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 the roots of the equation $x ^ { 2 } + p x + q = 0$ differ from the roots of the equation $x ^ { 2 } + q x + p = 0$ by the same quantity. then $p + q$
  1. $-4$
  2. $-2$
  3. $-3$

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

Let roots of the first be a, b and second be c, d. Given |a-b| = |c-d|. Squaring gives (a+b)^2 - 4ab = (c+d)^2 - 4cd. Substituting coefficients: p^2 - 4q = q^2 - 4p. Rearranging: p^2 - q^2 + 4p - 4q = 0, which is (p-q)(p+q) + 4(p-q) = 0. Assuming p != q, we get p+q = -4.

Multiple choice

If $p$ and $q$ are non-zero real numbers and $\alpha^{3}+\beta^{3}=-p, \alpha \beta=q$, then a quadratic equation whose roots are $\dfrac{\alpha^{2}}{\beta}, \dfrac{\beta^{2}}{\alpha}$ is :
  1. $px^{2}-qx+p^{2}=0$
  2. $qx^{2}+px+q^{2}=0$
  3. $px^{2}+qx+p^{2}=0$
  4. $qx^{2}-px+q^{2}=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given alpha^3 + beta^3 = -p and alpha*beta = q. We know alpha^3 + beta^3 = (alpha+beta)(alpha^2 - alpha*beta + beta^2) = (alpha+beta)((alpha+beta)^2 - 3*alpha*beta). Let S = alpha+beta. S(S^2 - 3q) = -p. The roots of the new equation are a = alpha^2/beta and b = beta^2/alpha. Sum = (alpha^3 + beta^3)/(alpha*beta) = -p/q. Product = alpha*beta = q. The equation is x^2 - (sum)x + product = 0, so x^2 + (p/q)x + q = 0, which is qx^2 + px + q^2 = 0.

Multiple choice

A root of the equation $( x - 1 ) ( x - 2 ) = \dfrac { 30 } { 49 }$ is
  1. $\dfrac { 17 } { 7 }$
  2. $\dfrac { 15 } { 7 }$
  3. $\dfrac { 13 } { 7 }$
  4. $\dfrac { 11 } { 7 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

(x - 1)(x - 2) = 30/49. x^2 - 3x + 2 - 30/49 = 0. x^2 - 3x + 68/49 = 0. Using quadratic formula: x = [3 +/- sqrt(9 - 4*68/49)] / 2 = [3 +/- sqrt(441/49 - 272/49)] / 2 = [3 +/- sqrt(169/49)] / 2 = [3 +/- 13/7] / 2. Roots are (21/7 + 13/7)/2 = 34/14 = 17/7 and (21/7 - 13/7)/2 = 8/14 = 4/7.

Multiple choice

Find the set of $'a'$ for which the equation has real roots $x^{2}+2ax+\dfrac{1}{16}=-a+\sqrt{a^{2}+x-\dfrac{1}{16}}$
  1. $a \in \left(-\infty, \dfrac{1}{4}\right] \cup \left[\dfrac{3}{4}, \infty\right)$
  2. $ a \in (-\infty, 1]$
  3. $a \in $\right[\dfrac{1}{4},\dfrac{3}{4}\right]$
  4. $a \in $\right[\dfrac{1}{4},\dfrac{1}{2}\right]$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the equation to have real roots, the discriminant of the quadratic must be non-negative and the square root term must be defined. Solving the inequality leads to the interval (-infinity, 1/4] union [3/4, infinity).

Multiple choice

The number of roots of the equation $x^2 - 2x - log_2 |1 - x| = 3$ is

  1. 4

  2. 2

  3. 1

  4. 0
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation x^2 - 2x - 3 = log_2 |1 - x| can be solved by graphing or checking intersection points. The function f(x) = x^2 - 2x - 3 is a parabola, and g(x) = log_2 |1 - x| has vertical asymptotes at x = 1. Analyzing the intersections shows 4 distinct roots.