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

Which of the following equations has real and equal roots?

  1. 9x2 – 24x – 16 = 0

  2. 9x2 + 24x – 16 = 0

  3. 9x2 – 24x + 16 = 0

  4. 9x2 + 12x + 16 = 0
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For real and equal roots, the discriminant D = b^2 - 4ac must be 0. For 9x^2 - 24x + 16 = 0, D = (-24)^2 - 4*9*16 = 576 - 576 = 0.

Multiple choice

The roots of the quadratic equation 5x2 - 14x - 3 = 0 are

  1. 1/5, 3

  2. 5, 1/3
    • 1/5, 3

  4. 1/5, - 3
    • 1/5, - 3
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using the quadratic formula x = (-b +/- sqrt(b^2 - 4ac)) / 2a, for 5x^2 - 14x - 3 = 0, we get x = (14 +/- sqrt(196 - 4*5*(-3))) / 10 = (14 +/- sqrt(256)) / 10 = (14 +/- 16) / 10. The roots are 30/10 = 3 and -2/10 = -1/5.

Multiple choice

If the roots of the equation x2 – 2ax + a2 + a – 3 = 0 are real and less than 3, then

  1. a < 2

  2. 2 ≤ a ≤ 3

  3. 3 < a ≤ 4

  4. a > 4
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Roots of x^2 - 2ax + a^2 + a - 3 = 0. D = 4a^2 - 4(a^2 + a - 3) = 4a^2 - 4a^2 - 4a + 12 = 12 - 4a. For real roots, D >= 0 => a <= 3. Roots are x = (2a +/- sqrt(12-4a))/2 = a +/- sqrt(3-a). Both roots < 3: a + sqrt(3-a) < 3 => sqrt(3-a) < 3-a. Let u = sqrt(3-a), u < u^2 => u^2 - u > 0 => u(u-1) > 0. u > 1 or u < 0. Since u >= 0, u > 1 => sqrt(3-a) > 1 => 3-a > 1 => a < 2.

Multiple choice

The condition that the roots of the equation px2 + qx + r = 0 are the reciprocals of the roots of equation ax2 + bx + c = 0 is

  1. acq2 = b2pr

  2. ac = pr

  3. b2 = q2

  4. b2ac = q2pr
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If roots of px^2 + qx + r = 0 are reciprocals of ax^2 + bx + c = 0, then the coefficients must be reversed. Thus, p=c, q=b, r=a. Substituting these into the original equation, we get cx^2 + bx + a = 0. Comparing the coefficients of the two equations, we find the condition acq^2 = b^2pr.

Multiple choice

What is the number of real roots of the equation A2/x + B2/(x - 1) = 1, where A and B are real numbers not equal to zero?

  1. 0

  2. 1

  3. 1 or 2

  4. 2
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Multiplying the equation by x(x - 1) yields the quadratic equation x^2 - (A^2 + B^2 + 1)x + A^2 = 0. The discriminant of this quadratic is ((A - 1)^2 + B^2)((A + 1)^2 + B^2), which is strictly positive because B is non-zero. Thus, the equation always has exactly two distinct real roots, neither of which can be 0 or 1.

Multiple choice

If one root of the equation x2 + mx + 12 = 0 is 4 and the equation x2 + mx + n = 0 has equal roots, then the value of n is

  1. 49/4

  2. 4/49

  3. 4

  4. ¼
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since 4 is a root of x^2 + mx + 12 = 0, substitute x=4 to get 16 + 4m + 12 = 0, which gives 4m = -28, so m = -7. The second equation x^2 - 7x + n = 0 has equal roots, meaning its discriminant D = b^2 - 4ac = 0; thus (-7)^2 - 4(1)(n) = 0, which leads to 49 - 4n = 0, or n = 49/4.