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 ${z}{1}$ and ${z}{2}$ are two complex roots of equal magnitude and their arguments differ by $\displaystyle \frac { \pi }{ 2 } $, of the quadratic equation $a{ x }^{ 2 }+bx+c=0\left( a\neq 0 \right) $ then $a($ in terms of $b$ and $c)$ is
  1. $\displaystyle \frac { { b }^{ 2 } }{ 2c } $
  2. $\displaystyle \frac { { b }^{ 2 } }{ c } $
  3. $\displaystyle \frac { { b } }{ 2c } $

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

Let roots be r*exp(i*theta) and r*exp(i*(theta + pi/2)). Product of roots c/a = r^2 * exp(i*(2*theta + pi/2)) = r^2 * i * exp(i*2*theta). Sum of roots -b/a = r*exp(i*theta) * (1 + i). Squaring: b^2/a^2 = r^2 * exp(i*2*theta) * (1 + i)^2 = r^2 * exp(i*2*theta) * 2i. Thus, b^2/a^2 = 2 * (c/a) = 2c/a. a = b^2 / 2c.

Multiple choice

Let ${z}{1}$ and ${z}{2}$ be roots of the equation ${z}^{2}+pz+q=0,$ where the coefficients $p$ and $q$ may be complex numbers. Let $A$ and $B$ represents ${z}{1}$ and ${z}{2}$ in the complex plane. If $\angle AOB=\alpha \neq 0$ and $OA=OB,$ where $O$ is the origin, then $\displaystyle{ p }^{ 2 }=k\cos ^{ 2 }{ \frac { \alpha }{ 2 } } ,$ where $k=$
  1. $q$
  2. $2q$
  3. $4q$

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

The roots are z1 = r*exp(i*theta1) and z2 = r*exp(i*theta2) with |theta1 - theta2| = alpha. Then p = -(z1 + z2) and q = z1*z2. Using the identity (z1 + z2)^2 = (z1 - z2)^2 + 4*z1*z2, and noting (z1 - z2)^2 = r^2 * (exp(i*theta1) - exp(i*theta2))^2 = r^2 * exp(i*(theta1+theta2)) * (exp(i*alpha/2) - exp(-i*alpha/2))^2 * exp(-i*alpha) = -4*r^2 * exp(i*(theta1+theta2)) * sin^2(alpha/2). Since q = r^2 * exp(i*(theta1+theta2)), p^2 = 4*q*sin^2(alpha/2) - 4*q*cos^2(alpha/2) is not quite right; the standard derivation leads to p^2 = 4q*cos^2(alpha/2).

Multiple choice

If the equation, $x^2+bx+45 = 0 (b \in R)$ has conjugate complex roots and they satisfy $|z + 1| = 2\sqrt{10}$, then:
  1. $b^2 + b = 72$
  2. $b^2 - b = 30$
  3. $b^2 + b = 12$
  4. $b^2 - b = 42$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For x^2 + bx + 45 = 0, roots are conjugate complex, so b^2 - 4*45 < 0. Let roots be p + iq and p - iq. Product of roots = p^2 + q^2 = 45. Sum of roots = 2p = -b. |z + 1| = 2*sqrt(10) => (p+1)^2 + q^2 = 40. p^2 + 2p + 1 + q^2 = 40. Since p^2 + q^2 = 45, 45 + 2p + 1 = 40 => 2p = -6 => p = -3. Then -b = 2p = -6 => b = 6. Check: b^2 - b = 36 - 6 = 30.

Multiple choice

Solve the equation $\displaystyle x^{4}-4x^{2}+8x+35= 0$ having given that one root is $\displaystyle 2+\sqrt{-3}.$
  1. The roots of the given equation are $\displaystyle -2\pm i,2\pm\sqrt{3}i.$
  2. The roots of the given equation are $\displaystyle 2\pm i,2\pm\sqrt{3}i.$
  3. The roots of the given equation are $\displaystyle \pm2+ i,2\pm\sqrt{3}i.$
  4. The roots of the given equation are $\displaystyle \pm2- i,2\pm\sqrt{3}i.$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given root 2 + i*sqrt(3). Since coefficients are real, 2 - i*sqrt(3) is also a root. Sum = 4, product = 4 + 3 = 7. Quadratic factor: x^2 - 4x + 7. Dividing x^4 - 4x^2 + 8x + 35 by x^2 - 4x + 7 gives x^2 + 4x + 5 = 0. Roots are (-4 +/- sqrt(16 - 20))/2 = -2 +/- i. So roots are 2 +/- i*sqrt(3) and -2 +/- i.

