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 roots $z_1$, $z_2$, $z_3$ of the equation $x^3 + 3ax^2 + 3bx + c = 0$ in which a, b, c are complex numbers, correspond to the points A,B, C on the complex plane. Then ABC is an equilateral triangle if
  1. $a^2=b$
  2. $a=b^2$
  3. $a+b^2=0$
  4. $a^2+b=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a cubic equation x^3 + 3ax^2 + 3bx + c = 0, the roots form an equilateral triangle in the complex plane if the derivative of the cubic has roots that are equal, or more specifically, if a^2 = b.

Multiple choice

Each of the following quadratic equations represents the graph shown. Which equation reveals the exact values of the $x-$intercepts of the graph?
  1. $y=\cfrac{1}{2}(2x-5)(x+1)$
  2. $y=2{x}^{2}-\cfrac{3}{2}x-{5}$
  3. $y={ \left( x-\cfrac { 3 }{ 4 } \right) }^{ 2 }$
  4. $y={ \left( x-\cfrac { 3 }{ 2 } \right) }^{ 2 }-\cfrac{9}{16}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The factored form y = a(x - r1)(x - r2) directly shows the x-intercepts as r1 and r2. Option A is in a form that can be easily factored to show intercepts.

Multiple choice

lf $ 0<\alpha<\beta<\gamma <\displaystyle \dfrac{\pi}{2}$ , then the equation $\displaystyle \dfrac{1}{x-\sin\alpha}+\dfrac{1}{x-\sin\beta}+\dfrac{1}{x-\sin\gamma}=0$ has

  1. imaginary roots

  2. real and equal roots

  3. real and unequal roots

  4. rational roots
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let f(x) = 1/(x-sin(alpha)) + 1/(x-sin(beta)) + 1/(x-sin(gamma)). Since sin(alpha) < sin(beta) < sin(gamma), the function has vertical asymptotes at these values. Between each asymptote, the function goes from -infinity to +infinity, ensuring three real roots.

Multiple choice

Let $\alpha,:\beta$ be roots of the equation $\displaystyle ax^{2}+bx+c=0$ and $\Delta=b^{2}-4ac$. If $\displaystyle \alpha+\beta,\alpha^{2}+\beta^{2},\alpha^{3}+\beta^{3}$ are in $GP$, then
  1. $\Delta \neq 0$
  2. $b\cdot \Delta = 0$
  3. $c\cdot \Delta = 0$
  4. $\Delta = 0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For terms to be in GP, the square of the middle term must equal the product of the outer terms: (alpha^2 + beta^2)^2 = (alpha + beta)(alpha^3 + beta^3). Expanding this leads to the condition c * Delta = 0.

Multiple choice

If roots of $x^3 + 5x^2 - 7x - 1 = 0$ are $\alpha, \beta, \gamma$, then the equation whose roots are $\alpha \beta, \beta \gamma, \gamma \alpha$, is
  1. $x^3 - 7x^2 + 5x + 1 = 0$
  2. $x^3 + 7x^2 - 5x - 1 = 0$
  3. $x^3 + 5x^2 + 7x + 1 = 0$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given x^3 + 5x^2 - 7x - 1 = 0, the roots are a, b, c. We have a+b+c = -5, ab+bc+ca = -7, and abc = 1. We want an equation with roots ab, bc, ca. Let y = ab = 1/c, y = bc = 1/a, y = ca = 1/b. Substituting c = 1/y into the original equation: (1/y)^3 + 5(1/y)^2 - 7(1/y) - 1 = 0. Multiplying by -y^3 gives y^3 + 7y^2 - 5y - 1 = 0.

Multiple choice

lf $\alpha$ is the repeated root of quadratic equation $f(x)=0$ and $A(x)$ , $B(x)$ and $C(x)$ are polynomials of degree 3,4 and 5 respectively, then $\phi(x)=$ $\begin{vmatrix} A\left ( x \right ) &B\left ( x \right ) &C\left ( x \right ) \ A\left ( \alpha \right )&B\left ( \alpha \right ) & C\left ( \alpha \right )\ A^{'}\left ( \alpha \right ) &B^{'}\left ( \alpha \right ) & C^{'}\left ( \alpha \right ) \end{vmatrix}$ is divisible by
  1. $f\left ( x \right )$
  2. $A\left ( x \right )$
  3. $B\left ( x \right )$
  4. $C\left ( x \right )$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If alpha is a repeated root of f(x)=0, then f(alpha)=0 and f'(alpha)=0. The determinant phi(x) has a row where all elements are evaluated at alpha, and another row where all derivatives are evaluated at alpha. Since alpha is a repeated root, the determinant will be zero at x=alpha, implying (x-alpha)^2 is a factor, which is f(x).

Multiple choice

If $\alpha$ and $\beta$ are the roots of the quadratic equation $ax^2 + bx + c = 0,$ then $\displaystyle\lim_{x\rightarrow \frac{1}{\alpha}}\sqrt{\dfrac{1-cos(cx^{2}+bx+a)}{2(1-\alpha x)^{2}}}$
  1. $\left | \dfrac{c}{2\alpha}\left ( \dfrac{1}{\alpha }-\dfrac{1}{\beta } \right ) \right |$
  2. $\left | \dfrac{c}{2\beta}\left ( \dfrac{1}{\alpha }-\dfrac{1}{\beta } \right ) \right |$
  3. $\left | \dfrac{c}{\alpha \beta}\left ( \dfrac{1}{\alpha }-\dfrac{1}{\beta } \right ) \right |$

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

The limit involves a quadratic expression inside a cosine. Using the Taylor expansion for cos(u) approx 1 - u^2/2, the expression simplifies to |c/2a * (1/a - 1/b)|.

Multiple choice

Consider a quadratic polynomial $f(x) = x^{2} - 4ax + 5a^{2} - 6a.$ $p :$ denotes smallest positive integral value of $a$, for which $f(x)$ is positive for every real $x$, and $q :$ denotes largest distance between the roots of the equation $f(x)= 0$ The value of $p + q$ is
  1. $9$
  2. $11$
  3. $13$
  4. $15$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For f(x) > 0 for all x, discriminant D < 0. 16a^2 - 4(5a^2 - 6a) < 0. 16a^2 - 20a^2 + 24a < 0. -4a^2 + 24a < 0. a(a-6) > 0. Smallest positive integer a = 7 (p=7). Roots distance = sqrt(D)/a_coeff = sqrt(16a^2 - 20a^2 + 24a) = sqrt(24a - 4a^2). Max distance at a=3, sqrt(72-36) = 6 (q=6). p+q = 13.