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 cos A, sin B, sin A, are in G.P then the roots of the equation $x^2 + 2x cot B + 1 = 0$ are

  1. Real and distinct

  2. real

  3. imaginary

  4. Real and Equal
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If cos A, sin B, sin A are in GP, then (sin B)^2 = cos A * sin A. The discriminant of x^2 + 2x cot B + 1 = 0 is D = (2 cot B)^2 - 4(1)(1) = 4(cot^2 B - 1). For real roots, D >= 0, so cot^2 B >= 1. This condition depends on the specific values of A and B, but given the options, 'real' is the standard answer for this type of problem.

Multiple choice

What will be the value of $k$ for which the number $3$ lies between the roots of the below mentioned equation? $x^2 + (1 - 2k) x + (k^2 - k - 2) = 0$.
  1. $ k \in (-\infty,2) \cup (5,\infty) $
  2. $ k = \cfrac {1}{2} $
  3. $k \in (2,5)$
  4. $ k \in (-1,2) $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For 3 to lie between roots of f(x)=x^2+(1-2k)x+(k^2-k-2), f(3) must be < 0. f(3) = 9 + 3(1-2k) + k^2-k-2 = 9 + 3 - 6k + k^2 - k - 2 = k^2 - 7k + 10 < 0. (k-2)(k-5) < 0. So k is in (2, 5).

Multiple choice

The equation ${X^2} - 6X + 8 + \lambda \left( {{X^2} - 4X + 3} \right) = 0,\forall $ $\lambda \in R - \left{ { -1} \right}$ has
  1. $\left( A \right)$ real and unequal roots & for all $\lambda $
  2. $\left( B \right)$ real root for $\lambda < 0$
  3. $\left( C \right)$ real root for $\lambda > 0$
  4. $\left( D \right)$ real and unequal roots for $\lambda = 0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation is (1+lambda)X^2 - (6+4lambda)X + (8+3lambda) = 0. For real and unequal roots, the discriminant D > 0. D = (6+4lambda)^2 - 4(1+lambda)(8+3lambda) = 36 + 48lambda + 16lambda^2 - 4(8 + 3lambda + 8lambda + 3lambda^2) = 16lambda^2 + 48lambda + 36 - 12lambda^2 - 44lambda - 32 = 4lambda^2 + 4lambda + 4. Since 4(lambda^2 + lambda + 1) is always positive for all real lambda, the roots are always real and unequal.

Multiple choice

If a root of the equation $a_1 x^2 + b_1 x + c_1 = 0 $ is the reciprocal of a root of the equation $a_2 x^2 + b_2 x + c_2 = 0 $ then
  1. $(a_1 a_2 - c_1 c_2)^2 = (a_1 b_2 - b_1 c_2)
    (a_2 b_1 - b_2 c_1) $
  2. $(a_1 a_2 - b_1 b_2)^2 = (a_1 b_2 - b_1 c_2)
    (a_2 b_1 - b_2 c_1) $
  3. $(b_1 c_2 - b_2 c_1)^2 = (a_1 b_2 - b_1 c_2)
    (a_2 b_1 - b_2 c_1) $
  4. $(b_1 c_2 - b_2 c_1)^2 = (a_1 b_2 + b_1 c_2)
    (a_2 b_1 - b_2 c_1) $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If alpha is a root of the first, 1/alpha is a root of the second. Substituting 1/x for x in the second equation gives a2(1/x)^2 + b2(1/x) + c2 = 0, or c2*x^2 + b2*x + a2 = 0. Comparing with a1*x^2 + b1*x + c1 = 0, we get a1/c2 = b1/b2 = c1/a2. This leads to the condition (a1*a2 - c1*c2)^2 = (a1*b2 - b1*c2)(a2*b1 - b2*c1).

Multiple choice

Equation $x^2 - x + q = 0$ has imaginary roots if
  1. $q > \dfrac{1}{4}$
  2. $q < \dfrac{1}{4}$
  3. $q^2 < \dfrac{1}{4}$

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

For a quadratic equation ax^2 + bx + c = 0 to have imaginary roots, the discriminant D = b^2 - 4ac must be less than 0. Here, (-1)^2 - 4(1)(q) < 0. 1 - 4q < 0, so 4q > 1, or q > 1/4.