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

Solve the following quadratic equation by factorization : $(a \, + \, b)^2x^2 \, - \, 4abx \, - \, (a \, - \, b)^2 \, = \, 0$ The roots are 1, $-\left(\dfrac{a \, - \, b}{a \, + \, b} \right )^2$

  1. True

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

Factoring (a+b)^2 x^2 - 4abx - (a-b)^2 = 0. The roots are x = 1 and x = -((a-b)/(a+b))^2 is incorrect. The roots are actually 1 and -((a-b)/(a+b))^2 is not the standard factorization result for this specific quadratic.

Multiple choice

Solve the following quadratic equation by factorization, the root is : $-1$ and $1$ $\dfrac{x \, - \, 1}{2x \, + \, 1} \, + \, \dfrac{2x \, + \, 1}{x \, - \, 1} \, = \, \dfrac{5}{2}, \, x \, \neq \, -\dfrac{1}{2}, \, 1$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation is (x-1)/(2x+1) + (2x+1)/(x-1) = 5/2. Let y = (x-1)/(2x+1). Then y + 1/y = 5/2, which leads to 2y^2 - 5y + 2 = 0. Solving gives y = 2 or y = 1/2. If (x-1)/(2x+1) = 2, x-1 = 4x+2, so 3x = -3, x = -1. If (x-1)/(2x+1) = 1/2, 2x-2 = 2x+1, which is impossible. Thus, x = -1 is a root. The statement says the roots are -1 and 1, but x=1 is excluded by the denominator. Therefore, the statement is false.

Multiple choice

The given quadratic equations have real roots and the roots are $-\sqrt2, \, \dfrac{-5}{\sqrt2}$ : $\sqrt{2}x^2 \, + \, 7x \, + \, 5\sqrt{2} \, = \, 0$

  1. True

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

For the equation sqrt(2)x^2 + 7x + 5sqrt(2) = 0, the roots are given by the quadratic formula: x = [-7 +/- sqrt(49 - 4*sqrt(2)*5sqrt(2))] / (2*sqrt(2)) = [-7 +/- sqrt(49 - 40)] / (2*sqrt(2)) = [-7 +/- 3] / (2*sqrt(2)). Roots are -4/(2*sqrt(2)) = -2/sqrt(2) = -sqrt(2) and -10/(2*sqrt(2)) = -5/sqrt(2). These match the given roots.

Multiple choice

In the following, determine whether the given quadratic equations have real roots and if so, find the roots : $16x^2$ = 24x + 1
  1. $\text{not real}$
  2. $real,\dfrac{3 \, \pm \, \sqrt{10}}{4}$
  3. $\dfrac{3 \, \pm \, \sqrt{10}}{2}$
  4. $\dfrac{3 \, \pm \, \sqrt{13}}{4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

16x^2 - 24x - 1 = 0. Discriminant D = b^2 - 4ac = (-24)^2 - 4(16)(-1) = 576 + 64 = 640. Roots = (-b +/- sqrt(D)) / 2a = (24 +/- sqrt(640)) / 32 = (24 +/- 8*sqrt(10)) / 32 = (3 +/- sqrt(10)) / 4.

Multiple choice

In the following, determine whether the given quadratic equation have real roots and if so, find the roots : $\sqrt{3}x^2 \, + \, 10x \, - \, 8\sqrt{3} \, = \, 0$
  1. $-4\sqrt3, \, \dfrac{2}{\sqrt 3}$
  2. $4\sqrt3, \, \dfrac{2}{\sqrt 3}$
  3. $-4\sqrt3, \, \dfrac{-2}{\sqrt 3}$

  4. not real
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the quadratic formula x = (-b +/- sqrt(b^2 - 4ac)) / 2a. Here a = sqrt(3), b = 10, c = -8sqrt(3). Discriminant = 100 - 4(sqrt(3))(-8sqrt(3)) = 100 + 96 = 196. x = (-10 +/- 14) / 2sqrt(3). Roots are 4/2sqrt(3) = 2/sqrt(3) and -24/2sqrt(3) = -12/sqrt(3) = -4sqrt(3).