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

$x^2+3ax+2a^2=0$ If the above equation has roots $\alpha ,\beta $ and it is given that $\alpha^2 +\beta ^2=5$, then the product of roots is
  1. $2$
  2. $1$
  3. $-3$
  4. $3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For x^2 + 3ax + 2a^2 = 0, the sum of roots alpha + beta = -3a and the product alpha*beta = 2a^2. Given alpha^2 + beta^2 = 5, we use (alpha + beta)^2 - 2(alpha*beta) = 5, which is (-3a)^2 - 2(2a^2) = 5, leading to 9a^2 - 4a^2 = 5, so 5a^2 = 5, a^2 = 1. The product of roots is 2a^2 = 2(1) = 2.

Multiple choice

The value of $m$ for which the equation $\cfrac{a}{x+a+m}+\cfrac{b}{x+b+m}=1$, has roots equal in magnitude but opposite in sign is
  1. $\cfrac{a+b}{a-b}$
  2. $0$
  3. $\cfrac{a-b}{a+b}$
  4. $\cfrac{2(a-b)}{a+b}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For roots to be equal in magnitude but opposite in sign, the sum of roots must be 0. The equation is a/(x+a+m) + b/(x+b+m) = 1. Rearranging: a(x+b+m) + b(x+a+m) = (x+a+m)(x+b+m). x(a+b) + ab + am + bx + ab + bm = x^2 + x(b+m) + x(a+m) + (a+m)(b+m). x(a+b) + 2ab + m(a+b) = x^2 + x(a+b+2m) + (a+m)(b+m). x^2 + x(2m) + (a+m)(b+m) - 2ab - m(a+b) = 0. For sum of roots to be 0, the coefficient of x must be 0, so 2m = 0, thus m = 0.

Multiple choice

If the ratio of the roots of the equation $x^2+bx+c=0$ is the same as that of $x^2 + qx +r=0$, then
  1. $r^2b = qc^2$
  2. $r^2c = qb^2$
  3. $c^2r = q^2b$
  4. $b^2r=q^2c$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let roots of the first be a, b. a/b = k. Roots of second are ka, kb. Sum of roots: a(1+k) = -b, a(1+k) = -q/k. Product: a^2 k = c, a^2 k^2 = r. Ratio of products: c/r = 1/k^2. k^2 = r/c. From sum: (a(1+k))^2 = b^2 => a^2(1+k)^2 = b^2. c/k * (1+k)^2 / k = b^2. Substituting k^2 = r/c leads to the relation b^2r = q^2c.

Multiple choice

If product of roots of the equation ${x^2} - 3kx + 2{e^{\log k}} - 1 = 0$ is $7,$ then

  1. roots are integers and positive

  2. roots are integers and negative

  3. roots are rational not integers

  4. roots are irrational
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The product of roots for ax^2 + bx + c = 0 is c/a. Here, c = 2*e^(log k) - 1 = 2k - 1. So (2k - 1)/1 = 7, which means 2k = 8, so k = 4. The equation becomes x^2 - 12x + 7 = 0. The discriminant is 144 - 28 = 116, which is not a perfect square, so the roots are irrational.

Multiple choice

If $\alpha$ and $\beta$ are the roots of the equation $x^{2}-p(x+1)-q=0$ then value of $\dfrac { { \alpha }^{ 2 }+2\alpha +1 }{ { \alpha }^{ 2 }+2\alpha +q } +\dfrac { { \beta }^{ 2 }+2{ \beta }+1 }{ { \beta }^{ 2 }+2{ \beta }+q } $
  1. $1$
  2. $2$
  3. $3$
  4. $0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given x^2 - px - p - q = 0, the roots alpha and beta satisfy alpha + beta = p and alpha*beta = -(p+q). Substituting these into the expression and simplifying using the properties of roots, the expression evaluates to 1.