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 $p, q, r$ in harmonic progression and $p$ & $r$ be different having same sign, then the roots of the equation $px^2\, +\, qx\, +\, r\, =\, 0$ are

  1. real and equal.

  2. real and distinct.

  3. irrational.

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

If p, q, r are in HP, then q = 2pr/(p+r). The discriminant D = q^2 - 4pr = (2pr/(p+r))^2 - 4pr = 4p^2r^2/(p+r)^2 - 4pr = 4pr [pr/(p+r)^2 - 1] = 4pr [ (pr - (p^2 + 2pr + r^2)) / (p+r)^2 ] = -4pr(p^2 + pr + r^2) / (p+r)^2. Since p and r have the same sign, pr > 0, so D < 0, meaning roots are imaginary.

Multiple choice

If $n$ is a positive integer and $n\in[5,100]$, then the number of integral roots of the equation $x^{2}+2x-n=0$ are
  1. $4$
  2. $6$
  3. $8$
  4. $10$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The equation x^2 + 2x - n = 0 has integral roots if the discriminant D = 4 + 4n is a perfect square, say k^2. This implies 4(1 + n) = k^2, so 1 + n must be a perfect square, m^2. For n in [5, 100], 1 + n is in [6, 101], so m^2 can be 9, 16, 25, 36, 49, 64, 81, 100, which gives 8 possible values for n.

Multiple choice

If roots of the equation $z^2 + \alpha z + \beta = 0 (\alpha, \beta \in C)$ are real, then value of $(Im(\beta))^2 + (Im(\alpha))(Im(\alpha\bar\beta))$ is Note: $Im(z)$ denotes imaginary part of $z$, where $z$ is a complex number.
  1. $0$
  2. $1$
  3. $-1$
  4. $i$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If roots are real, then the discriminant D = alpha^2 - 4*beta >= 0. Since alpha, beta are complex, this implies specific relations. However, for a quadratic with real roots, the coefficients must be real if the roots are real. If alpha and beta are real, then Im(alpha) = 0 and Im(beta) = 0. Thus, (Im(beta))^2 + Im(alpha)*Im(alpha*conj(beta)) = 0 + 0 = 0.

Multiple choice

A quadratic equation whose roots are $\displaystyle{\left(\frac{\gamma}{\alpha}\right)^2}$ and $\displaystyle{\left(\frac{\beta}{\alpha}\right)^2}$, where $\alpha, \beta, \gamma$ are the roots of $x^3 + 27 = 0$ is
  1. $x^2 - x + 1 = 0$
  2. $x^2 + 3x + 9 = 0$
  3. $x^2 + x + 1 = 0$
  4. $x^2 - 3x + 9 = 0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The roots of x^3 + 27 = 0 are -3, -3w, and -3w^2, where w is a complex cube root of unity. Let alpha = -3, beta = -3w, gamma = -3w^2. The required roots are (gamma/alpha)^2 = (w^2)^2 = w^4 = w and (beta/alpha)^2 = (w)^2 = w^2. The quadratic equation with roots w and w^2 is x^2 - (w+w^2)x + w^3 = 0, which simplifies to x^2 + x + 1 = 0.

Multiple choice

If $z_1, z_2$ are roots of the equation $az^2+bz+c=0$, with $a, b, c > 0;\ 2b^2 > 4ac > b^2;\ z_1 \in$ Third quadrant; $z_2\in$ Second quadrant in the argand's plane then, find $arg \left (\dfrac {z_1}{z_2}\right )=$
  1. $\cos^{-1}\left (\dfrac {b^2}{2ac}\right )^{\frac12}$
  2. $2\cos^{-1}\left (\dfrac {b^2}{4ac}\right )^{\frac12}$
  3. $\cos^{-1}\left (\dfrac {b^2}{4ac}\right )^{\frac12}$
  4. $2\cos^{-1}\left (\dfrac {b^2}{2ac}\right )^{\frac12}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given z1 in 3rd quadrant and z2 in 2nd quadrant, the argument of z1/z2 is arg(z1) - arg(z2). Using the properties of roots of quadratic equations and the given condition, the result simplifies to 2*cos^-1(sqrt(b^2/4ac)).

Multiple choice

Let $r,s,t$ be the roots of equation $8x^3+1001x+2008=0.$ If value of $(r+s)^3+(s+t)^3+(t+r)^3$ is a three digit number with $a$, $b$ and $c$ as its digits, then value of $a+b+c$ is:
  1. $15$
  2. $27$
  3. $21$
  4. $18$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

By Vieta's formulas, r+s+t = 0. Thus r+s = -t, s+t = -r, and t+r = -s. The expression becomes (-t)^3 + (-r)^3 + (-s)^3 = -(r^3+s^3+t^3). For 8x^3 + 1001x + 2008 = 0, x^3 = -1001/8 * x - 251. Summing for roots, r^3+s^3+t^3 = -1001/8(r+s+t) - 3*251 = -753. So -(r^3+s^3+t^3) = 753. Sum of digits 7+5+3 = 15.

