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 the zeroes of a quadratic polynomial $ax^{2}+bx+c$ are both negative, then which of the following is always correct
  1. $\dfrac{-b}{a}>0$
  2. $\dfrac{c}{a}=0$
  3. $a,b$ and $c$ all are zero
  4. $a,b$ and $c$ all have same sign
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

If both roots are negative, their sum (-b/a) must be negative, so b/a > 0. Their product (c/a) must be positive. For both to be positive, a, b, and c must have signs such that b/a > 0 and c/a > 0, implying a, b, and c share the same sign.

Multiple choice

If exactly one root of the quadratic equation ${ x }^{ 2 }-(k+\frac { 11 }{ 3 } )x-({ k }^{ 2 }+k+1)=0$ lies in (0, 3) then which one of the following relation is correct

  1. -3 < k < -1

  2. -8 < k < -4

  3. 1 < k < 4

  4. -7 < k < 5
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For exactly one root to lie in (0, 3), the product of the function values at the endpoints must be negative, f(0)*f(3) < 0. f(0) = -(k^2+k+1), f(3) = 9 - 3(k+11/3) - (k^2+k+1) = 9 - 3k - 11 - k^2 - k - 1 = -k^2 - 4k - 3. So (k^2+k+1)(k^2+4k+3) < 0. Since k^2+k+1 is always positive, we need k^2+4k+3 < 0, which is (k+1)(k+3) < 0, so -3 < k < -1.

Multiple choice

If one root of the quadratic equation $x^2+px+q=0$ is $2-\sqrt{3}$; where p, q $\in$ Q. Then which of the following is true?
  1. $p^2-4q+12=0$
  2. $p^2-4q-12=0$
  3. $q^2-4p+12=0$
  4. $q^2-4p-12=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since the coefficients p and q are rational, if 2-sqrt(3) is a root, then its conjugate 2+sqrt(3) must also be a root. The sum of roots is -p = (2-sqrt(3)) + (2+sqrt(3)) = 4, so p = -4. The product of roots is q = (2-sqrt(3))(2+sqrt(3)) = 4-3 = 1. Substituting into p^2-4q-12 gives (-4)^2 - 4(1) - 12 = 16 - 4 - 12 = 0.

Multiple choice

If the sum of the two roots of the equation $\displaystyle \frac {1}{x+a}+\frac {1}{x+b}=\frac {1}{c}$ is zero, then the product of the two roots is
  1. $0$
  2. $\displaystyle \frac {a^{2}+b^{2}}{2}$
  3. $\displaystyle \frac {a+b}{2}$
  4. $\displaystyle -\frac {(a^{2}+b^{2})}{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

1/(x+a) + 1/(x+b) = 1/c => (2x + a + b) / (x^2 + (a+b)x + ab) = 1/c. 2cx + c(a+b) = x^2 + (a+b)x + ab. x^2 + (a+b-2c)x + (ab - c(a+b)) = 0. Sum of roots = -(a+b-2c) = 0 => 2c = a+b. Product of roots = ab - c(a+b) = ab - ((a+b)/2)(a+b) = ab - (a^2 + 2ab + b^2)/2 = (2ab - a^2 - 2ab - b^2)/2 = -(a^2 + b^2)/2.