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
What is Brahmagupta's formula for solving a quadratic equation?
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$x = (-b \pm \sqrt{b^2 - 4ac}) / 2a$
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$x = (-b \pm \sqrt{b^2 + 4ac}) / 2a$
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$x = (-b \pm \sqrt{b^2 - 4ac}) / a$
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$x = (-b \pm \sqrt{b^2 + 4ac}) / a$
A
Correct answer
Explanation
Brahmagupta's formula for solving a quadratic equation is $x = (-b \pm \sqrt{b^2 - 4ac}) / 2a$.
What is the value of (a) in the equation (ax^2 + bx + c = 0) if the roots of the equation are equal?
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\(\frac{-b}{2a}\)
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\(\frac{b}{2a}\)
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\(\frac{-b}{a}\)
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\(\frac{b}{a}\)
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.
Brahmagupta's formula for solving quadratic equations is:
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$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
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$x = \frac{-b \pm \sqrt{b^2 + 4ac}}{2a}$
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$x = \frac{-b \pm \sqrt{b^2 - 2ac}}{2a}$
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$x = \frac{-b \pm \sqrt{b^2 + 2ac}}{2a}$
A
Correct answer
Explanation
Brahmagupta's formula for solving quadratic equations is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
Which of the following is an example of an algebraic equation?
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2x + 3 = 5
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x^2 - 4x + 3 = 0
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sin(x) = 0.5
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log(x) = 2
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.
What is the general form of a quadratic equation?
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$$ax^2 + bx + c = 0$$
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$$ax^2 - bx + c = 0$$
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$$ax^2 + bx - c = 0$$
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$$ax^2 - bx - c = 0$$
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 $$a \ne 0$$.
What is Brahmagupta's Formula for solving quadratic equations?
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$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
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$$x = \frac{-b \pm \sqrt{b^2 + 4ac}}{2a}$$
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$$x = \frac{-b \pm \sqrt{b^2 - 2ac}}{2a}$$
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$$x = \frac{-b \pm \sqrt{b^2 + 2ac}}{2a}$$
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.
What is the discriminant of a quadratic equation?
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$$b^2 - 4ac$$
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$$b^2 + 4ac$$
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$$b^2 - 2ac$$
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$$b^2 + 2ac$$
A
Correct answer
Explanation
The discriminant of a quadratic equation is $$b^2 - 4ac$$. It determines the nature of the roots of the equation.
What is the nature of the roots of a quadratic equation if the discriminant is positive?
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Real and distinct
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Real and equal
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Imaginary
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None of the above
A
Correct answer
Explanation
If the discriminant is positive, the roots of the quadratic equation are real and distinct.
What is the nature of the roots of a quadratic equation if the discriminant is negative?
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Real and distinct
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Real and equal
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Imaginary
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None of the above
C
Correct answer
Explanation
If the discriminant is negative, the roots of the quadratic equation are imaginary.
What is the nature of the roots of a quadratic equation if the discriminant is zero?
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Real and distinct
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Real and equal
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Imaginary
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None of the above
B
Correct answer
Explanation
If the discriminant is zero, the roots of the quadratic equation are real and equal.
Solve the quadratic equation $$x^2 + 4x + 3 = 0$$ using Brahmagupta's Formula.
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$$x = -1 \pm \sqrt{2}$$
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$$x = -2 \pm \sqrt{2}$$
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$$x = -3 \pm \sqrt{2}$$
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$$x = -4 \pm \sqrt{2}$$
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})).
What is the formula provided by Brahmagupta for solving quadratic equations?
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$x = (-b \pm \sqrt{b^2 - 4ac}) / 2a$
-
$x = (-b \pm \sqrt{b^2 + 4ac}) / 2a$
-
$x = (-b \pm \sqrt{b^2 - 2ac}) / 2a$
-
$x = (-b \pm \sqrt{b^2 + 2ac}) / 2a$
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.
What is the solution to the equation x^2 - 4x + 3 = 0?
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x = 1, x = 3
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x = 2, x = 3
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x = 1, x = 4
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x = 2, x = 4
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. 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. Therefore, the solutions to the equation are x = 1 and x = 3.