Mathematics ยท Quantitative Aptitude

Polynomial and Quadratic Equations

239 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 general form of an indeterminate equation of the first degree?

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

The general form of an indeterminate equation of the first degree is $ax + by = c$, where $a$, $b$, and $c$ are integers and $x$ and $y$ are the unknowns.

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

What is Brahmagupta's formula for solving a quadratic equation?

  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 a quadratic equation is $x = (-b \pm \sqrt{b^2 - 4ac}) / 2a$.

Multiple choice

What is the value of (a) in the equation (ax^2 + bx + c = 0) if the roots of the equation are equal?

  1. \(\frac{-b}{2a}\)
  2. \(\frac{b}{2a}\)
  3. \(\frac{-b}{a}\)
  4. \(\frac{b}{a}\)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For equal roots, the discriminant (b^2 - 4ac) must be equal to zero. Substituting (a) with (\frac{-b}{2a}) in the equation (ax^2 + bx + c = 0) satisfies this condition.

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

Which of the following is an example of an algebraic equation?

  1. 2x + 3 = 5

  2. x^2 - 4x + 3 = 0

  3. sin(x) = 0.5

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

An algebraic equation is an equation that contains variables and constants. The equation 2x + 3 = 5 is an example of an algebraic equation, as it contains the variable x and the constants 2, 3, and 5.

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 Brahmagupta's Formula for solving quadratic equations?

  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 $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$, where a, b, and c are the coefficients of the quadratic equation.

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 negative?

  1. Real and distinct

  2. Real and equal

  3. Imaginary

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

If the discriminant is negative, the roots of the quadratic equation are imaginary.

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

What is the formula provided by Brahmagupta for solving quadratic equations?

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

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