ≥ y

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

Equation I: x^2 + 12x + 32 = 0 factors to (x+8)(x+4) = 0, so x = -8, -4. Equation II: y^2 + 17y + 72 = 0 factors to (y+8)(y+9) = 0, so y = -8, -9. Comparing the values, x is always greater than or equal to y (e.g., -4 > -8, -4 > -9, -8 = -8, -8 > -9).

Multiple choice

Directions: In the following question, equations numbered I and II are given. You have to solve both the equations and mark the correct option. I. x2 + 13x + 42 = 0 II. y2 + 19y + 90 = 0

  1. x < y

  2. x ≤ y

  3. x > y

  4. x ≥ y

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

Equation I: x^2 + 13x + 42 = 0 => (x+6)(x+7) = 0 => x = -6, -7. Equation II: y^2 + 19y + 90 = 0 => (y+9)(y+10) = 0 => y = -9, -10. Comparing values: -6 > -9, -6 > -10, -7 > -9, -7 > -10. Thus, x > y.

Multiple choice

Directions: In the following question, equations numbered I and II are given. You have to solve both the equations and mark the correct option. I. x2 - 15x + 56 = 0 II. y2 - 23y + 132 = 0

  1. x < y

  2. x ≤ y

  3. x > y

  4. x ≥ y

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

Solving x^2 - 15x + 56 = 0 gives (x-7)(x-8)=0, so x is 7 or 8. Solving y^2 - 23y + 132 = 0 gives (y-11)(y-12)=0, so y is 11 or 12. Since all values of x are less than all values of y, x < y.

Multiple choice

Directions: In the following question, equations numbered I and II are given. You have to solve both the equations and mark the correct option. I. x2 - 22x + 120 = 0 II. y2 - 26y + 168 = 0

  1. x < y

  2. x ≤ y

  3. x > y

  4. x ≥ y

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

Eq I: x^2 - 22x + 120 = 0 -> (x-10)(x-12) = 0 -> x = 10, 12. Eq II: y^2 - 26y + 168 = 0 -> (y-12)(y-14) = 0 -> y = 12, 14. Comparing roots: 10 < 12, 10 < 14, 12 = 12, 12 < 14. Thus x <= y.

Multiple choice

Directions: In the following question, two equations numbered I and II are given. You have to solve both the equations and then choose the correct option. I. 4X2 + 36X + 80 = 0 II. 7Y2 + 49Y + 84 = 0

  1. X > Y

  2. X ≥ Y

  3. X < Y

  4. X ≤ Y

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

Equation I: 4(x^2 + 9x + 20) = 0 -> 4(x+4)(x+5) = 0, so x = -4, -5. Equation II: 7(y^2 + 7y + 12) = 0 -> 7(y+3)(y+4) = 0, so y = -3, -4. Comparing values, -4 <= -3 and -5 <= -3, so X <= Y.

Multiple choice

Directions: In the following question, two equations numbered I and II are given. You have to solve both the equations and then choose the correct option. I. 90x2 + 90x + 20 = 0 II. 36y2 – 66y + 24 = 0

  1. x < y

  2. x ≤ y

  3. x = y, or the relationship cannot be established.

  4. x ≥ y

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

I: 90x^2 + 90x + 20 = 0 -> 9x^2 + 9x + 2 = 0 -> (3x+1)(3x+2) = 0. x = -1/3, -2/3. II: 36y^2 - 66y + 24 = 0 -> 6y^2 - 11y + 4 = 0 -> (2y-1)(3y-4) = 0. y = 1/2, 4/3. Comparing values, x < y.

Multiple choice

Directions: In the following question, two equations numbered I and II are given. You have to solve both the equations and find out the relationship between x and y. I. 3x2 - 14x + 15 = 0 II. 6y2 - 5y + 1 = 0

  1. x > y

  2. x ≥ y

  3. x < y

  4. x ≤ y

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

I: 3x^2 - 14x + 15 = 0 -> (3x-5)(x-3) = 0 -> x = 5/3 (1.66), 3. II: 6y^2 - 5y + 1 = 0 -> (3y-1)(2y-1) = 0 -> y = 1/3 (0.33), 1/2 (0.5). Since all x values are greater than all y values, x > y.

Multiple choice

Directions: In the following question, two equations numbered I and II are given. You have to solve both the equations and find out the relationship between x and y. I. 3x2 + 11x + 10 = 0 II. 2y2 + 13y + 21 = 0

  1. x > y

  2. x ≥ y

  3. x < y

  4. x ≤ y

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

I: 3x^2 + 11x + 10 = 0 => (3x+5)(x+2) = 0 => x = -5/3, -2. II: 2y^2 + 13y + 21 = 0 => (2y+7)(y+3) = 0 => y = -3.5, -3. Comparing values: -1.66 > -3 and -2 > -3. Thus x > y.