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

Consider the cubic equation $x^3+px^2+qx+r=0$, where p, q, r are real numbers. Which of the following statements is correct?
  1. If $p^2-2q < 0$, then the equation has one real and two imaginary root.
  2. If $p^2-2q\geq 0$, then the equation has all real roots.
  3. If $p^2-2q > 0$, then the equation has all real and distinct roots.
  4. If $4p^3-27q^2 > 0$, then the equation has real and distinct roots.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a cubic equation x^3 + px^2 + qx + r = 0, the nature of roots is determined by the discriminant. The condition p^2 - 2q < 0 relates to the derivative and the existence of local extrema, which implies one real root and two complex roots.

Multiple choice

If a root of the equation $ax^2+bx+c=0$ be reciprocal of a root of the equation $a'x^2+b'x+c'=0$, then?
  1. $(cc'-aa')^2=(ba'-cb')(ab'-bc')$
  2. $(bb'-aa')^2=(ca'-bc')(ab'-bc')$
  3. $(cc'-aa')^2=(ba'+cb')(ab'+bc')$

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

Let alpha be a root of the first equation. Then 1/alpha is a root of the second. Substituting 1/alpha into the second equation and using the first equation's root property leads to the condition (cc'-aa')^2 = (ba'-cb')(ab'-bc').

Multiple choice

The range of values of $a$ so that all the roots of the equation $2{ x }^{ 3 }-3{ x }^{ 2 }-12x+a=0$ are real and distinct, belongs to
  1. $\left( 7,20 \right) $
  2. $\left( -7,20 \right) $
  3. $\left( -20,7 \right) $
  4. $\left( -7,7 \right) $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let f(x) = 2x^3 - 3x^2 - 12x + a. For roots to be real and distinct, the local maximum and minimum must have opposite signs. f'(x) = 6x^2 - 6x - 12 = 6(x-2)(x+1). Critical points at x = 2, -1. f(2) = 16 - 12 - 24 + a = a - 20. f(-1) = -2 - 3 + 12 + a = a + 7. For opposite signs, (a-20)(a+7) < 0, so -7 < a < 20.

Multiple choice

If a and b are the zeros of the quadratic polynomial $f(x) = 6{x^2} + x - 2$, find the value of $\frac{a}{b} + \frac{b}{a}$.
  1. $\dfrac12$
  2. $\dfrac32$
  3. $\dfrac52$
  4. $-\dfrac{25} {12} $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For 6x^2 + x - 2 = 0, sum of roots (a+b) = -1/6, product (ab) = -2/6 = -1/3. We need (a/b + b/a) = (a^2 + b^2) / ab = ((a+b)^2 - 2ab) / ab. Substituting: ((-1/6)^2 - 2(-1/3)) / (-1/3) = (1/36 + 2/3) / (-1/3) = (1/36 + 24/36) / (-1/3) = (25/36) * (-3) = -25/12.

Multiple choice

Find a quadratic equation each with the given numbers as the sum and product of its zeroes respectively. $\dfrac {-10}{\sqrt 3}, 7$
  1. $3x^2-11\sqrt3x+21=0$
  2. $3x^2-17\sqrt2x+21=0$
  3. $3x^2+10 \sqrt3 x+21=0$
  4. $3x^2-15x+21=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a quadratic equation with sum of roots S and product of roots P, the equation is x^2 - Sx + P = 0. Here S = -10/sqrt(3) and P = 7, so x^2 - (-10/sqrt(3))x + 7 = 0. Multiplying by 3 gives 3x^2 + 10sqrt(3)x + 21 = 0.

Multiple choice

Find a quadratic equation each with the given numbers as the sum and product of its zeroes respectively. $2\sqrt 3-1, 3-\sqrt 3$
  1. $x^2-(2\sqrt 5-1)x+(3-\sqrt 3)$
  2. $x^2-(2\sqrt 11-1)x+(3-\sqrt 3)$
  3. $x^2-(2\sqrt 3-1)x+(3-\sqrt 3)$
  4. $x^2-(2\sqrt 7-1)x+(3-\sqrt 3)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A quadratic equation is x^2 - (sum of zeroes)x + (product of zeroes). Given sum = 2*sqrt(3) - 1 and product = 3 - sqrt(3), the equation is x^2 - (2*sqrt(3) - 1)x + (3 - sqrt(3)).

Multiple choice

Find the zeros of the quadratic polynomial $9x^2-5$.
  1. $\cfrac {  { 5 }  }{ 9 } , -\cfrac { 5  }{ 9 } $
  2. $\cfrac {  { 5 }  }{ 3 } , -\cfrac { 5  }{ 3 } $
  3. $\cfrac { \sqrt { 5 }  }{ 3 } , -\cfrac { \sqrt { 5 }  }{ 3 } $
  4. $\cfrac { \sqrt { 3 }  }{ 5 } , -\cfrac { \sqrt { 3 }  }{ 5 } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Set 9x^2 - 5 = 0. Then x^2 = 5/9, so x = +/- sqrt(5)/3.

Multiple choice

Find a quadratic polynomial whose sum and product respectively of the zeros, are $\dfrac {-3}{2\sqrt 5}, -\dfrac {1}{2}$ .
  1. Polynomial: $x^2+3\sqrt{5}x-7\sqrt{5}$
  2. Polynomial: $2\sqrt{5}x^2+3x-\sqrt{5}$
  3. Polynomial: $2x^2-7\sqrt{5}x-3\sqrt{5}$

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

A quadratic polynomial is given by x^2 - (sum of zeros)x + (product of zeros). Here, x^2 - (-3/(2*sqrt(5)))x + (-1/2) = x^2 + 3/(2*sqrt(5))x - 1/2. Multiplying by 2*sqrt(5) gives 2*sqrt(5)x^2 + 3x - sqrt(5).