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 one root of a quadratic equation is 3-√2, find the equation.

  1. x2+ 6x+5 = 0

  2. x2 - 9x + 7 = 0

  3. x2 - 6x + 7 = 0

  4. x2 + 6x - 7 = 0
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If one root is 3-sqrt(2), the other must be 3+sqrt(2). Sum of roots = 6. Product of roots = (3-sqrt(2))(3+sqrt(2)) = 9-2 = 7. The equation is x^2 - (sum)x + (product) = 0, which is x^2 - 6x + 7 = 0.

Multiple choice

If the quadratic equation (b2 + c2) x2 - 2(a + b)cx + (c2 + a2) = 0 has equal roots, then

  1. a, b, c are in G.P

  2. a, b, c are in A.P

  3. a, c, b are in G.P

  4. a, c, b are in A.P
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a quadratic equation Ax^2 + Bx + C = 0 to have equal roots, the discriminant D = B^2 - 4AC must be zero. Substituting the given coefficients, we get 4(a+b)^2 c^2 - 4(b^2+c^2)(c^2+a^2) = 0, which simplifies to a^2 c^2 + b^2 c^2 - b^2 c^2 - b^2 a^2 - c^4 - c^2 a^2 = 0. Rearranging gives b^2 a^2 + c^4 = 0, which implies c^2 = ab, meaning a, c, b are in G.P.

Multiple choice

The cubic equation which have the roots 2, 3, 4 is

  1. x3 - 26x2 - 9x + 9 = 0

  2. x3 + 9x2 + 26x + 24 = 0

  3. x3 - 9x2 + 26x - 24 = 0

  4. x3 + 5x2 + 24 = 0
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Equation with roots 2, 3, 4 is (x-2)(x-3)(x-4) = 0. (x^2 - 5x + 6)(x-4) = x^3 - 4x^2 - 5x^2 + 20x + 6x - 24 = x^3 - 9x^2 + 26x - 24 = 0.

Multiple choice

Which of the following equations has equal roots?

  1. x2 + 4x + 4 = 0

  2. x2 + 4x - 4 = 0
    • x2 + 4x + 4 = 0

  4. x2 + 6x + 6 = 0
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A quadratic equation ax^2 + bx + c = 0 has equal roots if the discriminant D = b^2 - 4ac = 0. For option A: 4^2 - 4(1)(4) = 16 - 16 = 0. This satisfies the condition.

Multiple choice

Amit and Alok attempted to solve a quadratic equation. Amit made a mistake in writing down the constant term and ended up with roots (4, 3). Alok made a mistake in writing down coefficient of x to get roots (3, 2). The correct roots of the equation are:

  1. -4, -3

  2. 6, 1

  3. 4, 3

  4. -6, -1
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Amit's roots (4,3) imply the equation is x^2 - 7x + k = 0. Alok's roots (3,2) imply x^2 - kx + 6 = 0. Comparing, the correct equation is x^2 - 7x + 6 = 0. Roots are 6 and 1.

Multiple choice

If -1 is one of the roots of the cubic equation x3 + (3s - 2)x2 + (3s + 1)x + 4 = 0, and the other two roots of equation are real, then the minimum positive integral value of s is:

  1. 2

  2. 3

  3. 4

  4. 5
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since -1 is a root, (-1)^3 + (3s-2)(-1)^2 + (3s+1)(-1) + 4 = 0. -1 + 3s - 2 - 3s - 1 + 4 = 0. This simplifies to 0=0, which is always true for any s. For the other two roots to be real, the discriminant of the quadratic factor must be >= 0. Dividing the cubic by (x+1) gives x^2 + (3s-3)x + 4 = 0. Discriminant D = (3s-3)^2 - 16 >= 0. (3s-3)^2 >= 16. 3s-3 >= 4 or 3s-3 <= -4. 3s >= 7 (s >= 2.33) or 3s <= -1. Smallest positive integer s is 3.

Multiple choice

Two positive roots of the equation y² - Sy + 45 = 0 are m and n, with m - n = 4. Quantity I: Value of 2S Quantity II: Value of n² + m + 1

  1. Quantity I > Quantity II

  2. Quantity I < Quantity II

  3. Quantity I ≥ Quantity II

  4. Quantity I ≤ Quantity II

  5. Quantity I = Quantity II or no relation
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

From the quadratic equation y^2 - Sy + 45 = 0, the roots m and n satisfy m+n = S and mn = 45. Given m-n = 4, we solve the system: (m+n)^2 - (m-n)^2 = 4mn, so S^2 - 16 = 180, S^2 = 196, S = 14. Then m+n=14 and m-n=4 gives m=9, n=5. Quantity I = 2S = 28. Quantity II = n^2 + m + 1 = 25 + 9 + 1 = 35. Since 28 < 35, Quantity I < Quantity II.