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 $\alpha$ and $\beta$ are the roots of the quadratic equation $4x^2+3x+7=0$, then the value of $\displaystyle\frac{1}{\alpha}+\frac{1}{\beta}$ is:
  1. $-\displaystyle\frac{3}{4}$
  2. $-\displaystyle\frac{3}{7}$
  3. $\displaystyle\frac{3}{7}$
  4. $\displaystyle\frac{4}{7}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For 4x^2 + 3x + 7 = 0, sum of roots alpha + beta = -3/4 and product alpha*beta = 7/4. The expression 1/alpha + 1/beta = (alpha + beta) / (alpha*beta) = (-3/4) / (7/4) = -3/7.

Multiple choice

If $\displaystyle \alpha $ and $\displaystyle \beta $ are the roots of the equation $\displaystyle { x }^{ 2 }-2x+4=0$, then the value of $\displaystyle { \alpha }^{ n }+{ \beta }^{ n }$ will be:
  1. $\displaystyle i{ 2 }^{ n+1 }\sin { \left( { n\pi }/{ 3 } \right) } $
  2. $\displaystyle { 2 }^{ n+1 }\cos { \left( { n\pi }/{ 3 } \right) } $
  3. $\displaystyle i{ 2 }^{ n-1 }\sin { \left( { n\pi }/{ 3 } \right) } $
  4. $\displaystyle { 2 }^{ n-1 }\cos { \left( { n\pi }/{ 3 } \right) } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The roots of x^2 - 2x + 4 = 0 are x = (2 +/- sqrt(4 - 16))/2 = 1 +/- i*sqrt(3). In polar form, these are 2(cos(pi/3) +/- i*sin(pi/3)). By De Moivre's Theorem, alpha^n + beta^n = 2^n * (cos(n*pi/3) + i*sin(n*pi/3)) + 2^n * (cos(n*pi/3) - i*sin(n*pi/3)) = 2 * 2^n * cos(n*pi/3) = 2^(n+1) * cos(n*pi/3).

Multiple choice

If $\alpha, \beta, \gamma$ are the roots of the equation $x^{3} - 3x^{2} + 2x - 1 = 0$, then the value of $(1 - \alpha) (1 - \beta) (1 - \gamma)$ is:
  1. $1$
  2. $2$
  3. $-1$
  4. $-2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The cubic polynomial can be factored as P(x) = (x - alpha)(x - beta)(x - gamma) = x^3 - 3x^2 + 2x - 1. To find the value of (1 - alpha)(1 - beta)(1 - gamma), we simply evaluate the polynomial at x = 1, which gives 1^3 - 3(1)^2 + 2(1) - 1 = -1.

Multiple choice

If $a, b, c, d$ are the roots of the equation $x^{4} + 2x^{3} + 3x^{2} + 4x + 5 = 0$, then $1 + a^{2} + b^{2} + c^{2} + d^{2}$ is equal to
  1. $-2$
  2. $-1$
  3. $2$
  4. $1$`
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For x^4 + 2x^3 + 3x^2 + 4x + 5 = 0, the sum of roots a+b+c+d = -2 and sum of roots taken two at a time ab+ac+ad+bc+bd+cd = 3. We know a^2+b^2+c^2+d^2 = (a+b+c+d)^2 - 2(ab+ac+ad+bc+bd+cd) = (-2)^2 - 2(3) = 4 - 6 = -2. Thus, 1 + a^2 + b^2 + c^2 + d^2 = 1 - 2 = -1.

Multiple choice

If $\alpha ,\beta ,\gamma $ are roots of the equation ${ x }^{ 3 }+125=0$, then the quadratic equation whose roots are ${ \left( \cfrac { \alpha }{ \beta } \right) }^{ 2 }$ and ${ \left( \cfrac { \alpha }{ \gamma } \right) }^{ 2 }$ is
  1. ${ x }^{ 2 }+5x+1=0$
  2. ${ x }^{ 2 }-x+1=0$
  3. ${ x }^{ 2 }+x-1=0$
  4. ${ x }^{ 2 }+x+1=0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The roots of x^3 + 125 = 0 are -5, -5w, -5w^2 where w is a complex cube root of unity. The ratios alpha/beta and alpha/gamma are w and w^2 (or similar permutations). Squaring these gives w^2 and w^4 = w. The quadratic equation with roots w and w^2 is x^2 - (w+w^2)x + w*w^2 = 0, which simplifies to x^2 + x + 1 = 0.

Multiple choice

Find the sum of the real roots of the equation $x^2+ 5|x|+ 6=0$
  1. $5$
  2. $10$
  3. $-5$

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

The equation x^2 + 5|x| + 6 = 0 can be written as |x|^2 + 5|x| + 6 = 0. Let u = |x|, then u^2 + 5u + 6 = 0, which factors to (u+2)(u+3) = 0. This gives |x| = -2 or |x| = -3. Since the absolute value cannot be negative, there are no real roots.

Multiple choice

The sum of the roots of quadratic equation $ax^2 + bx + c = 0 \,(a, b, \neq 0)$ is equal to the sum of squares of their reciprocals, then $\dfrac {a}{c}, \dfrac{b}{a}$ and $ \dfrac {c}{b}$ are in

  1. AP

  2. GP

  3. HP

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

Let the roots be alpha and beta. Using Vieta's relations and the given condition involving the reciprocals, the reciprocals of a/c, b/a, and c/b form an arithmetic progression. Therefore, the original three quantities are in harmonic progression.