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

Find the value of a for which the sum of the squares of the roots of the equation $x^2 - (a - 2)x - a- 1 = 0$ assumes the least value.
  1. $a = 1$
  2. $a = -1$
  3. $a = 0$
  4. $a = 2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The sum of the squares of the roots of x^2 - (a-2)x - (a+1) = 0 is (sum of roots)^2 - 2(product of roots). This is (a-2)^2 - 2(-(a+1)) = a^2 - 4a + 4 + 2a + 2 = a^2 - 2a + 6. This quadratic in 'a' reaches its minimum at a = -(-2)/(2*1) = 1.

Multiple choice

The sum of non-real roots of the polynomial equation $x^3 + 3x^2 + 3x + 3 =0$ :
  1. Equal $0$
  2. Lies between $0$ and $1$
  3. Lies between $-1$ and $0$
  4. Has absolute value bigger than $1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The equation is x^3 + 3x^2 + 3x + 3 = 0, which is (x+1)^3 + 2 = 0. So (x+1)^3 = -2. The roots are x = -1 + cube_root(-2), -1 + cube_root(-2) * omega, -1 + cube_root(-2) * omega^2. The real root is -1 - 2^(1/3). The sum of all roots is -3. The sum of non-real roots is -3 - (-1 - 2^(1/3)) = -2 + 2^(1/3). Since 2^(1/3) is approx 1.26, the sum is approx -0.74, which lies between -1 and 0.

Multiple choice

If $\alpha, \beta, \gamma$ be the roots of the equation $2x^3 + 3x^2 - 12x + 3 = 0$ and $A (\alpha, \beta, \gamma), B (\beta, \gamma, \alpha) C(\gamma, \alpha, \beta)$ represent vertices of a triangle ABC then the centroid of the triangle lies upon the line

  1. x = y = z

  2. x = 2y = 3z

  3. x = - 2y =3z

  4. x =y = - 2z
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The centroid of triangle ABC with vertices (alpha, beta, gamma), (beta, gamma, alpha), and (gamma, alpha, beta) is ((alpha+beta+gamma)/3, (beta+gamma+alpha)/3, (gamma+alpha+beta)/3). Since alpha+beta+gamma = -3/2 (from Vieta's formulas), the centroid is (-1/2, -1/2, -1/2). This point satisfies the line x = y = z.

Multiple choice

Let ${ \alpha }{ 1 },{ \alpha }{ 2 }$ and ${ \beta }{ 1 },{ \beta }{ 2 }$ be the roots of $a{ x }^{ 2 }+bx+c=0$ and $p{ x }^{ 2 }+qx+r=0$ respectively. If the system of equations ${ \alpha }{ 1 }y+{ \alpha }{ 2 }z=0$ and ${ \beta }{ 1 }y+{ \beta }{ 2 }z=0$ has a non-trivial solution, then
  1. ${ b }^{ 2 }pr={ q }^{ 2 }ac$
  2. $bp{ r }^{ 2 }=qa{ c }^{ 2 }$
  3. $b{ p }^{ 2 }r=q{ a }^{ 2 }c$

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

The system has a non-trivial solution if the determinant of the coefficients is zero: alpha1*beta2 - alpha2*beta1 = 0. This implies alpha1/alpha2 = beta1/beta2. Using properties of roots, this leads to the condition b^2pr = q^2ac.

Multiple choice

If $b < 0$, then the roots ${ x }{ 1 }$ and ${ x }{ 2 }$ of the equation $2{ x }^{ 2 }+6x+b=0$, satisfy the condition $\left( \dfrac { { x }{ 1 } }{ { x }{ 2 } } \right) +\left( \dfrac { { x }{ 2 } }{ { x }{ 1 } } \right) < K$, where $K$ is equal to
  1. $2$
  2. $-2$
  3. $0$
  4. $4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given 2x^2 + 6x + b = 0. x1 + x2 = -3, x1*x2 = b/2. Expression = (x1/x2) + (x2/x1) = (x1^2 + x2^2) / (x1*x2) = ((x1+x2)^2 - 2x1x2) / (x1*x2) = (9 - b) / (b/2) = 18/b - 2. Since b < 0, 18/b is negative, so 18/b - 2 < -2.

