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 roots of the equation $x^{2}(1+m^{2})+2mcx+c^{2}-a^{2}=0$ are equal, then value of $c$ is
  1. $a\ \sqrt{(1+m^{2})}$
  2. $a\ \sqrt{(1-m^{2})}$
  3. $m\ \sqrt{(1+a^{2})}$
  4. $m\ \sqrt{(1-a^{2})}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a quadratic equation to have equal roots, the discriminant D = b^2 - 4ac must be 0. Here, (2mc)^2 - 4(1+m^2)(c^2 - a^2) = 0. Simplifying this leads to 4m^2c^2 - 4(c^2 - a^2 + m^2c^2 - m^2a^2) = 0, which reduces to c^2 = a^2(1+m^2). Thus, c = a*sqrt(1+m^2).

Multiple choice

If $ \alpha \neq \beta $ but $ \alpha ^ { 2 } = 2 \alpha - 3 ; \beta ^ { 2 } = 2 \beta - 3 $ then the equation whose roots are $ \dfrac { \alpha } { \beta } $ and $ \dfrac { \beta } { \alpha } $ is
  1. $2x^{2}+3x+2=0$
  2. $3x^{2}+2x+3=0$
  3. $2x^{2}-3x+2=0$
  4. $3x^{2}-2x+3=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given alpha^2 - 2alpha + 3 = 0 and beta^2 - 2beta + 3 = 0, alpha and beta are roots of x^2 - 2x + 3 = 0. Thus, alpha + beta = 2 and alpha*beta = 3. The new roots are alpha/beta + beta/alpha = (alpha^2 + beta^2)/(alpha*beta) = ((alpha+beta)^2 - 2alpha*beta)/(alpha*beta) = (4 - 6)/3 = -2/3. The product is 1. The equation is x^2 - (sum)x + product = 0, which is x^2 + 2/3x + 1 = 0, or 3x^2 + 2x + 3 = 0.

Multiple choice

The roots of the equation $x^{2}+6x+a=0$ are real and distinct and they differ by atmost $4$, then find the range of values of $a.$
  1. $(5, 9]$
  2. $[5, 9)$
  3. $[5, \infty)$
  4. $[3, 9)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For real and distinct roots, the discriminant D = 36 - 4a must be > 0, so a < 9. If the roots are r1 and r2, then |r1 - r2| = sqrt(D)/|a_coeff| = sqrt(36 - 4a). Given the difference is at most 4, sqrt(36 - 4a) <= 4 implies 36 - 4a <= 16, so 4a >= 20, or a >= 5. Combining these, 5 <= a < 9.

Multiple choice

If $a,b,c$ are postive rational numbers such that $a > b > c$ and the quadratic equation $(a+b-2c)x^{2}+(b+c -2a)x+(c+a-2b)=0$ has a root in the interval $(-1,0)$, then
  1. $b-c > a$
  2. $c+a <2b$

  3. both roots of given equation are rational
  4. the equation $ax^{2}+2bx+c=0$ has both negative real roots
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let f(x) be the quadratic. Its leading coefficient is positive, and f(-1) = 2(2a - b - c) > 0 because a > b > c. A root in (-1,0) therefore requires f(0) = c + a - 2b < 0, so c + a < 2b.

Multiple choice

If $\alpha$ & $\beta$ are roots of the equation $x^{2}+6x+\lambda=0$ & $3\alpha+2\beta=-20$ find $\lambda$..
  1. $16$
  2. $-8$
  3. $-16$
  4. $8$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Roots alpha, beta satisfy alpha + beta = -6 and alpha * beta = lambda. Given 3alpha + 2beta = -20. From alpha + beta = -6, beta = -6 - alpha. Substitute: 3alpha + 2(-6 - alpha) = -20 => 3alpha - 12 - 2alpha = -20 => alpha = -8. Then beta = -6 - (-8) = 2. lambda = alpha * beta = -8 * 2 = -16.