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

Find the discriminant of the equation and the nature of roots. Also find the roots , if they are real: $3x^2 - 2x + \frac{1}{3} = 0$

  1. Roots are imaginary
  2. D $=$ 0, Roots are real and equal $\displaystyle \dfrac{1}{3}, \frac{1}{3}$
  3. D$ =$ $\dfrac{2}{5}$, Roots are real and unequal $\displaystyle \frac{1}{5}, \frac{1}{2}$

  4. Cannot be determined
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Equation: 3x^2 - 2x + 1/3 = 0. Multiply by 3: 9x^2 - 6x + 1 = 0. This is (3x-1)^2 = 0. Discriminant D = (-6)^2 - 4*9*1 = 36 - 36 = 0. Roots are 1/3, 1/3.

Multiple choice

The quadratic equation whose roots are the A.M. and H.M. between the roots of the equation, $2x^2- 3x + 5 = 0$ is
  1. $4x^2 - 25x + 10 = 0$
  2. $12x^2 - 49x + 30 = 0$
  3. $14x^2 - 12x + 35 = 0$
  4. $2x^2 + 3x + 5 = 0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For 2x^2 - 3x + 5 = 0, sum of roots alpha+beta = 3/2 and product alpha*beta = 5/2. AM = (alpha+beta)/2 = 3/4. HM = 2*alpha*beta/(alpha+beta) = 5/(3/2) = 10/3. The equation is x^2 - (AM+HM)x + (AM*HM) = 0, which is x^2 - (3/4 + 10/3)x + (3/4 * 10/3) = 0, leading to 12x^2 - 49x + 30 = 0.

Multiple choice

If the roots of the quadratic equation $ax^2 + bx + c = 0$ are imaginary, then for all values of $a, b, c$ and $x \in R$, the expression $a^2x^2 + abx + ac$ is

  1. positive

  2. non-negative

  3. negative

  4. may be positive, zero and negative
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If roots are imaginary, the discriminant b^2 - 4ac < 0. The expression a^2x^2 + abx + ac can be rewritten by completing the square or analyzing the discriminant of the quadratic in x, which is (ab)^2 - 4(a^2)(ac) = a^2(b^2 - 4ac). Since b^2 - 4ac < 0 and a^2 > 0, the discriminant is negative, meaning the expression maintains the same sign as a^2, which is positive.

Multiple choice

Comment upon the nature of roots of the quadratic equation $\displaystyle x^{2}+2x=k+\int_{0}^{1}\left | t+k \right |dx$ depending on the value of $\displaystyle k\in R$
  1. real & distinct $\displaystyle \forall k\in R$
  2. only distinct $\displaystyle \forall k\notin R$
  3. only real $\displaystyle \forall k\in R$

  4. none of above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The discriminant is 4(1 + k + integral from 0 to 1 of |t + k| dt), which remains positive for every real k. Therefore, the quadratic always has two real and distinct roots.

Multiple choice

If one root of the quadratic equation $px^2 + qx + r = 0 (p \neq 0)$ is a surd $\displaystyle \frac{\sqrt{a}}{\sqrt{a} + \sqrt{a-b}}$, where $p, q, r, a, b$ are all rationals, then the other root is
  1. $\displaystyle \frac{\sqrt{b}}{\sqrt{a} - \sqrt{a-b}}$
  2. $\displaystyle a + \frac{\sqrt{a(a-b)}}{b}$
  3. $\displaystyle \frac{a + \sqrt{a(a-b)}}{b}$
  4. $\displaystyle \frac{\sqrt{a} - \sqrt{a-b}}{\sqrt{b}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The given root is sqrt(a)/(sqrt(a) + sqrt(a-b)). Rationalizing the denominator by multiplying by (sqrt(a) - sqrt(a-b)) gives (sqrt(a^2) - sqrt(a(a-b))) / (a - (a-b)) = (a - sqrt(a(a-b))) / b. Since the coefficients are rational, the conjugate root must be (a + sqrt(a(a-b))) / b.

Multiple choice

The roots of the equation $\displaystyle 2x^{2}-11x+15=0$
  1. $\displaystyle 3,\frac{5}{2}$
  2. $\displaystyle 5,\frac{3}{2}$
  3. $\displaystyle -3,-\frac{5}{2}$

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

2x^2 - 11x + 15 = 0. Roots are (11 +/- sqrt(121 - 120)) / 4 = (11 +/- 1) / 4. Roots are 12/4 = 3 and 10/4 = 2.5.

Multiple choice

The roots of the equation $a^2x^2 + (a + b)x - b^2 = 0$ are

  1. real and different

  2. real and equal

  3. imaginary

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

The discriminant D = (a + b)^2 - 4(a^2)(-b^2) = a^2 + 2ab + b^2 + 4a^2b^2. Since a^2, b^2, and 4a^2b^2 are non-negative, D is positive for any non-zero real a and b. Thus, the roots are real and different.