Multiple choice

If the roots of the equation $\left( a-1 \right) { \left( { x }^{ 2 }+x+1 \right) }^{ 2 }=\left( a+1 \right) \left( { x }^{ 4 }+{ x }^{ 2 }+1 \right) $ are real and distinct then the value of $a\in $
  1. $\left( -\infty ,3 \right] $
  2. $\left( -\infty ,-2 \right) \cup \left( 2,\infty \right) $
  3. $\left[ -2,2 \right] $
  4. $\left[ -3,\infty \right) $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

By factoring x^4 + x^2 + 1 as (x^2 + x + 1)(x^2 - x + 1) and dividing both sides by the non-zero term x^2 + x + 1, the equation simplifies to x^2 - ax + 1 = 0. For this quadratic equation to have real and distinct roots, its discriminant must be strictly positive, which yields a^2 - 4 > 0, or a in (-infinity, -2) U (2, infinity).

Multiple choice

A root of the equation $17x^2+17x\tan\left(\displaystyle 2\tan^{-1}\frac{1}{5}-\frac{\pi}{4}\right)-10=0$ is?
  1. $\displaystyle\frac{10}{17}$
  2. $\displaystyle -1$
  3. $\displaystyle -\frac{7}{17}$
  4. $\displaystyle 1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let theta = 2*tan^-1(1/5) - pi/4. tan(theta) = tan(2*tan^-1(1/5) - pi/4). tan(2*tan^-1(1/5)) = (2*(1/5)) / (1-(1/25)) = (2/5) / (24/25) = 5/12. tan(theta) = (5/12 - 1) / (1 + 5/12) = (-7/12) / (17/12) = -7/17. Equation: 17x^2 + 17x(-7/17) - 10 = 0 => 17x^2 - 7x - 10 = 0. Roots: (17x+10)(x-1) = 0. x=1 or x=-10/17.

Multiple choice

Let $\alpha$ and $\beta$ be the roots of $x^2+bx+1=0$. Then, the equation whose roots are $-\left(\displaystyle \alpha +\frac{1}{\beta}\right)$ and $-\left(\displaystyle \beta +\frac{1}{\alpha}\right)$, is?
  1. $x^2=0$
  2. $x^2+2bx+4=0$
  3. $x^2-2bx+4=0$
  4. $x^2-bx+1=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Roots of x^2 + bx + 1 = 0 are alpha, beta. alpha + beta = -b, alpha*beta = 1. New roots are S = -(alpha + 1/beta) and P = -(beta + 1/alpha). S = -( (alpha*beta + 1) / beta ) = -2/beta. P = -( (beta*alpha + 1) / alpha ) = -2/alpha. Sum of new roots = -2(alpha + beta) / (alpha*beta) = -2(-b)/1 = 2b. Product of new roots = 4 / (alpha*beta) = 4. Equation: x^2 - (sum)x + (product) = x^2 - 2bx + 4 = 0.

Multiple choice

If $\alpha ,\beta $ are the roots of the equation ${ x }^{ 2 }+px-\cfrac { 1 }{ 2{ p }^{ 2 } } =0$, where $p\in R$. Then, the minimum value of ${ \alpha }^{ 4 }+{ \beta }^{ 4 }$ is
  1. $2$
  2. $2+\sqrt { 2 } $
  3. $2\sqrt { 2 } $
  4. $2-\sqrt { 2 } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Roots alpha, beta satisfy alpha+beta = -p and alpha*beta = -1/(2p^2). alpha^4 + beta^4 = (alpha^2 + beta^2)^2 - 2(alpha*beta)^2 = ((alpha+beta)^2 - 2alpha*beta)^2 - 2(alpha*beta)^2. Substituting: (p^2 + 1/p^2)^2 - 2(1/(4p^4)) = p^4 + 2 + 1/p^4 - 1/(2p^4) = p^4 + 1/(2p^4) + 2. By AM-GM, p^4 + 1/(2p^4) >= 2*sqrt(1/2) = sqrt(2). Thus, min value is 2 + sqrt(2).

Multiple choice

The arithmetic mean of the roots of the equation $4\cos ^{ 2 }{ x } -4\cos ^{ 2 }{ x } -\cos { (315\pi +x) } =1$ in the interval $(0,315)$ is
  1. $50\pi $
  2. $51\pi $
  3. $100\pi $
  4. $315\pi $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The two 4cos^2(x) terms cancel, leaving -cos(315*pi + x) = 1. Since 315 is odd, cos(315*pi + x) = -cos(x), so cos(x) = 1. Solutions in (0, 315) are x = 2*pi, 4*pi, ..., 100*pi (50 roots). Their mean is 2*pi(1+2+...+50)/50 = 2550*pi/50 = 51*pi.