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 equation: $x^2-3x=10$

  1. 5, -2

  2. 4, 2

  3. 3, 6

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

Rearrange the equation to x^2 - 3x - 10 = 0. Factor the quadratic: (x - 5)(x + 2) = 0. The solutions are x = 5 and x = -2.

Multiple choice

Rational roots of the equation $2x^4+x^3-11x^2+x+2=0$ are
  1. $\dfrac {1}{2}, 2$
  2. $\dfrac {1}{2}, 2, \dfrac {1}{4}, -2$
  3. $\dfrac {1}{2}, 2, 3, 4$
  4. $\dfrac {1}{2}, 2, \dfrac {3}{4}, -2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the Rational Root Theorem, possible roots are factors of 2 divided by factors of 2, i.e., +/- 1, +/- 2, +/- 1/2. Testing x = 1/2: 2(1/16) + 1/8 - 11/4 + 1/2 + 2 = 1/8 + 1/8 - 22/8 + 4/8 + 16/8 = 0. Testing x = 2: 2(16) + 8 - 11(4) + 2 + 2 = 32 + 8 - 44 + 4 = 0. Both are roots.

Multiple choice

The product of the real roots of the equation, ${ x }^{ 2 } + 18x + 30 = 2 \sqrt { { x }^{ 2 } + 18x + 45 } $, is
  1. $20$
  2. $24$
  3. $16$
  4. $18$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let y = sqrt(x^2 + 18x + 45). Then x^2 + 18x + 30 = y^2 - 15. The equation becomes y^2 - 15 = 2y, or y^2 - 2y - 15 = 0. Factoring gives (y - 5)(y + 3) = 0. Since y must be positive, y = 5. Thus, sqrt(x^2 + 18x + 45) = 5, so x^2 + 18x + 45 = 25, or x^2 + 18x + 20 = 0. The product of the roots is c/a = 20/1 = 20.

Multiple choice

Let $-\dfrac{\pi}{6}< \theta <-\dfrac{\pi}{12}$. Suppose $\alpha_1$ and $\beta_1$ are the roots of the equation $x^2 - 2x \sec\theta+ 1 = 0 $, and $\alpha_2$ and $\beta_2$ are the roots of the equation $x^2 + 2 x\tan\theta - 1 = 0 $. If $\alpha_1 > \beta _2 $ , then $\alpha_1 + \beta_2$ equals
  1. $2(\sec\theta - \tan\theta)$
  2. $2\sec\theta$
  3. $-2\tan\theta$
  4. $0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For the first equation, the roots are sec(theta) +/- tan(theta). Given the range of theta, alpha1 = sec(theta) + tan(theta). For the second equation, the roots are -tan(theta) +/- sec(theta), so beta2 = -tan(theta) - sec(theta). Summing these gives alpha1 + beta2 = -2*tan(theta).

Multiple choice

If $a > b > c$, then equation $2 ( x - a ) ( x - b ) - ( x - c ) ^ { 2 } = 0$ has-
  1. exactly one root in $(b,a) $
  2. exactly one root in $(c,b)$
  3. both in $(c ,a)$
  4. one root in $( - \infty , b )$ and other in $( a , \infty )$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let f(x) = 2(x-a)(x-b) - (x-c)^2. f(c) = 2(c-a)(c-b) = 2(negative)(negative) > 0. f(b) = -(b-c)^2 < 0. Since f(c) > 0 and f(b) < 0, there is a root in (c,b). f(a) = -(a-c)^2 < 0. Since f(b) < 0 and f(a) < 0, we check limits. As x -> infinity, f(x) -> x^2 > 0. Thus there is another root in (a, infinity).

Multiple choice

Roots(s) of the equation $9x^{2}-18|x|+5=0$ belonging to the domain of definition of the $f(x)=\log (x^{2}-x-2)$ is (are)
  1. $-\dfrac {5}{3}, -\dfrac {1}{3}$
  2. $\dfrac {5}{3}, \dfrac {1}{3}$
  3. $-\dfrac {5}{3}$
  4. $-\dfrac {1}{3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let |x| = y. 9y^2 - 18y + 5 = 0. (3y-1)(3y-5) = 0. y = 1/3 or 5/3. |x| = 1/3 => x = +/- 1/3. |x| = 5/3 => x = +/- 5/3. Domain of log(x^2-x-2): x^2-x-2 > 0 => (x-2)(x+1) > 0. x > 2 or x < -1. Only -5/3 < -1.