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

If $\alpha$ and $\beta$ are the roots of the equation $x^2\, -\, 2x\, +\, 4\, =\, 0$, then $\alpha ^9\, +\, \beta ^9\, =$
  1. $-2^8$
  2. $2^9$
  3. $-2^{10}$
  4. $2^{10}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The roots of x^2 - 2x + 4 = 0 are 1 +/- i*sqrt(3), which can be written as 2(cos(pi/3) +/- i*sin(pi/3)). By De Moivre's theorem, alpha^9 + beta^9 = (2*e^(i*pi/3))^9 + (2*e^(-i*pi/3))^9 = 2^9 * (e^(i*3*pi) + e^(-i*3*pi)) = 2^9 * (cos(3*pi) + i*sin(3*pi) + cos(-3*pi) + i*sin(-3*pi)) = 2^9 * (-1 + 0 - 1 + 0) = -2 * 2^9 = -2^10.

Multiple choice

If $\alpha, \beta, \gamma$ are the roots of the equation $x^{3} + px^{2} + qx + r = 0$ then the coefficient of $x$ in the cubic equation whose roots are $\alpha (\beta + \gamma), \beta (\gamma + \alpha)$ and $\gamma (\alpha + \beta)$ is
  1. $2q$
  2. $q^{2} + pr$
  3. $p^{2} - qr$
  4. $r(pq - r)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Roots are a, b, c. a+b+c = -p, ab+bc+ca = q, abc = -r. New roots: a(b+c), b(c+a), c(a+b). Note a(b+c) = a(-p-a) = -ap - a^2. This is getting complicated. Let's use the property that the sum of new roots is -2(ab+bc+ca) = -2q. The sum of products taken two at a time is the coefficient of x. The calculation leads to q^2 + pr.

Multiple choice

If $\tan A$ and $\tan B$ are the roots of the quadratic equation $x^2-px+q=0$, then $\sin^2(A+B)=$
  1. $\dfrac{p^2}{p^2+q^2}$
  2. $\dfrac{p^2}{(p+q)^2}$
  3. $1-\dfrac{p}{(1-q)^2}$
  4. $\dfrac{p^2}{p^2+(1-q)^2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given tan A + tan B = p and tan A * tan B = q. Then tan(A+B) = (tan A + tan B) / (1 - tan A * tan B) = p / (1 - q). Using sin^2(theta) = tan^2(theta) / (1 + tan^2(theta)), we get sin^2(A+B) = (p/(1-q))^2 / (1 + (p/(1-q))^2) = p^2 / ((1-q)^2 + p^2).

Multiple choice

Let $p$ and $q$ be real numbers such that $p\neq 0,\ p^{3}\neq q$ and $p^{3}\neq-q$. If $\alpha$ and $\beta$ are non zero complex numbers satisfying $\alpha+\beta=-p$ and $\alpha^{3}+\beta^{3}=q$, then a quadratic equation having $\dfrac{\alpha}{\beta}$ and $\dfrac{\beta}{\alpha}$ as its roots is
  1. $(p^{3}+q)x^{2}-(p^{3}+2q)x+(p^{3}+q)=0$
  2. $(p^{3}+q)x^{2}-(p^{3}-2q)x+(p^{3}+q)=0$
  3. $(p^{3}-q)x^{2}-(5p^{3}-2q)x+(p^{3}-q)=0$
  4. $(p^{3}-q)x^{2}-(5p^{3}+2q)x+(p^{3}-q)=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given alpha + beta = -p and alpha^3 + beta^3 = q. We know alpha^3 + beta^3 = (alpha + beta)((alpha + beta)^2 - 3alpha*beta). So q = -p(p^2 - 3alpha*beta). q = -p^3 + 3p*alpha*beta, so alpha*beta = (p^3 + q) / 3p. The roots are x1 = alpha/beta and x2 = beta/alpha. Sum = (alpha^2 + beta^2) / (alpha*beta) = ((alpha+beta)^2 - 2alpha*beta) / (alpha*beta) = (p^2 - 2alpha*beta) / (alpha*beta). Product = 1. The quadratic is x^2 - (sum)x + 1 = 0.

Multiple choice

If $\alpha, \beta$ are the roots of the equations $x^{2} - 2x - 1 = 0$, then what is the value of $\alpha^{2} \beta^{-2} + \alpha^{-2}\beta^{2}$.
  1. $-2$
  2. $0$
  3. $30$
  4. $34$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given x^2 - 2x - 1 = 0, alpha+beta = 2, alpha*beta = -1. We want (alpha^2/beta^2) + (beta^2/alpha^2) = (alpha^4 + beta^4) / (alpha*beta)^2. alpha^2+beta^2 = (alpha+beta)^2 - 2*alpha*beta = 4 - 2(-1) = 6. alpha^4+beta^4 = (alpha^2+beta^2)^2 - 2(alpha*beta)^2 = 36 - 2(1) = 34. Result = 34 / (-1)^2 = 34.

Multiple choice

The root of the equation $2\left( 1+i \right) { x }^{ 2 }-4(2-i)x-5-3i=0 $ which has greater modulus is
  1. $\cfrac { 3-5i }{ 2 } $
  2. $\cfrac { 5-3i }{ 2 } $
  3. $\cfrac { 3-i }{ 2 } $

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

Solving 2(1+i)x^2 - 4(2-i)x - (5+3i) = 0 using the quadratic formula x = [-b +/- sqrt(b^2 - 4ac)] / 2a. Discriminant D = [4(2-i)]^2 - 4(2+2i)(-5-3i) = 16(4-1-4i) - 8(1+i)(-5-3i) = 16(3-4i) - 8(-5-3i-5i+3) = 48-64i - 8(-2-8i) = 48-64i + 16 + 64i = 64. Roots are [4(2-i) +/- 8] / [4(1+i)] = (8-4i +/- 8) / (4+4i). Root 1: (16-4i)/(4+4i) = (4-i)/(1+i) = (4-i)(1-i)/2 = (4-4i-i-1)/2 = (3-5i)/2. Root 2: (-4i)/(4+4i) = -i/(1+i) = -i(1-i)/2 = (-i-1)/2. Modulus of (3-5i)/2 is sqrt(9+25)/2 = sqrt(34)/2. Modulus of (-1-i)/2 is sqrt(2)/2. The first root has the greater modulus.