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 number of real roots of the equation: (x2 + 4x)2 - (x + 2)2 - 26 = 0 is

  1. 0

  2. 1

  3. 2

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

Let y = x^2 + 4x. The equation is y^2 - (x+2)^2 - 26 = 0. This does not simplify easily. Instead, expand: (x^2+4x)^2 - (x^2+4x+4) - 26 = 0. Let u = x^2+4x. Then u^2 - u - 30 = 0. Factoring gives (u-6)(u+5) = 0. Case 1: x^2+4x-6=0 has discriminant 16 - 4(1)(-6) = 40 > 0 (2 roots). Case 2: x^2+4x+5=0 has discriminant 16 - 4(1)(5) = -4 < 0 (0 real roots). Total real roots = 2.

Multiple choice

Find the real roots of the quadratic equation x2 – 9x + 18 = 0 and for the two roots x and x1, find a and b such that Log10 a = x and Log10 b = x1.

  1. 18, 6

  2. 103, 109

  3. 106, 103

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

Factoring x^2 - 9x + 18 gives (x - 3)(x - 6) = 0, so the roots are 3 and 6. Since log10 a = x and log10 b = x1, the corresponding values are powers of 10, namely 10^3 and 10^6. Option C lists these values in reverse order, which is allowed because the roots can be labeled either way.

Multiple choice

Let z1 and z2 be the roots of the equation z2 + pz + q = 0, where p and q are real. The points represented by z1, z2 and the origin form an equilateral triangle, if

  1. p2 = 3q

  2. p2 > 3q

  3. p2 < 3q

  4. p2 = 2q

  5. p = 3q
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For roots z1, z2 and the origin to form an equilateral triangle, the condition is z1^2 + z2^2 - z1*z2 = 0. Since z1+z2 = -p and z1*z2 = q, we substitute: (z1+z2)^2 - 3*z1*z2 = 0, which gives p^2 - 3q = 0, or p^2 = 3q.

Multiple choice

The roots of the equation x2 - x + 1 = 0 are

  1. imaginary and unequal

  2. real and unequal

  3. real and equal

  4. imaginary and equal
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For x^2 - x + 1 = 0, the discriminant D = b^2 - 4ac = (-1)^2 - 4(1)(1) = 1 - 4 = -3. Since the discriminant is negative, the roots are imaginary and unequal.