Mathematics · Quantitative Aptitude

Polynomial and Quadratic Equations

223 Questions

Polynomial and quadratic equations involve finding the roots of expressions with varying degrees. The focus is on the relationship between coefficients and roots, symmetric functions, and solving higher degree polynomials. This topic is a core component of algebra in competitive testing.

Roots of quadratic equationsSymmetric root functionsCubic polynomialsRoot approximation methodsSum and product of roots

Polynomial and Quadratic Equations Questions

Multiple choice

What is the formula for solving a quadratic equation $ax^2 + bx + c = 0$ according to Bhaskara I?

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

Bhaskara I's formula for solving a quadratic equation is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Multiple choice

What is the solution to the equation x^2 - 4x + 3 = 0?

  1. x = 1, x = 3

  2. x = 2, x = 4

  3. x = 3, x = 5

  4. x = 4, x = 6
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We can solve this equation using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a. Substituting a = 1, b = -4, and c = 3, we get x = (-(-4) ± √((-4)^2 - 4(1)(3))) / 2(1) = (4 ± √(16 - 12)) / 2 = (4 ± √4) / 2 = (4 ± 2) / 2. Therefore, the solutions are x = 1 and x = 3.

Multiple choice

What is the solution to the equation x^2 - 4x + 3 = 0?

  1. x = 1, x = 3

  2. x = -1, x = -3

  3. x = 2, x = 3

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

We can solve this equation using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a. Substituting the values of a, b, and c, we get x = (-(-4) ± √((-4)^2 - 4(1)(3))) / 2(1). Simplifying this equation, we get x = (4 ± √(16 - 12)) / 2. Further simplifying, we get x = (4 ± √4) / 2. Therefore, the solutions to the equation are x = 1 and x = 3.

Multiple choice

What is the name of the equation that states that the product of the two roots of a quadratic equation is equal to the constant term?

  1. Vieta's formula

  2. Aryabhata's formula

  3. Brahmagupta's formula

  4. Bhaskara's formula
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Vieta's formula is a well-known formula in algebra that states that the product of the two roots of a quadratic equation is equal to the constant term.

Multiple choice

Brahmagupta's formula for solving quadratic equations is:

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

Brahmagupta's formula for solving quadratic equations is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Multiple choice

Bhaskara II's work on algebra includes the study of quadratic equations. What is the formula for solving a quadratic equation of the form (ax^2 + bx + c = 0) using the method described by Bhaskara II?

  1. \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
  2. \(x = \frac{-b \pm \sqrt{b^2 + 4ac}}{2a}\)
  3. \(x = \frac{-b \pm \sqrt{b^2 - 2ac}}{2a}\)
  4. \(x = \frac{-b \pm \sqrt{b^2 + 2ac}}{2a}\)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Bhaskara II's formula for solving a quadratic equation is (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}), where (a), (b), and (c) are the coefficients of the quadratic equation.

Multiple choice

Brahmagupta's formula for solving quadratic equations is given by: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. What is the discriminant of this quadratic equation?

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

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

Multiple choice

What is the solution to the equation x^2 - 4x + 3 = 0?

  1. x = 1, 3

  2. x = -1, -3

  3. x = 2, 3

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

We can solve this equation using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a. Substituting a = 1, b = -4, and c = 3, we get x = (-(-4) ± √((-4)^2 - 4(1)(3))) / 2(1) = (4 ± √(16 - 12)) / 2 = (4 ± √4) / 2 = (4 ± 2) / 2. Therefore, x = 1 or x = 3.

Multiple choice

What was Bhaskara II's formula for solving quadratic equations?

  1. $x = (-b \pm \sqrt{b^2 - 4ac}) / 2a$
  2. $x = (-b \pm \sqrt{b^2 + 4ac}) / 2a$
  3. $x = (-b \pm \sqrt{b^2 - 4ac}) / a$
  4. $x = (-b \pm \sqrt{b^2 + 4ac}) / a$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Bhaskara II's formula for solving quadratic equations is $x = (-b \pm \sqrt{b^2 - 4ac}) / 2a$.

Multiple choice

What was Bhaskara II's formula for solving cubic equations?

  1. $x = (-b \pm \sqrt{b^2 - 4ac}) / 2a$
  2. $x = (-b \pm \sqrt{b^2 + 4ac}) / 2a$
  3. $x = (-b \pm \sqrt{b^2 - 4ac}) / a$
  4. $x = (-b \pm \sqrt{b^2 + 4ac}) / a$
Reveal answer Fill a bubble to check yourself
Correct answer
Explanation

Bhaskara II did not develop a formula for solving cubic equations.

Multiple choice

What is the solution to the equation x^2 - 4x + 3 = 0?

  1. x = 1, x = 3

  2. x = 2, x = 4

  3. x = 3, x = 5

  4. x = 4, x = 6
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We can solve this equation using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a. Substituting the values of a, b, and c, we get x = (-(-4) ± √((-4)^2 - 4(1)(3))) / 2(1). Simplifying this equation, we get x = 1, x = 3.

Multiple choice

What is the solution to the equation x^2 - 4x + 3 = 0?

  1. x = 1, x = 3

  2. x = 2, x = 3

  3. x = 1, x = 4

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

To solve the equation x^2 - 4x + 3 = 0, we can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a. In this case, a = 1, b = -4, and c = 3. Substituting these values into the formula, we get x = (-(-4) ± √((-4)^2 - 4(1)(3))) / 2(1) = (4 ± √(16 - 12)) / 2 = (4 ± √4) / 2 = (4 ± 2) / 2. Therefore, the solutions to the equation are x = 1 and x = 3.

Multiple choice

What is Brahmagupta's formula for solving quadratic equations?

  1. $x = (-b \pm \sqrt{b^2 - 4ac}) / 2a$
  2. $x = (-b \pm \sqrt{b^2 + 4ac}) / 2a$
  3. $x = (-b \pm \sqrt{b^2 - 4ac}) / a$
  4. $x = (-b \pm \sqrt{b^2 + 4ac}) / a$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Brahmagupta's formula for solving quadratic equations is $x = (-b \pm \sqrt{b^2 - 4ac}) / 2a$.

Multiple choice

What was the Babylonian method for solving quadratic equations?

  1. The Babylonian Method

  2. The Quadratic Formula

  3. The Completing the Square Method

  4. The Factoring Method
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The Babylonians had a method for solving quadratic equations that involved using a series of approximations to find the solution.

Multiple choice

Which of the following is not a method for solving quadratic equations?

  1. Completing the square

  2. Quadratic formula

  3. Vieta's formulas

  4. Lagrange's method
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Lagrange's method is a method for solving higher-degree polynomial equations, including quadratic equations. It is not a specific method for solving quadratic equations like completing the square, quadratic formula, or Vieta's formulas.