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 all values of parameter $a$ for which the quadratic equation $ \left( a+1 \right) { x }^{ 2 }+2\left( a+1 \right) x+a-2=0 $ has two distinct roots.
  1. $a \in (-1,\infty)$
  2. $a \in (-\infty,-1)$
  3. $a \in (-1,1)$
  4. $a \in (-\infty,\infty)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For two distinct roots, the discriminant D > 0 and the coefficient of x^2 must not be zero. D = (2(a+1))^2 - 4(a+1)(a-2) = 4(a+1)^2 - 4(a+1)(a-2) = 4(a+1)(a+1 - (a-2)) = 4(a+1)(3) = 12(a+1). 12(a+1) > 0 => a > -1. Also a+1 != 0 => a != -1.

Multiple choice

If $ { a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 }-3abc=0$ then the roots of the equation $ \left( { a }^{ 2 }-bc \right) { x }^{ 2 }+2\left( { b }^{ 2 }-ac \right) x+{ c }^{ 2 }-ab=0$ are

  1. imaginary

  2. real and unequal

  3. real and equal

  4. Cannot say
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The condition a^3 + b^3 + c^3 - 3abc = 0 implies either a+b+c = 0 or a=b=c. If a=b=c, the quadratic equation becomes (a^2-a^2)x^2 + 2(a^2-a^2)x + (a^2-a^2) = 0, which is 0=0, but more generally, the discriminant of the given quadratic equation simplifies to zero under these conditions, leading to real and equal roots.

Multiple choice

Let $\alpha,\,\beta$ be the roots of the equation ${x}^{2}-5x+r=0$ and $\displaystyle \frac{\alpha}{3},3\beta$ be the roots of the equation ${x}^{2}-3x+r=0$, then the value of $r$ is -
  1. $\displaystyle \frac{9}{4}$
  2. $\displaystyle \frac{4}{9}$
  3. $\displaystyle \frac{5}{4}$
  4. $\displaystyle \frac{4}{5}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

From the first equation, alpha + beta = 5 and alpha * beta = r. From the second, (alpha/3) + 3beta = 3 and (alpha/3) * 3beta = r, so alpha * beta = r. Substituting alpha = 3(3 - 3beta) into alpha + beta = 5: 9 - 9beta + beta = 5 => 8beta = 4 => beta = 0.5. Then alpha = 4.5. r = alpha * beta = 4.5 * 0.5 = 2.25 = 9/4.

Multiple choice

If the equation $\displaystyle \lambda x^{2}-2x+3= 0$ has positive roots for some real $\displaystyle \lambda $, then
  1. $\displaystyle \lambda > 0$
  2. $\displaystyle \lambda > \frac{1}{3}$
  3. $\displaystyle 0 < \lambda \leq \frac{1}{3}$
  4. $\displaystyle \lambda \leq \frac{1}{3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For roots to be positive, sum of roots = 2/lambda > 0 (so lambda > 0) and product of roots = 3/lambda > 0 (so lambda > 0). Discriminant D = 4 - 12*lambda >= 0. 4 >= 12*lambda, so lambda <= 1/3. Combined: 0 < lambda <= 1/3.