Quantitative Aptitude

Algebra

3,715 Questions
All Polynomial Roots and Coefficients 2451 Zero-Product Root Finding 667 Linear System Manipulation 162 Polynomial Theorems and Reduction 85 Algebraic Identity Simplification 82 Surds and Indices 70 Modulus and Absolute Value 54 Algebraic Expression Substitution 46 Algebraic Fraction Simplification 33 Logarithmic Equations 25 Repeated Series Sum 12 Linear Equation Word Problems 9

Algebra Questions

Multiple choice

Solve the equation: x + 5 = 12.

  1. 5

  2. 6

  3. 7

  4. 8
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

To solve the equation, isolate the variable x on one side: x = 12 - 5 = 7. Therefore, the answer is 7.

Multiple choice

Solve the equation: 3(x - 4) = 15.

  1. 5

  2. 7

  3. 9

  4. 11
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

To solve the equation, isolate the variable x on one side: 3(x - 4) = 15 => x - 4 = 5 => x = 9. Therefore, the answer is 11.

Multiple choice

Solve the equation for (x): (2x^2 - 5x + 2 = 0)

  1. \(x = 1, 2\)
  2. \(x = -1, -2\)
  3. \(x = 1, -2\)
  4. \(x = -1, 2\)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To solve the equation, we can use the quadratic formula: (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}). Here, (a = 2, b = -5, c = 2). Plugging these values into the formula, we get: (x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(2)}}{2(2)} = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm \sqrt{9}}{4} = \frac{5 \pm 3}{4}). Therefore, the solutions are (x = 1) and (x = 2).

Multiple choice

Solve the quadratic equation: 2x^2 + 7x - 4 = 0

  1. x = 1, -2

  2. x = -1, 2

  3. x = 1, 4

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

To solve this equation, we can use factoring. We can factor the left-hand side as follows: (2x - 1)(x + 4) = 0. Setting each factor equal to 0, we get: 2x - 1 = 0 and x + 4 = 0. Solving these equations, we get: x = 1/2 and x = -4. Therefore, the solutions to the quadratic equation are x = 1 and x = -2.

Multiple choice

Solve the quadratic equation: 3x^2 - 2x - 8 = 0

  1. x = 2, -4/3

  2. x = -2, 4/3

  3. x = 1, -8

  4. x = -1, 8
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To solve this equation, we can use the quadratic formula. Plugging in the values of a, b, and c, we get: x = (-(-2) ± √((-2)^2 - 4(3)(-8))) / 2(3). Simplifying this, we get: x = (2 ± √(4 + 96)) / 6. Further simplifying, we get: x = (2 ± √100) / 6. Therefore, x = 2 or x = -4/3.

Multiple choice

Solve the quadratic equation: 4x^2 + 12x + 9 = 0

  1. x = -3/2, -3/2

  2. x = 3/2, 3/2

  3. x = 1, 9

  4. x = -1, -9
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To solve this equation, we can use the quadratic formula. Plugging in the values of a, b, and c, we get: x = (-12 ± √(12^2 - 4(4)(9))) / 2(4). Simplifying this, we get: x = (-12 ± √(144 - 144)) / 8. Further simplifying, we get: x = (-12 ± 0) / 8. Therefore, x = -3/2 or x = -3/2.

Multiple choice

Solve the quadratic equation: 2x^2 - 5x + 2 = 0

  1. x = 1, 2

  2. x = -1, -2

  3. x = 1, -2

  4. x = -1, 2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To solve this equation, we can use factoring. We can factor the left-hand side as follows: (2x - 1)(x - 2) = 0. Setting each factor equal to 0, we get: 2x - 1 = 0 and x - 2 = 0. Solving these equations, we get: x = 1/2 and x = 2. Therefore, the solutions to the quadratic equation are x = 1 and x = 2.

Multiple choice

Solve the quadratic equation: 3x^2 + 7x + 2 = 0

  1. x = -1, -2

  2. x = 1, 2

  3. x = -1, 2

  4. x = 1, -2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To solve this equation, we can use the quadratic formula. Plugging in the values of a, b, and c, we get: x = (-7 ± √(7^2 - 4(3)(2))) / 2(3). Simplifying this, we get: x = (-7 ± √(49 - 24)) / 6. Further simplifying, we get: x = (-7 ± √25) / 6. Therefore, x = -1 or x = -2.

Multiple choice

Solve the quadratic equation: 5x^2 - 2x - 3 = 0

  1. x = 1, -3

  2. x = -1, 3

  3. x = 1, 3

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

To solve this equation, we can use factoring. We can factor the left-hand side as follows: (5x + 3)(x - 1) = 0. Setting each factor equal to 0, we get: 5x + 3 = 0 and x - 1 = 0. Solving these equations, we get: x = -3/5 and x = 1. Therefore, the solutions to the quadratic equation are x = 1 and x = -3.

Multiple choice

What is the general form of a quadratic equation?

  1. $$ax^2 + bx + c = 0$$
  2. $$ax^2 - bx + c = 0$$
  3. $$ax^2 + bx - c = 0$$
  4. $$ax^2 - bx - c = 0$$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The general form of a quadratic equation is $$ax^2 + bx + c = 0$$, where a, b, and c are constants and $$a \ne 0$$.

Multiple choice

What is the discriminant of a quadratic equation?

  1. $$b^2 - 4ac$$
  2. $$b^2 + 4ac$$
  3. $$b^2 - 2ac$$
  4. $$b^2 + 2ac$$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The discriminant of a quadratic equation is $$b^2 - 4ac$$. It determines the nature of the roots of the equation.

Multiple choice

What is the nature of the roots of a quadratic equation if the discriminant is positive?

  1. Real and distinct

  2. Real and equal

  3. Imaginary

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If the discriminant is positive, the roots of the quadratic equation are real and distinct.

Multiple choice

What is the nature of the roots of a quadratic equation if the discriminant is zero?

  1. Real and distinct

  2. Real and equal

  3. Imaginary

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If the discriminant is zero, the roots of the quadratic equation are real and equal.

Multiple choice

Solve the quadratic equation $$x^2 + 4x + 3 = 0$$ using Brahmagupta's Formula.

  1. $$x = -1 \pm \sqrt{2}$$
  2. $$x = -2 \pm \sqrt{2}$$
  3. $$x = -3 \pm \sqrt{2}$$
  4. $$x = -4 \pm \sqrt{2}$$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using Brahmagupta's Formula, we have $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. Substituting the values of a, b, and c, we get $$x = \frac{-4 \pm \sqrt{4^2 - 4(1)(3)}}{2(1)}$$. Simplifying this, we get $$x = -1 \pm \sqrt{2}$$. Therefore, the solution set is ({-1 \pm \sqrt{2})).

Multiple choice

Solve the quadratic equation $$2x^2 - 5x + 2 = 0$$ using Brahmagupta's Formula.

  1. $$x = \frac{5 \pm \sqrt{21}}{4}$$
  2. $$x = \frac{5 \pm \sqrt{29}}{4}$$
  3. $$x = \frac{5 \pm \sqrt{37}}{4}$$
  4. $$x = \frac{5 \pm \sqrt{41}}{4}$$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using Brahmagupta's Formula, we have $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. Substituting the values of a, b, and c, we get $$x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(2)}}{2(2)}$$. Simplifying this, we get $$x = \frac{5 \pm \sqrt{21}}{4}$$. Therefore, the solution set is ({\frac{5 \pm \sqrt{21}}{4})).