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 roots of the following equation: $\displaystyle\,35y^2\,+\,\frac{12}{y^2}\,=\,44$, then
  1. $\displaystyle\,y\,=\,\pm\,\sqrt\frac{6}{7},\,\,\pm\,\sqrt\frac{2}{5}$
  2. $\displaystyle\,y\,=\,\pm\,\sqrt\frac{6}{11},\,\,\pm\,\sqrt\frac{2}{5}$
  3. $\displaystyle\,y\,=\,\pm\,\sqrt\frac{6}{7},\,\,\pm\,\sqrt\frac{2}{7}$
  4. $\displaystyle\,y\,=\,\pm\,\sqrt\frac{6}{7},\,\,\pm\,\sqrt\frac{3}{5}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let u = y^2. The equation is 35u + 12/u = 44. Multiply by u: 35u^2 - 44u + 12 = 0. Using the quadratic formula, u = (44 +/- sqrt(1936 - 1680)) / 70 = (44 +/- 16) / 70. u1 = 60/70 = 6/7, u2 = 28/70 = 2/5. Thus y = +/- sqrt(6/7) and +/- sqrt(2/5).

Multiple choice

Find the roots of the following equation: $\displaystyle\,2\left(x^2\,+\,\frac{1}{x^2}\right)\,-\,9\,\left(x\,+\,\frac{1}{x}\right)\,+\,14\,=\,0$, then
  1. $\displaystyle\,x\,=\,-1, \,2,\, \frac{1}{2}$
  2. $\displaystyle\,x\,=\,1, \,2,\, \frac{1}{2}$
  3. $\displaystyle\,x\,=\,1, \,-2,\, \frac{1}{2}$
  4. $\displaystyle\,x\,=\,1, \,2,\, \frac{1}{4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let y = x + 1/x. Then x^2 + 1/x^2 = y^2 - 2. Equation: 2(y^2 - 2) - 9y + 14 = 0. 2y^2 - 9y + 10 = 0. (2y - 5)(y - 2) = 0. y = 2 or y = 2.5. If x + 1/x = 2, x = 1. If x + 1/x = 2.5, x = 2 or 1/2.

Multiple choice

Find the roots of the following equation $\displaystyle\,2y^2\,+\,\frac{15}{y^2}\,=\,12$, then
  1. $\displaystyle\,y\,=\,\pm\,\sqrt\frac{6\,+\,\sqrt6}{4},\,y\,=\,\pm\,\sqrt\frac{6\,-\,\sqrt6}{2}$
  2. $\displaystyle\,y\,=\,\pm\,\sqrt\frac{6\,+\,\sqrt6}{2},\,y\,=\,\pm\,\sqrt\frac{6\,-\,\sqrt5}{2}$
  3. $\displaystyle\,y\,=\,\pm\,\sqrt\frac{6\,+\,\sqrt6}{2},\,y\,=\,\pm\,\sqrt\frac{6\,-\,\sqrt6}{2}$
  4. $\displaystyle\,y\,=\,\pm\,\sqrt\frac{6\,+\,\sqrt5}{2},\,y\,=\,\pm\,\sqrt\frac{6\,-\,\sqrt6}{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let u = y^2. 2u + 15/u = 12 => 2u^2 - 12u + 15 = 0. Using quadratic formula: u = [12 +/- sqrt(144 - 120)] / 4 = [12 +/- sqrt(24)] / 4 = [12 +/- 2*sqrt(6)] / 4 = (6 +/- sqrt(6)) / 2. Since y^2 = u, y = +/- sqrt((6 +/- sqrt(6))/2).

Multiple choice

Find the roots of the following equation: $ (x^2+x)(x^2+x-7)+10=0 $, then ( x 2 + x ) ( x 2 + x − 7 ) + 10 = 0 (x2+x)(x2+x−7)+10= , then
  1. $\displaystyle\,x\,=\,\frac{-1\,\pm\,\sqrt{21}}{2},\,1,\,-2$
  2. $\displaystyle\,x\,=\,\frac{-1\,\pm\,\sqrt{23}}{2},\,1,\,-2$
  3. $\displaystyle\,x\,=\,\frac{-1\,\pm\,\sqrt{21}}{2},\,-1,\,-2$
  4. $\displaystyle\,x\,=\,\frac{-1\,\pm\,\sqrt{21}}{2},\,1,\,2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let y = x^2 + x. Then y(y - 7) + 10 = 0, so y = 5 or y = 2. These give x = (-1 +/- sqrt(21))/2 and x = 1, -2, respectively.

Multiple choice

Find the roots of the following equation $3x^4\,-\,13x^2\,+\,10\,=\,0$, then
  1. $\displaystyle\,x\,=\,\pm\,\sqrt\frac{10}{3},\,\pm\,3$
  2. $\displaystyle\,x\,=\,\pm\,\sqrt\frac{10}{7},\,\pm\,1$
  3. $\displaystyle\,x\,=\,\pm\,\sqrt\frac{10}{3},\,\pm\,1$
  4. $\displaystyle\,x\,=\,\pm\,\sqrt\frac{10}{3},\,\pm\,2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let y = x^2. The equation becomes 3y^2 - 13y + 10 = 0. Factoring: (3y - 10)(y - 1) = 0. So y = 10/3 or y = 1. Therefore, x^2 = 10/3 or x^2 = 1. Roots are x = +/- sqrt(10/3) and x = +/- 1.