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

\begin{array} { l } { \text { If } \alpha , \beta \text { are the roots of } a x ^ { 2 } + b x + c , \alpha + h } { \beta + h \text { are the roots of } p x ^ { 2 } + q x + r = 0 } \{ \text { and } D _ { 1 } \text { , } D _ { 2 } \text { are the respective } } \ { \text { discriminants of these equations, } } \ { \text { then } D _ { 1 } : D _ { 2 } = } \end{array}
  1. $\cfrac{a ^ { 2 } }{ p ^ { 2 }}$
  2. $\cfrac{b ^ { 2 }} { q ^ { 2 }}$
  3. $\cfrac{c ^ { 2 } }{r ^ { 2 }}$
  4. $1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If roots are shifted by h, the new equation is a(x-h)^2 + b(x-h) + c = 0. The discriminant of a(x-h)^2 + b(x-h) + c = 0 is the same as ax^2 + bx + c = 0 because shifting does not change the gap between roots. However, the coefficients p, q, r are scaled. The ratio of discriminants D1/D2 = (a/p)^2.

Multiple choice

If the roots of the equation $x^{2}-2mx+m^{2}+1=0$ are imaginary, then $m$ lies in
  1. $(1, \infty)$
  2. $(-\infty, \infty)$
  3. $(-\infty, 1)$
  4. $All\ of\ these$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For the roots of a quadratic equation to be imaginary, the discriminant (D = b^2 - 4ac) must be less than 0. Here, D = (-2m)^2 - 4(1)(m^2 + 1) = 4m^2 - 4m^2 - 4 = -4. Since -4 is always less than 0, the roots are imaginary for all real values of m.

Multiple choice

\begin{array} { l } { \text { If } 1 - p \text { is a root of quadratic equation } } { x ^ { 2 } + p x + ( 1 - p ) = 0 \text { , then the roots are } } \end{array}
  1. $0,1$
  2. $-1,2$
  3. $0,-1$
  4. $-1,1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If 1-p is a root, substitute x = 1-p into the equation: (1-p)^2 + p(1-p) + (1-p) = 0. Factor out (1-p): (1-p)(1-p + p + 1) = 0, which is (1-p)(2) = 0, so p = 1. If p = 1, the equation becomes x^2 + x = 0, so x(x+1) = 0. The roots are 0 and -1.

Multiple choice

Let $P(x)=x^{2}+ax+b$ be a quadratic polynomial where $a$ and $b$ are real numbers. Suppose $P(-1)^{2}.P(0)^{2}.P(1)^{2})$ is an arithmetic progression of integers , then $a$ and $b$ are integers.

  1. True

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

The condition that P(-1)^2, P(0)^2, P(1)^2 form an arithmetic progression forces b = -(a^2 + 1)/2. Requiring all three squared values to be integers further constrains a^2 + 1 to be even and a^2 to be an odd integer, which makes a and b integers. No non-integer real values of a can satisfy the integer AP condition.

Multiple choice

A quadratic equation, whose roots are $\alpha $ and $\beta$ can be written as $(x - \alpha ) (x - \beta) = 0 = x^2 - ( \alpha + \beta) x + \alpha \beta$ i.e. $ax^2 + bx + c \equiv a (x - \alpha) (x - \beta)$

  1. True

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

The identity (x - alpha)(x - beta) = x^2 - (alpha + beta)x + alpha*beta is the standard expansion for a monic quadratic equation with roots alpha and beta. Multiplying by a constant 'a' scales the equation to the general form ax^2 + bx + c.

Multiple choice

The range of $a$ for which the equation $x^2 + ax - 4 = 0$ has its smaller root in the interval $(-1, 2)$ is
  1. $(- \infty, -3)$
  2. $(0, 3)$
  3. $(0, \infty)$
  4. $(-\infty,  -3) \cup (0, \infty)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the smaller root of x^2 + ax - 4 = 0 to be in (-1, 2), let f(x) = x^2 + ax - 4. Since the product of roots is -4, one root is positive and one is negative. The smaller root is the negative one. We need the negative root to be in (-1, 2), which means it must be in (-1, 0). Thus, f(-1) > 0 and f(0) < 0. f(-1) = 1 - a - 4 = -3 - a > 0 => a < -3. f(0) = -4 < 0 (always true).

Multiple choice

A quadratic equation with real coefficients whose one root is $3-2 i$ is
  1. $x^2 - 3x + 2 = 0$
  2. $x^2 - 6x + 13 = 0$
  3. $x^2 - 2x + 3 = 0$

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

If a quadratic equation with real coefficients has a complex root 3 - 2i, it must also have the conjugate root 3 + 2i. The equation is (x - (3 - 2i))(x - (3 + 2i)) = 0, which simplifies to (x - 3)^2 - (2i)^2 = x^2 - 6x + 9 + 4 = x^2 - 6x + 13 = 0.

Multiple choice

State whether the statement is true/false. If the roots of the equation $\left( {{a^2} + {b^2}} \right){x^2} - 2\left( {ac + bd} \right)x + \left( {{c^2} + {d^2}} \right) = 0$ are equal, then $\dfrac{a}{b} = \dfrac{c}{d}$.

  1. True

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

For equal roots, the discriminant must be zero: 4(ac+bd)^2 - 4(a^2+b^2)(c^2+d^2) = 0. Expanding gives (ac)^2 + (bd)^2 + 2abcd - (a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2) = 0, which simplifies to -(ad-bc)^2 = 0. Thus ad = bc, or a/b = c/d.