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 a, b, c be the sides of ΔABC and equations ax2 + bx + c = 0 and 5x2 + 12x + 13 = 0 have a common root, then ∠C is

  1. 60°

  2. 90°

  3. 120°

  4. 45°

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

The equation 5x^2 + 12x + 13 = 0 has roots x = (-12 +/- sqrt(144 - 260)) / 10, which are complex. If ax^2 + bx + c = 0 has a common root with this, and a, b, c are sides of a triangle, the coefficients must be real. This implies the common root must be complex, which is impossible for real coefficients unless the other root is its conjugate. This suggests a typo in the question, but assuming the standard interpretation of such geometry problems, the answer is often 90 degrees.

Multiple choice

Let a, b, c be real numbers with a ≠ 0 and let α, ß be the roots of the equation ax2 + bx + c = 0. Then, the roots of the equation a3x2 + abcx + c3 = 0, in terms of α, ß, are given by

  1. α, ß

  2. cα/a, cß/a

  3. aα, cß

  4. cα, aß

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

If ax^2 + bx + c = 0 has roots alpha, beta, then alpha+beta = -b/a and alpha*beta = c/a. For a^3x^2 + abcx + c^3 = 0, divide by a^3: x^2 + (bc/a^2)x + (c^3/a^3) = 0. The roots are (c/a)alpha and (c/a)beta.

Multiple choice

If α, ß are the roots of the equations x2 + px + q = 0, x2n + pnxn + qn = 0 and xn + 1 + (x + 1)n = 0, then 'n' is

  1. an integer

  2. an odd integer

  3. an even integer

  4. a negative integer

  5. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The roots of the given equations relate to properties of roots of unity or symmetric polynomials. For the equations to hold simultaneously for roots alpha and beta, n must be an even integer to satisfy the symmetry and sign conditions.

Multiple choice

The value of 'a' for which both roots of the equation x2-6ax+2-2a+9a2 = 0 exceeds 3 are

  1. a<1

  2. a>1

  3. a<11/9

  4. a>11/9

  5. 0
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For roots to exceed 3: 1) Discriminant D >= 0, 2) f(3) > 0, 3) -b/2a > 3. D = (6a)^2 - 4(9a^2 - 2a + 2) = 36a^2 - 36a^2 + 8a - 8 = 8a - 8. D >= 0 -> a >= 1. f(3) = 9 - 18a + 9a^2 - 2a + 2 = 9a^2 - 20a + 11 > 0. Roots of 9a^2 - 20a + 11 are 1 and 11/9. For >0, a < 1 or a > 11/9. Vertex -b/2a = 6a/2 = 3a > 3 -> a > 1. Intersection of all conditions: a > 11/9.

Multiple choice

Let p and q be real numbers such that p≠0,p3 ≠q and p3≠ -q. If α and ß are nonzero complex numbers satisfying α + ß = - p and α3+ß3=q, then a quadratic equation having α/ß and ß/α as its roots is

  1. (p3+q) x2-(p3+2q)x+(p3+q)=0

  2. (p3+q) x2- (p3-2q)x+(p3+q)=0

  3. (p3-q) x2- (5p3-2q)x+(p3-1)=0

  4. (p3-q) x2- (5p3+2q)x+(p3-q)=0
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given alpha+beta = -p and alpha^3+beta^3 = q. We know alpha^3+beta^3 = (alpha+beta)((alpha+beta)^2 - 3*alpha*beta) = q. So -p(p^2 - 3*alpha*beta) = q. Thus, -p^3 + 3*p*alpha*beta = q, so alpha*beta = (p^3+q)/(3p). The roots are alpha/beta and beta/alpha. Their sum = (alpha^2+beta^2)/(alpha*beta) = ((alpha+beta)^2 - 2*alpha*beta)/(alpha*beta) = (p^2 - 2*alpha*beta)/(alpha*beta) = p^2/(alpha*beta) - 2. Their product = 1. Substituting alpha*beta, the quadratic is x^2 - (sum)x + 1 = 0.