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 equation $\begin{vmatrix} a-x & c & b \ c & b-x & a \ b & a & c-x \end{vmatrix}=0$ where $a+b+c\neq 0$.
  1. $x=a+b+c,\pm \sqrt { \cfrac { 1 }{ 2 } \left\{ { \left( a-b \right) }^{ 2 }-{ \left( b-c \right) }^{ 2 }-{ \left( c-a \right) }^{ 2 } \right\} } $
  2. $x=a+b+c,\pm \sqrt { \cfrac { 1 }{ 2 } \left\{ { \left( a-b \right) }^{ 2 }+{ \left( b-c \right) }^{ 2 }+{ \left( c-a \right) }^{ 2 } \right\} } $
  3. $0$
  4. $1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The determinant of this circulant matrix is -(a+b+c)(x^2 - (a+b+c)x + (ab+bc+ca - a^2-b^2-c^2)). Setting to 0 gives x = a+b+c or x^2 - (a+b+c)x + (ab+bc+ca - a^2-b^2-c^2) = 0. The roots of the quadratic part are (a+b+c)/2 +/- sqrt((a+b+c)^2/4 - (ab+bc+ca - a^2-b^2-c^2)). This simplifies to the expression in option B.

Multiple choice

If $a, b, c$ are roots of the equation $Ax^{3}+Bx^{2}+C=0$ then the value of determinant $\begin{vmatrix} ab & bc & ca\ bc & ca & ab\ ca & ab & bc \end{vmatrix}$ equals to=?
  1. $A$
  2. $B$
  3. $C$
  4. $0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The determinant is a circulant determinant. For roots a, b, c of Ax^3 + Bx^2 + C = 0, the product abc = -C/A. The determinant evaluates to -(a^3 + b^3 + c^3 - 3abc)(abc). Since a, b, c are roots, the expression is related to the symmetric polynomials. Specifically, for this structure, the determinant is 0.

Multiple choice

If $x = -1$ is a root of the equation $\begin{vmatrix} 2-x &3 & 3\ 3& 4-x&5 \ 3 & 5 & 4-x \end{vmatrix}=0$ then the other roots are

  1. 0,11

  2. 11,12

  3. 0, 12

  4. 1,11
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Expanding the determinant with x = -1: The determinant becomes |3 3 3; 3 5 5; 3 5 5|. Since two rows are identical, the determinant is 0, confirming x = -1 is a root. Solving the characteristic equation or testing the options, we find the roots are -1, 0, and 11.

Multiple choice

Let $\Delta =\begin{vmatrix} -bc & { b }^{ 2 }+bc & { c }^{ 2 }+bc \ { a }^{ 2 }+ac & -ac & { c }^{ 2 }+ac \ { a }^{ 2 }+ab & { b }^{ 2 }+ab & -ab \end{vmatrix}$ and the equation $p{ x }^{ 3 }+q{ x }^{ 2 }+rx+s=0$ has roots $a,b,c$, where $a,b,c\in { R }^{ +\quad }$. The value of $\Delta$ is
  1. $\le 9{ r }^{ 2 }/{ p }^{ 2 }$
  2. $\ge 27{ s }^{ 2 }/{ p }^{ 2 }$
  3. $\le 27{ s }^{ 3 }/{ p }^{ 3 }$

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

The determinant Delta evaluates to (abc)^2 * (a+b+c) * (1/a + 1/b + 1/c). Given roots a, b, c of px^3 + qx^2 + rx + s = 0, we have a+b+c = -q/p, ab+bc+ca = r/p, and abc = -s/p. The expression simplifies using these relations to show Delta >= 27s^2/p^2 based on AM-GM inequality.

Multiple choice

The number of real roots of equation $\log_{e}x + ex = 0$
  1. $0$ (zero)
  2. $1$
  3. $2$
  4. $3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let f(x) = log(x) + ex. The derivative f'(x) = 1/x + e. Since x must be positive for log(x) to be defined, f'(x) is always positive for x > 0. Thus, the function is strictly increasing. As x approaches 0, f(x) approaches -infinity, and as x approaches infinity, f(x) approaches infinity. By the intermediate value theorem, there is exactly one root.

Multiple choice

If $\alpha,\ \beta$ are roots of the quadratic equation $\mathrm{x}^{2}-\mathrm{x}(\mathrm{a}-2)-(\mathrm{a}+1)=0$ when $\mathrm{a}$ is a variable then the least value of $\alpha^{2}+\beta^{2}$
  1. $5$
  2. $3$
  3. $4$
  4. $1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

alpha + beta = a-2, alpha*beta = -(a+1). alpha^2 + beta^2 = (alpha+beta)^2 - 2*alpha*beta = (a-2)^2 + 2(a+1) = a^2 - 4a + 4 + 2a + 2 = a^2 - 2a + 6. This is a parabola opening upward with vertex at a = -(-2)/2 = 1. Min value = 1^2 - 2(1) + 6 = 5.