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

The pure quadratic equation in the following is
  1. $ x+\cfrac { 1 }{ x } =4$
  2. $ x+\cfrac { 1 }{ x } =0$
  3. $x-\cfrac { 3 }{ 4 } =2x$
  4. $3x(x-1)=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A pure quadratic equation is of the form ax^2 + c = 0 (no x term). x + 1/x = 0 becomes x^2 + 1 = 0.

Multiple choice

The roots of the equation $2\sqrt{x} + 2 \dfrac{1}{\sqrt{x}} = 5$ can be found by solving:
  1. $16x^2 - 92x + 1 = 0$
  2. $4x^2 - 25x + 4 = 0$
  3. $4x^2 - 17x + 4 = 0$
  4. $2x^2 - 21x + 2 = 0$
  5. $4x^2 - 25x - 4 = 0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let u = sqrt(x). 2u + 2/u = 5 => 2u^2 - 5u + 2 = 0. Roots for u are 2 and 1/2. So sqrt(x) = 2 or 1/2. x = 4 or 1/4. The quadratic equation with these roots is (x - 4)(x - 1/4) = 0 => x^2 - 4.25x + 1 = 0 => 4x^2 - 17x + 4 = 0.

Multiple choice

The number of real roots of the equation $\dfrac{A^{2}}{x}+\dfrac{B^{2}}{x-1}=1$, where A and B are real numbers not equal to zero simultaneously is:

  1. None
  2. $1$
  3. $2$
  4. $1$ or $2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Equation: A^2(x-1) + B^2x = x(x-1) => A^2x - A^2 + B^2x = x^2 - x => x^2 - (1+A^2+B^2)x + A^2 = 0. This is a quadratic equation. The discriminant D = (1+A^2+B^2)^2 - 4A^2. Since A, B are real and not both zero, this can be positive or zero, leading to 1 or 2 real roots.