Multiple choice

Let $z_k(k = 0, 1, 2, ............ 6)$ be the roots of the equation $(z + 1)^7 + z^7 = 0$, then $\sum { k=0 }^{ 6 }{ Re({ z }{ k }) } $ is equal to
  1. $0$
  2. $\dfrac{3}{2}$
  3. $-\dfrac{7}{2}$
  4. $\dfrac{7}{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The equation can be rewritten as ((z + 1)/z)^7 = -1. This implies that the modulus of (z + 1)/z is 1, which geometrically means that z is equidistant from 0 and -1, so its real part must be -1/2. Since the equation is of degree 7, there are 7 roots, and the sum of their real parts is 7 * (-1/2) = -7/2.

Multiple choice

State whether the following statement is true or false. If $\beta$ is an imaginary root of the equation $z^n \, - \, 1 \, = \, 0$, then $1 \, + \, \beta \, + \, \beta^2 \, + \, ..... \, + \, \beta^{n \, - \, 1} \, = \, 0$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The expression 1 + beta + beta^2 + ... + beta^(n-1) is a geometric series sum equal to (beta^n - 1) / (beta - 1). Since beta is a root of z^n - 1 = 0, beta^n = 1. Thus, the sum is 0 / (beta - 1) = 0.

Multiple choice

Let $x_1, x_2,$ are the roots of quadratic equation $x^2 + ax + b = 0$, Where $ a, b$ are complex numbers and $y_1, y_2$ are the roots of the quadratic equation $y^2 + \left|a\right| y + \left|b\right| =0$. If $\left|x_1\right| = \left|x_2\right| = 1$, then
  1. $\left|y_1\right| = \left|y_2\right|= 1$.
  2. $\left|y_1\right| = \left|y_2\right|\neq 1$.
  3. $\left|y_1\right| \neq 1, \left|y_2\right|= 1$.
  4. $\left|y_1\right| =1, \left|y_2\right|\neq 1$.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given |x1| = |x2| = 1, the product of roots |x1*x2| = |b| = 1. The equation for y is y^2 + |a|y + 1 = 0. Since the roots of y^2 + |a|y + 1 = 0 have a product of 1, if one root has magnitude 1, the other must also have magnitude 1 (since |y1*y2| = 1).

Multiple choice

Consider the equation $az^2 + z + 1 = 0$ having purely imaginary root where $a = \cos \theta + i \sin \theta, i = \sqrt{-1}$ and function $f(x) = x^3 - 3x^2 + 3(1 + \cos \theta)x + 5,$ then answer the following questions. Which of the following is true?
  1. $f(x) = 0$ has three real distinct roots
  2. $f(x) = 0$ has one positive real root
  3. $f(x) = 0$ has one negative real root
  4. $f(x) = 0$ has three but not distinct roots
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The root of az^2 + z + 1 = 0 is purely imaginary (let it be ki). a(k^2 i^2) + ki + 1 = 0 => -ak^2 + ki + 1 = 0. Since a = cos(theta) + i sin(theta), this leads to cos(theta) = 0, so theta = pi/2. f(x) = x^3 - 3x^2 + 3(1+0)x + 5 = x^3 - 3x^2 + 3x + 5. f(-1) = -1 - 3 - 3 + 5 = -2. f(0) = 5. By IVT, there is a root between -1 and 0.

Multiple choice

If $\displaystyle ^{7}C_{r} + 3 ^{7}C_{r+1} + 3 ^{7}C_{r+2} + ^{7}C_{r+3} > ^{10}C_{4}$, then the quadratic equation whose roots are $\displaystyle \alpha, : \beta$ and $\displaystyle \alpha^{r-1}, : \beta^{r-1}$ have

  1. no common roots

  2. only one common root

  3. two common root

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

Using the binomial identity, the left side becomes C(10, r + 3). The inequality requires C(10, r + 3) > C(10, 4), which gives r = 2. Hence α^(r-1) = α and β^(r-1) = β, so both roots are common.

Multiple choice

Consider a rational function $f(x)= \dfrac {x^{2} - 3x - 4}{x^{2} - 3x + 4}$ and a quadratic function $g(x) = x^{2} - (b + 1) x + b - 1$, where $b$ is a parameter. The sum of integers in the range of $f(x)$, is
  1. $-5$
  2. $-6$
  3. $-9$
  4. $-10$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The function f(x) = (x^2 - 3x - 4) / (x^2 - 3x + 4) can be rewritten as 1 - 8 / (x^2 - 3x + 4). The quadratic x^2 - 3x + 4 has a minimum value of 7/4 at x = 1.5, so the expression 8 / (x^2 - 3x + 4) ranges from 8/(infinity) to 8/(7/4) = 32/7. Thus, f(x) ranges from 1 - 32/7 = -25/7 to 1 - 0 = 1. The integers in this range [-3.57, 1] are -3, -2, -1, 0, 1. Their sum is -5.