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

cosec 18° is a root of the equation:

  1. x2 - 2x - 4 = 0

  2. x2 - 2x + 4 = 0

  3. 4x2 + 2x - 1 = 0

  4. x2 + 2x - 4 = 0
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

sin 18 = (sqrt(5)-1)/4. cosec 18 = 4/(sqrt(5)-1) = sqrt(5)+1. Let x = sqrt(5)+1, then x-1 = sqrt(5), (x-1)^2 = 5, x^2 - 2x + 1 = 5, x^2 - 2x - 4 = 0.

Multiple choice

Directions: In the following question, two equations (I) and (II) are given. You have to solve both the equations and give answer. (I) 13x2 - 28x + 4 = 0 (II) -4 = 5y2 + 12y

  1. x > y

  2. x < y

  3. x ≥ y

  4. x ≤ y

  5. x = y or no relation can be established between 'x' and 'y'
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

I: 13x^2 - 28x + 4 = 0. Roots: x = [28 +/- sqrt(784 - 208)] / 26 = [28 +/- sqrt(576)] / 26 = [28 +/- 24] / 26. x = 52/26 = 2 or x = 4/26 = 2/13. II: 5y^2 + 12y + 4 = 0. Roots: y = [-12 +/- sqrt(144 - 80)] / 10 = [-12 +/- 8] / 10. y = -4/10 = -0.4 or y = -20/10 = -2. Since both roots of x are positive and both roots of y are negative, x > y.

Multiple choice

Choose the quadratic equation with real roots.

  1. x 2 3 + x 4 − 3 = 0

  2. x 2 4 + 16 = 0

  3. (x + 4)2 + 20 = 0

  4. 7x2 + 1 = 0

  5. 4x2 - 3x + 2 = 0
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Option A: x^2/3 + x/4 - 3 = 0. Discriminant D = (1/4)^2 - 4(1/3)(-3) = 1/16 + 4 > 0, so roots are real. Other options have negative discriminants or lead to imaginary roots.

Multiple choice

Let α and β be the roots of the equation x2 + px + q = 0. If α3 and β3 are the roots of the equation x2 + mx + n = 0, then what is the value of m + n?

  1. p3 + q3 + pq

  2. p3 + q3 - pq

  3. p3 + q3 + 3pq

  4. p3 + q3 - 3pq
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Roots alpha, beta satisfy alpha+beta = -p, alpha*beta = q. New roots alpha^3, beta^3 satisfy sum = -m, product = n. alpha^3 + beta^3 = (alpha+beta)^3 - 3*alpha*beta*(alpha+beta) = (-p)^3 - 3*q*(-p) = -p^3 + 3pq. So m = p^3 - 3pq. Product = (alpha*beta)^3 = q^3 = n. m+n = p^3 - 3pq + q^3.

Multiple choice

Let α and β (α > β) be the roots of the equation x2 - 8x + q = 0. If α2 - β2 = 16, then what is the value of q?

  1. -15

  2. -10

  3. 10

  4. 15
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Roots alpha, beta satisfy alpha + beta = 8 and alpha * beta = q. We have alpha^2 - beta^2 = (alpha + beta)(alpha - beta) = 16. So 8(alpha - beta) = 16, which means alpha - beta = 2. Solving alpha + beta = 8 and alpha - beta = 2 gives alpha = 5, beta = 3. Then q = alpha * beta = 15.

Multiple choice

If α and β are the roots of the quadratic equation x2 + αx + β = 0, where β ≠ 0, then what is the value of α - β?

  1. 4

  2. 3

  3. -1

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

For x^2 + ax + b = 0, the sum of roots is -a and product is b. So alpha + beta = -a and alpha * beta = b. Given alpha and beta are roots, alpha^2 + a*alpha + b = 0 and beta^2 + a*beta + b = 0. Substituting a = -(alpha+beta), we get alpha^2 - (alpha+beta)alpha + beta = 0 => alpha^2 - alpha^2 - alpha*beta + beta = 0 => beta(1-alpha) = 0. Since beta is not 0, alpha = 1. Then 1 + a + b = 0 and beta^2 + a*beta + b = 0. Solving these leads to alpha - beta = 3.

Multiple choice

If the sum of the roots of the equation x2 - k2 x + 30kx - 161x - 64 = 0 is zero, then what is the difference of the roots?

  1. 15

  2. 16

  3. 17

  4. 18
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For the quadratic x^2 - (k^2 - 30k + 161)x - 64 = 0, the sum of roots is zero, meaning the coefficient of x must be zero. Thus, k^2 - 30k + 161 = 0. The roots of this quadratic are (x - a)(x + a) = 0, so the roots are a and -a. The product of roots is -a^2 = -64, so a = 8. The difference of roots is a - (-a) = 2a = 16.

Multiple choice

The equation (x2 + 1)2 - x2 = 0 has

  1. four real roots

  2. two real roots

  3. no real root

  4. one real root

  5. cannot be determined
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Expanding (x^2 + 1)^2 - x^2 = 0 gives x^4 + 2x^2 + 1 - x^2 = 0, which simplifies to x^4 + x^2 + 1 = 0. Since x^4 and x^2 are always non-negative for real x, x^4 + x^2 + 1 is at least 1, meaning there are no real roots.

Multiple choice

The sum and the product of the roots of a quadratic equation are 7 and 12 respectively. If the bigger root is halved and the smaller root is doubled, then what is the resulting quadratic equation?

  1. x2 - 6x + 12 = 0

  2. x2 - 8x + 12 = 0

  3. x2 + 8x + 12 = 0

  4. x2 - 10x + 12 = 0
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Roots a, b satisfy a+b=7, ab=12. Roots are 3 and 4. Bigger root halved = 4/2 = 2. Smaller root doubled = 3*2 = 6. New roots 2 and 6. New equation: x^2 - (2+6)x + (2*6) = 0, which is x^2 - 8x + 12 = 0.