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

The equation $(\dfrac{x}{x+1})^{2}+(\dfrac{x}{x-1})^{2}=a(a-1)$ has
  1. $four \ real \ roots \ if \ a>2 $
  2. $four \ real \ roots \ if \ a<-1$
  3. $two \  real \  roots \  if \  a= 1$
  4. $no \ real \ root \ if a<-1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let f(x) = (x/(x+1))^2 + (x/(x-1))^2. This simplifies to 2(x^4 + x^2)/(x^2-1)^2. Analysis of the range of this function shows it has four real roots for specific values of a.

Multiple choice

If $\alpha,\beta$ are the roots of the equation $\left( x-a \right) \left( x-b \right) +c=0\left( c\neq 0 \right) ,$ then the roots of the equation $\left( x-c-\alpha \right) \left( x-c-\beta \right) =c$ are
  1. $a$ and $b+c$
  2. $a^2+c^2$ and $b^2+c^2$
  3. $a+c$ and $a-c$
  4. $a+c$ and $b+c$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The roots of (x-a)(x-b) + c = 0 are alpha and beta. This means x^2 - (a+b)x + ab + c = 0. The roots of (x-c-alpha)(x-c-beta) = c are x = c+alpha and x = c+beta. Since alpha and beta are roots of the original equation, adding c to them shifts the roots by c. Thus, the new roots are a+c and b+c.

Multiple choice

If one root of the equation $x ^ { 2 } + p x + q = 0$ is $2 + \sqrt { 3 },$ then the values of $p$ and $q$ are respectively.
  1. $- 4,1$
  2. $4 , - 1$
  3. $2 , \sqrt { 3 }$
  4. $- 2 - \sqrt { 3 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a quadratic equation with rational coefficients, if 2+sqrt(3) is a root, the other must be 2-sqrt(3). The sum of roots = -p = (2+sqrt(3)) + (2-sqrt(3)) = 4, so p = -4. The product of roots = q = (2+sqrt(3))(2-sqrt(3)) = 4-3 = 1.

Multiple choice

Two non - integer roots of the equation $(x^{2}+3x)^{2}-(x^{2}+3x)-6= 0$ are
  1. $\frac{1}{2}(-3+\sqrt{11}),\frac{1}{2}(-3-\sqrt{11})$
  2. $\frac{1}{2}(-3+\sqrt{7}),\frac{1}{2}(-3-\sqrt{7})$
  3. $\frac{1}{2}(-3+\sqrt{21}),\frac{1}{2}(-3-\sqrt{21})$

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

Let u = x^2 + 3x. The equation becomes u^2 - u - 6 = 0, which factors to (u-3)(u+2) = 0. So x^2 + 3x = 3 or x^2 + 3x = -2. Solving x^2 + 3x - 3 = 0 gives x = (-3 +/- sqrt(9 - 4(1)(-3)))/2 = (-3 +/- sqrt(21))/2.

Multiple choice

Solve the following equation. $3\, \left(x^{2}\, +\, \displaystyle \frac{1}{x^{2}}\right)\, -\, 4\, \left(x\, -\, \displaystyle \frac{1}{x}\right)\, -\, 6\, =\, 0$
  1. $x\, =\, 1,\, -1,\, \displaystyle \frac{-2\, \pm\, \sqrt{13}}{3}$
  2. $x\, =\, 1,\, -1,\, \displaystyle \frac{2\, \pm\, \sqrt{13}}{3}$
  3. $x\, =\, 1,\, -1,\, \displaystyle \frac{4\, \pm\, \sqrt{13}}{6}$

  4. None of these 
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: 3(y^2 + 2) - 4y - 6 = 0 => 3y^2 - 4y = 0. y(3y - 4) = 0. y=0 or y=4/3. If x - 1/x = 0, x^2 = 1, x = +/- 1. If x - 1/x = 4/3, 3x^2 - 4x - 3 = 0, x = (4 +/- sqrt(16 + 36))/6 = (4 +/- sqrt(52))/6 = (4 +/- 2*sqrt(13))/6 = (2 +/- sqrt(13))/3.