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

The number of roots of the equation $|x-x^{2}-2|=|2x-3-x^{2}|$ is
  1. $0$
  2. $1$
  3. $2$

  4. infinitely
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation is |x - x^2 - 2| = |2x - 3 - x^2|. Since both sides are absolute values, we solve x - x^2 - 2 = 2x - 3 - x^2 or x - x^2 - 2 = -(2x - 3 - x^2). The first case gives -x - 2 = -3, so x = 1. The second case gives x - x^2 - 2 = -2x + 3 + x^2, leading to 2x^2 - 3x + 5 = 0, which has no real roots.

Multiple choice

If $\alpha ,\beta $ are the roots of the quadratic equation ${ x }^{ 2 }+px+q=0$, then the values of ${ \alpha }^{ 3 }+{ \beta }^{ 3 }$ and ${ \alpha }^{ 4 }+{ \alpha }^{ 2 }{ \beta }^{ 2 }+{ \beta }^{ 4 }$ are respectively
  1. $3pq-{ p }^{ 3 }$ and ${ p }^{ 4 }-3{ p }^{ 2 }q+3{ q }^{ 2 }$
  2. $-p\left( 3q-{ p }^{ 2 } \right) $ and $\left( { p }^{ 2 }-q \right) \left( { p }^{ 2 }+3q \right) $
  3. $pq-4$ and ${ p }^{ 4 }-{ q }^{ 4 }$
  4. $3pq-{ p }^{ 3 }$ and $\left( { p }^{ 2 }-q \right) \left( { p }^{ 2 }-3q \right) $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

alpha + beta = -p, alpha * beta = q. alpha^3 + beta^3 = (alpha + beta)^3 - 3*alpha*beta*(alpha + beta) = (-p)^3 - 3*q*(-p) = 3pq - p^3. alpha^4 + alpha^2*beta^2 + beta^4 = (alpha^2 + beta^2)^2 - alpha^2*beta^2 = ((-p)^2 - 2q)^2 - q^2 = (p^2 - 2q)^2 - q^2 = p^4 - 4p^2q + 4q^2 - q^2 = p^4 - 4p^2q + 3q^2 = (p^2 - q)(p^2 - 3q).

Multiple choice

The quadratic expression $(2X+1)^2-pX+q\neq 0$ for any real $X$, if
  1. $p^2-16p-8q < 0$
  2. $p^2-8p+16q < 0$
  3. $p^2-8p-16q < 0$
  4. $p^2-16p+8q < 0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For the expression to never be zero, the discriminant of the quadratic (2X+1)^2 - pX + q = 0 must be negative. 4X^2 + 4X + 1 - pX + q = 4X^2 + (4-p)X + (1+q) = 0. D = (4-p)^2 - 4 * 4 * (1+q) < 0. 16 - 8p + p^2 - 16 - 16q < 0 => p^2 - 8p - 16q < 0.

Multiple choice

If $\alpha, \beta, \gamma $ are the roots of the equation $2x^3-3x^2+6x+1=0$, then $\alpha^2 +\beta^2+ \gamma^2$ is equal to
  1. $-\dfrac{15}{4}$
  2. $\dfrac{15}{4}$
  3. $\dfrac{9}{4}$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a cubic equation ax^3 + bx^2 + cx + d = 0, the sum of roots alpha + beta + gamma = -b/a = 3/2. The sum of roots taken two at a time alpha*beta + beta*gamma + gamma*alpha = c/a = 6/2 = 3. We use the identity alpha^2 + beta^2 + gamma^2 = (alpha + beta + gamma)^2 - 2(alpha*beta + beta*gamma + gamma*alpha). Substituting values: (3/2)^2 - 2(3) = 9/4 - 6 = 9/4 - 24/4 = -15/4.

Multiple choice

lf $\alpha,\beta,\gamma,\delta$ are the roots of the equation $3x^{4}-8x^{3}+2x^{2}-9=0$, then $\Sigma\alpha\beta=$
  1. $\displaystyle \frac{2}{3}$
  2. $\displaystyle \frac{-2}{3}$
  3. $\displaystyle \frac{8}{3}$
  4. $\displaystyle \frac{-8}{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a polynomial equation of the form ax^4 + bx^3 + cx^2 + dx + e = 0, the sum of the products of the roots taken two at a time is given by c/a. In the given equation, a = 3 and c = 2, so the sum of the products of the roots is 2/3.

Multiple choice

If the roots of the quadratic equation $m{ x }^{ 2 }-nx+k=0$ are $\tan { { 33 }^{ o } } $ and $\tan { { 12 }^{ o } } $, then the value of $\cfrac { 2m+n+k }{ m } $ is equal to
  1. $0$
  2. $1$
  3. $3$
  4. $2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Roots are tan(33) and tan(12). Sum of roots = n/m = tan(33)+tan(12). Product = k/m = tan(33)tan(12). Using tan(A+B) formula, tan(45) = (tan(33)+tan(12))/(1-tan(33)tan(12)) = 1. So (n/m) = 1 - (k/m), thus n/m + k/m = 1. The expression (2m+n+k)/m = 2 + (n+k)/m = 2 + 1 = 3.