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 what real values of $k$, other than $k=0$, does the equation ${ x }^{ 2 }+kx+{ k }^{ 2 }=0$ have real roots? (The symbol $x\ge a$ means that $x$ can take on all values greater than $a$ and the value $a$ itself; $x\le a$ has the corresponding meaning with "less than")
  1. $k< 0$
  2. $k> 0$
  3. $k\ge 1$
  4. all values of $k$
  5. no values of $k$
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

For the quadratic equation x^2 + kx + k^2 = 0 to have real roots, the discriminant D = b^2 - 4ac must be greater than or equal to 0. Here, D = k^2 - 4(1)(k^2) = -3k^2. Since k is not 0, -3k^2 is always negative, meaning there are no real roots for any non-zero k.

Multiple choice

Consider the equation ${x^2} + 2x - n = 0$, where $n \in N$ and $n \in \left[ {5,100} \right]$. Total number of different values of 'n' so that the given equation has integral roots, is

  1. 4

  2. 6

  3. 8

  4. 3
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For x^2 + 2x - n = 0 to have integral roots, the discriminant D = 4 + 4n must be a perfect square, say k^2. So 4(1+n) = k^2, meaning 1+n must be a perfect square. For n in [5, 100], 1+n is in [6, 101]. Perfect squares in this range are 9, 16, 25, 36, 49, 64, 81, 100. There are 8 such values.

Multiple choice

The condition that the equation $x^2 + px + q = 0$, whose one root is the cube of the other root is :
  1. $p = { q }^{ { 1 }/{ 4 } } [1 - {q}^{{1}/{2}}]$
  2. $-p = q^{{1}/{2}} [1 - {q}^{{1}/{4}}]$
  3. $-p = q^{{1}/{4}} [1 + {q}^{{1}/{2}}]$
  4. $p = q^{{1}/{2}} [1 + {q}^{{1}/{4}}]$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If roots are a and a^3, then a + a^3 = -p and a * a^3 = a^4 = q. Thus a = q^(1/4). Substituting into the first equation: -p = q^(1/4) + (q^(1/4))^3 = q^(1/4) + q^(3/4) = q^(1/4)(1 + q^(1/2)).

Multiple choice

If the roots of the equation $(a-2b+c)x^2+(b+a-2c)x+(b+c-2a)=0$ be equal then which of the following is true
  1. $a=2b$
  2. $a=b$
  3. $a=2c$
  4. $a=b=c$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For equal roots, the discriminant D = b^2 - 4ac = 0. Here A = (a-2b+c), B = (b+a-2c), C = (b+c-2a). Note that A+B+C = 0. If A+B+C = 0, then one root is 1. If roots are equal, both roots are 1. Product of roots C/A = 1, so C=A. (b+c-2a) = (a-2b+c) => b-2a = a-2b => 3b = 3a => a=b.

Multiple choice

If the quadratic equation $ax^{2}+(b-c)x+a-b-c=0$ has unequal real roots for every $c\ \epsilon R$ (the set of all real numbers), then $ba>0$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the quadratic ax^2 + (b-c)x + (a-b-c) = 0 to have real roots, the discriminant D must be > 0. D = (b-c)^2 - 4a(a-b-c) = b^2 - 2bc + c^2 - 4a^2 + 4ab + 4ac. This expression must be positive for all c, which implies the coefficient of c^2 is positive and the discriminant of this quadratic in c is negative. The provided statement is a known property for this specific structure.

Multiple choice

Find the roots of the equation: :$25a^2-35a+12$.
  1. $(5a-3)(5a-4)$
  2. $(5a-3)(5a+4)$
  3. $(5a+3)(5a-4)$
  4. $(5a+3)(5a+4)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The expression factors as 25a^2 - 35a + 12 = (5a - 3)(5a - 4). Therefore, the roots are a = 3/5 and a = 4/5. The listed correct option gives the factorization rather than the roots themselves.