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

For the equation ${ \left| x \right| }^{ 2 }+\left| x \right| -6=0$

  1. There are four roots
  2. The sum of the root is $-1$
  3. The product of the roots is $-4$
  4. The product of the roots is $-6$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let u = |x|. The equation becomes u^2 + u - 6 = 0, which factors to (u + 3)(u - 2) = 0. Since |x| cannot be negative, |x| = 2, meaning x = 2 or x = -2. The product of these roots is 2 * (-2) = -4.

Multiple choice

Find the sum of the cubes of the roots of the equation $3y^2- 14y + 8 = 0$:
  1. $\cfrac{1728}{27}$
  2. $\cfrac{1736}{27}$
  3. $\cfrac{1730}{27}$
  4. $\cfrac{1732}{27}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For 3y^2 - 14y + 8 = 0, sum of roots alpha + beta = 14/3 and product alpha * beta = 8/3. The sum of cubes is (alpha + beta)^3 - 3 * alpha * beta * (alpha + beta) = (14/3)^3 - 3 * (8/3) * (14/3) = 2744/27 - 336/9 = 2744/27 - 1008/27 = 1736/27.

Multiple choice

If $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2-12x+35 = 0$, then find the value of $2\begin{pmatrix}\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}\end{pmatrix} :$
  1. $\dfrac{74}{35}$
  2. $\dfrac{148}{35}$
  3. $\dfrac{12}{35}$
  4. $\dfrac{34}{35}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For x^2 - 12x + 35 = 0, alpha + beta = 12 and alpha * beta = 35. The expression is 2 * ((alpha^2 + beta^2) / (alpha * beta)). Since alpha^2 + beta^2 = (alpha + beta)^2 - 2*alpha*beta = 144 - 70 = 74. The value is 2 * (74 / 35) = 148 / 35.

Multiple choice

If $ \alpha + \beta = 3$ and $ \alpha^3 + \beta^3 = 9$, find the quadratic equation whose roots are $ \alpha$ or $\beta$ :
  1. $x^2-3x+3=0$
  2. $x + \cfrac{2}{x}+ 3 = 0$
  3. $x^2- 2x + 3 = 0$
  4. $x + \cfrac{2}{x} = 3$
  5. $x^2-3x+2=0$
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Given alpha + beta = 3 and alpha^3 + beta^3 = 9. Using (alpha + beta)^3 = alpha^3 + beta^3 + 3*alpha*beta*(alpha + beta), we get 3^3 = 9 + 3*alpha*beta*(3), so 27 = 9 + 9*alpha*beta, meaning alpha*beta = 2. The quadratic equation is x^2 - (sum)x + (product) = 0, which is x^2 - 3x + 2 = 0.

Multiple choice

Solve the following quadratic equation using quadratic formula: $9x^2-9x+2=0$
  1. $\left \{ \dfrac{2}{3}, \dfrac{1}{3} \right \}$
  2. $\left \{ \dfrac{2}{3}, \dfrac{-1}{3} \right \}$
  3. $\left \{ \dfrac{-2}{3}, \dfrac{1}{3} \right \}$

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

Using the quadratic formula x = (-b +/- sqrt(b^2 - 4ac)) / 2a: x = (9 +/- sqrt(81 - 72)) / 18 = (9 +/- 3) / 18. Roots are 12/18 = 2/3 and 6/18 = 1/3.

Multiple choice

Solve the following quadratic equation using quadratic formula: $a(x^2+1)=x(a^2+1)$
  1. $\left \{ \dfrac{1}{a}, a \right \}$
  2. $\left \{ \dfrac{1}{a}, 1 \right \}$
  3. $\left \{ \dfrac{1}{a}, a^2 \right \}$

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

Expand the equation: ax^2 + a = a^2x + x. Rearrange to standard form: ax^2 - (a^2 + 1)x + a = 0. Using the quadratic formula or factoring: (ax - 1)(x - a) = 0. The roots are 1/a and a.

Multiple choice

Solve the following quadratic equation using quadratic formula: $3a^2x^2+abx-2b^2=0$
  1. $\left \{ \dfrac{2b}{3a},- \dfrac{b}{a} \right \}$
  2. $\left \{ \dfrac{2b}{3a}, \dfrac{b}{a} \right \}$
  3. $\left \{ -\dfrac{2b}{3a}, \dfrac{-b}{a} \right \}$

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

3a^2x^2 + abx - 2b^2 = 0. Using quadratic formula x = (-b +/- sqrt(b^2 - 4ac)) / 2a. x = (-ab +/- sqrt((ab)^2 - 4(3a^2)(-2b^2))) / (2*3a^2). x = (-ab +/- sqrt(a^2b^2 + 24a^2b^2)) / 6a^2. x = (-ab +/- sqrt(25a^2b^2)) / 6a^2. x = (-ab +/- 5ab) / 6a^2. Roots: (4ab)/6a^2 = 2b/3a and (-6ab)/6a^2 = -b/a.