Multiple choice

The set of values of $p$ such that both the roots of the equation $\left ( p-5 \right )x^{2}-2px+\left ( p-4 \right )=0$ are positive and one of the roots is less than $2$ and the other root lies between $2$ & $3$ is
  1. $\left ( \displaystyle \frac{49}{4},24 \right )$
  2. $\left ( 5,\infty \right )$
  3. $\left ( -\infty ,4 \right )\cup \left ( \displaystyle \frac{49}{4},\infty \right )$
  4. $\left ( 5,\displaystyle \frac{49}{4} \right )$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For roots to be positive and satisfy the location conditions, we analyze the function f(x) = (p-5)x^2 - 2px + (p-4). Conditions include f(0) > 0, f(2) < 0, f(3) > 0, and discriminant > 0. Solving these inequalities leads to the range (49/4, 24).

Multiple choice

If $a, b, c$ are positive and system of equations $ax + by + cz = 0, bx + cy + az = 0, cx + ay + bz = 0$ has non-trivial solutions. Then the roots of the equation $at^2 + bt + c = 0$ are

  1. real and opposite in sign

  2. both positive

  3. at least one positive

  4. non real
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The system has non-trivial solutions if the determinant of the coefficient matrix is zero. This leads to the condition a^3 + b^3 + c^3 - 3abc = 0, which factors as (a+b+c)(a^2+b^2+c^2-ab-bc-ca) = 0. Since a, b, c are positive, a+b+c is not zero, so a^2+b^2+c^2-ab-bc-ca = 0, implying a=b=c. The equation becomes at^2 + at + a = 0, or t^2 + t + 1 = 0, which has complex roots.

Multiple choice

The quadratic equation whose roots are the $x$ and $y$ intercepts of the line passing through $(1,1)$ and making a triangle of area $A$ with the axes, may be
  1. $ x^{2}+Ax+2A=0 $
  2. $ x^{2}-2Ax+2A=0 $
  3. $ x^{2}-Ax+2A=0 $

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

Let intercepts be a and b. Line is x/a + y/b = 1. Passes through (1,1) => 1/a + 1/b = 1 => (a+b)/ab = 1 => a+b = ab. Area A = 0.5 * |ab| = A => ab = 2A. Since a+b = ab, a+b = 2A. The quadratic equation with roots a and b is x^2 - (a+b)x + ab = 0, which is x^2 - 2Ax + 2A = 0.

Multiple choice

If $\alpha ,\beta $ are the roots of the equation $ax^{2}+bx+c=0$, then the value of $\displaystyle \frac{1}{a\alpha +b}+\frac{1}{a\beta +b}$ is
  1. $\displaystyle \frac{a}{bc}$
  2. $\displaystyle \frac{{b} }{{a} c}$
  3. $\displaystyle \frac{c}{ab}$
  4. $\displaystyle \frac{ab}{c}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given roots alpha, beta, alpha+beta = -b/a and alpha*beta = c/a. The expression is (a*beta+b + a*alpha+b) / ((a*alpha+b)(a*beta+b)). Numerator: a(alpha+beta) + 2b = a(-b/a) + 2b = -b + 2b = b. Denominator: a^2(alpha*beta) + ab(alpha+beta) + b^2 = a^2(c/a) + ab(-b/a) + b^2 = ac - b^2 + b^2 = ac. Result is b/(ac).

Multiple choice

If the equation $z^4+a_1z^3+a_2z^2+a_3z+a_4=0$, where $a_1, a_2, a_4$ are real coefficients different from zero, has a purely imaginary root, then the expression $\dfrac{a_3}{(a_1a_2)}+\dfrac{a_1a_4}{a_2a_3}$ has the value equal to
  1. $0$
  2. $1$
  3. $-2$
  4. $2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the imaginary root be ki. Substituting into the equation and separating real and imaginary parts allows one to solve for the coefficients. The resulting relation between the coefficients satisfies the given expression, yielding a value of 1.

Multiple choice

If $2$ and $3$ are the roots of cubic equation $2x^3+mx^2-13x+n=0$, then the values of $m$ and $n$ are:
  1. $-5,-30$
  2. $-5,30$
  3. $5,30$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If 2 and 3 are roots, then (x-2) and (x-3) are factors. The third root r can be found using the product of roots: 2 * 3 * r = -n/2, and sum of roots 2 + 3 + r = -m/2. Using the coefficient of x, the sum of roots taken two at a time is (2*3 + 2*r + 3*r) = -13/2, which gives 6 + 5r = -6.5, so 5r = -12.5 and r = -2.5. Then m = -2*(2+3-2.5) = -5 and n = -2*(2*3*-2.5) = 30.