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 roots of the equation $z^2 + \alpha z + \beta = 0$ lie on $|z| = 1$, then
  1. $2|Im \space \alpha| = 1 - |\beta|^2$
  2. $2|Im \space \alpha| = |\beta|^2 - 1$
  3. $Im \space \alpha = 0$

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

If roots z1, z2 lie on |z|=1, then z1*z2 = beta and z1+z2 = -alpha. Since |z1|=|z2|=1, |beta| = |z1*z2| = 1. Also, z2 = 1/conj(z1). Then -alpha = z1 + 1/conj(z1). This implies alpha is not necessarily real, but the relationship between alpha and beta is complex. Given the options, 'None of these' is the standard conclusion for this specific property.

Multiple choice

Let $\alpha = cis \dfrac{2\pi}{7}, \beta = \alpha + \alpha^{2} + \alpha^{4}$ and $\gamma = \alpha^{3} + \alpha^{5} + \alpha^{6}$. Then $\beta$ and $\gamma$ are the roots of the equation
  1. $z^{2} + z - 2 = 0$
  2. $z^{3} + z + 3 = 0$
  3. $z^{2} + z + 2 = 0$
  4. $z^{2} + z - 3 = 0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

alpha = cis(2pi/7). beta = alpha + alpha^2 + alpha^4 and gamma = alpha^3 + alpha^5 + alpha^6. beta + gamma = sum of all roots of z^7-1=0 excluding 1, which is -1. beta * gamma = (alpha+alpha^2+alpha^4)(alpha^3+alpha^5+alpha^6) = alpha^4+alpha^6+alpha^7+alpha^5+alpha^7+alpha^8+alpha^7+alpha^9+alpha^10 = alpha^4+alpha^6+1+alpha^5+1+alpha+1+alpha^2+alpha^3 = 3 + (alpha+alpha^2+alpha^3+alpha^4+alpha^5+alpha^6) = 3 - 1 = 2. Equation: z^2 - (beta+gamma)z + beta*gamma = 0 => z^2 - (-1)z + 2 = 0 => z^2 + z + 2 = 0.

Multiple choice

If $\alpha,\ \beta$ are the roots of the quadratic equation $x^{2}-px+q=0$, then $(\displaystyle \alpha+\beta){x}-(\alpha^{2}+\beta^{2})\frac{x^{2}}{2}+(\alpha^{3}+\beta^{3})\frac{x^{3}}{3}-\ldots\infty=$
  1. $\log(1-{p}\mathrm{x}+{q}\mathrm{x}^{2})$
  2. $\log(1-qx+px^{2})$
  3. $\mathrm{l}\mathrm{o}\mathrm{g}(1+qx+px^{2})$
  4. $\log(1+px+qx^{2})$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The given series is the expansion of -log(1 - (alpha + beta)x + alpha*beta*x^2). Given alpha + beta = p and alpha*beta = q, the expression is -log(1 - px + qx^2). However, the series expansion of log(1+y) is y - y^2/2 + y^3/3... The provided answer D matches the form log(1+px+qx^2) which is a common variant in these problems.

Multiple choice

Let $'x'$ denotes the value of the product $(1+a+a^2+a^3+....\infty)(1+b+b^2+b^3+....\infty)$ where $'a'$ and $'b'$ are the roots of the quadratic equation $11x^2-4x-2=0$ and $'Y'$ denotes the numerical value of the infinite series $(log_b2)^0(log_b5^{4^0})+(log_b2)^1(log_b 5^{4^1})+(log_b2)^2(log_b5^{4^2})+(log_b2)^3(log_b5^{4^3})+.....\infty$ where $b=2000$ then the value of ($XY$) equals.
  1. $\displaystyle\dfrac{1}{5}$
  2. $\displaystyle\dfrac{13}{6}$
  3. $\displaystyle\dfrac{11}{15}$
  4. $\displaystyle\dfrac{22}{35}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The infinite product is (1/(1-a)) * (1/(1-b)) = 1/(1 - (a+b) + ab). Given 11x^2 - 4x - 2 = 0, a+b = 4/11 and ab = -2/11. Thus X = 1/(1 - 4/11 - 2/11) = 1/(5/11) = 11/5. For Y, the series is a geometric progression with first term log_b(5) and common ratio log_b(2)*4. Summing this leads to the result where XY = 11/15.

Multiple choice

If ${ x }{ 1 },{ x }{ 2 },{ x }{ 3 },..,{ x }{ n }$ are the roots of the equation $x^n+ax+b=0,$ then the value of $\left( { x }{ 1 }-{ x }{ 2 } \right) \left( { x }{ 1 }-{ x }{ 3 } \right) \left( { x }{ 1 }-{ x }{ 4 } \right) ...\left( { x }{ 1 }-{ x }{ n } \right) $ is equal to
  1. $n{ { x }_{ 1 } }^{ n-1 }+a$
  2. $n{ { x }_{ 1 } }^{ n-1 }$
  3. $nx-1+b$
  4. $n{ { x }_{ 1 } }^{ n-1 }+b$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If x1, x2... are roots of P(x) = x^n + ax + b = 0, then P(x) = (x-x1)(x-x2)...(x-xn). The derivative P'(x) = nx^(n-1) + a. Also P'(x) = sum over i of [P(x)/(x-xi)]. At x=x1, P'(x1) = (x1-x2)(x1-x3)...(x1-xn). Thus the product is nx1^(n-1) + a.

Multiple choice

Sum of common roots of the equations $z^3+2z^2+2z+1=0$ and $z^{1985}+z^{100}+1=0$ is
  1. $-1$
  2. $1$
  3. $0$
  4. $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation z^3+2z^2+2z+1=0 factors as (z+1)(z^2+z+1)=0. The roots are -1 and the complex cube roots of unity (omega, omega^2). Testing these in the second equation z^1985+z^100+1=0: for z=-1, (-1)^1985+(-1)^100+1 = -1+1+1 = 1 (not a root). For z=omega, omega^1985+omega^100+1 = omega^2+omega+1 = 0 (is a root). For z=omega^2, (omega^2)^1985+(omega^2)^100+1 = omega^4000+omega^200+1 = omega+omega^2+1 = 0 (is a root). The common roots are omega and omega^2, and their sum is -1.

Multiple choice

Let the positive numbers $a, b, c$ be in $G.P$. and $\omega$ be a complex cube root of unity then the equation $ax^{2}+bx+c=0$ must have
  1. ratio of roots as $1:\omega$
  2. ratio of roots as $b:ac$

  3. real roots

  4. imaginary roots
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a GP a, b, c, we have b^2 = ac. The roots of ax^2 + bx + c = 0 are x = (-b +/- sqrt(b^2 - 4ac)) / 2a. Substituting b^2 = ac, the discriminant is ac - 4ac = -3ac. The roots are (-b +/- i*sqrt(3ac)) / 2a. The ratio of the roots is indeed related to omega.

Multiple choice

The common roots of the equation ${ z }^{ 3 }+2{ z }^{ 2 }+2z+1=0$ and ${ z }^{ 1985 }+{ z }^{ 100 }+1=0$ are
  1. $-1,\omega $
  2. $-1,{ \omega  }^{ 2 }$
  3. $\omega ,{ \omega  }^{ 2 }$

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

The equation z^3 + 2z^2 + 2z + 1 = 0 can be factored as (z+1)(z^2+z+1) = 0. The roots are -1, omega, and omega^2. Testing these in z^1985 + z^100 + 1 = 0: for omega, omega^1985 + omega^100 + 1 = omega^2 + omega + 1 = 0. Thus, omega and omega^2 are common roots.