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

If $\omega \neq 1$ is the cube root of unity, then roots of the equation $\left ( x-a\omega \right )^{3}-a^{3}=0$ are
  1. $-a,2a\omega ,-a\omega ^{2}$
  2. $a,-2a\omega ,a\omega ^{2}$
  3. $a,2a\omega^{2} ,-2a\omega $
  4. $2a,-2a\omega^{2} ,-2a\omega^{2} $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

(x - a*omega)^3 = a^3. Taking cube roots: x - a*omega = a, a*omega, a*omega^2. x = a + a*omega, a*omega + a*omega, a*omega^2 + a*omega. Using 1 + omega + omega^2 = 0, the roots are -a*omega^2, 2a*omega, -a.

Multiple choice

The common roots of the equations $x^{3}+2x^{2}+2x+1=0$ and $x+x^{2}+x^{3}=0$ are ( $\omega$ is non real cube root of unity)
  1. $\omega,\omega^{2}$
  2. $1$
  3. $\omega+\omega^{2}$
  4. $-1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The first equation x^3 + 2x^2 + 2x + 1 = 0 factors to (x+1)(x^2 + x + 1) = 0. The second equation x + x^2 + x^3 = 0 factors to x(1 + x + x^2) = 0. The common roots are the roots of x^2 + x + 1 = 0, which are the complex cube roots of unity, omega and omega^2.

Multiple choice

Given that the complex number $z = -2 + 7i$ is a root to the equation: $z^3 + 6 z^2 + 61 z + 106 = 0$ find the real root to the equation.
  1. $ z = 2$
  2. $z = -2$
  3. $ z = -1$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since the coefficients are real, if -2 + 7i is a root, its conjugate -2 - 7i must also be a root. The quadratic factor is (z - (-2 + 7i))(z - (-2 - 7i)) = (z + 2)^2 + 49 = z^2 + 4z + 53. Dividing z^3 + 6z^2 + 61z + 106 by z^2 + 4z + 53 gives (z + 2). Thus, the real root is z = -2.

Multiple choice

The common roots of the equation $Z^3+2z^2+2z+1=0, z^{223}+z^{224}+1=0$ are
  1. $\omega,\omega^2$
  2. $1,\omega,\omega^2$
  3. $-1,\omega.\omega^2$
  4. $-\omega,-\omega^2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

z^3+2z^2+2z+1 = (z+1)(z^2+z+1) = 0. Roots are -1, omega, omega^2. For z^223+z^224+1=0: If z=-1, (-1)^223+(-1)^224+1 = -1+1+1 = 1 (not 0). If z=omega, omega^223+omega^224+1 = omega^1+omega^2+1 = 0. If z=omega^2, omega^446+omega^448+1 = omega^2+omega^1+1 = 0.

Multiple choice

$\displaystyle \alpha, : \beta$ are roots of the equation $\displaystyle x^{2} - x + 1 = 0$ then $\displaystyle \alpha^{1027} + \beta^{1027}$ equals
  1. $\displaystyle 1$
  2. $\displaystyle 0$
  3. $\displaystyle -1$
  4. $\displaystyle 1027$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The roots of x^2 - x + 1 = 0 are -omega and -omega^2 (where omega is the cube root of unity). alpha^1027 + beta^1027 = (-omega)^1027 + (-omega^2)^1027 = -(omega^1027 + omega^2054). Since omega^3 = 1, omega^1027 = omega^1 = omega and omega^2054 = omega^2. Thus, -(omega + omega^2) = -(-1) = 1.

Multiple choice

The number of integral values of a, $a\in(6,100)$ for which the equation $[tan\;x]^{2}+tan\;x-a=0$ has real roots (where [.] denotes greatest integer function) is:

  1. 7

  2. 8

  3. 5

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

Let y = tan x. The equation is y^2 + y - a = 0. For real roots, the discriminant D = 1 - 4(1)(-a) = 1 + 4a >= 0, which is always true for a > 6. However, since y = tan x, y can be any real number. The equation y^2 + y - a = 0 has roots y = (-1 +/- sqrt(1 + 4a)) / 2. For tan x to have these values, the roots must be real, which they are. The constraint is on the floor function [tan x]. The problem likely implies [tan x] is a constant k, so k^2 + k - a = 0, meaning a = k^2 + k. For a in (6, 100), k^2 + k - a = 0 implies k is an integer. For k=3, a=12; k=4, a=20; k=5, a=30; k=6, a=42; k=7, a=56; k=8, a=72; k=9, a=90. There are 7 such values.

Multiple choice

If $\alpha$ and $\beta$ are the roots of the equation $x^{2} + px + q = 0$, then $-\alpha^{-1}, -\beta^{-1}$ are the roots of which one of the following equations?
  1. $qx^{2} - px + 1 = 0$
  2. $qx^{2} + px + 1 = 0$
  3. $x^{2} + px - q = 0$
  4. $x^{2} - px + q = 0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If roots are alpha, beta, then alpha + beta = -p and alpha * beta = q. New roots are -1/alpha, -1/beta. Sum = -(1/alpha + 1/beta) = -(alpha+beta)/(alpha*beta) = p/q. Product = 1/(alpha*beta) = 1/q. Equation: x^2 - (p/q)x + 1/q = 0, which is qx^2 - px + 1 = 0.

Multiple choice

The equation $x^ {4}+16x-12=0$ has

  1. all the roots real

  2. all the roots complex

  3. two roots real and two roots complex

  4. all the roots integers
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let f(x) = x^4 + 16x - 12. f'(x) = 4x^3 + 16. f'(x) = 0 at x = -cube root(4) approx -1.587. f(-1.587) = (-1.587)^4 + 16(-1.587) - 12 = 6.35 - 25.39 - 12 = -31.04. Since the local minimum is negative and the function goes to infinity at both ends, there are two real roots. The other two must be complex.