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 $a,b,c \in R$ and $a+b+c=0$, then the quadratic equation $4ax^{2}+3bx+2c=0$ has

  1. one positive and one negative

  2. imaginary roots

  3. real roots

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Discriminant D = (3b)^2 - 4(4a)(2c) = 9b^2 - 32ac. Since a+b+c=0, b = -(a+c). D = 9(a+c)^2 - 32ac = 9a^2 + 18ac + 9c^2 - 32ac = 9a^2 - 14ac + 9c^2. This is always positive for real a, c (as it's a quadratic in a/c with negative discriminant), so roots are real.

Multiple choice

If $p, q, r$ are positive and are in A.P., the roots of quadratic equation $px^2+qx+r=0$ are all real for
  1. $\left |\dfrac {r}{p}-7 \right |\geq 4\sqrt 3$
  2. $\left |\dfrac {p}{r}-7\right |\geq 4\sqrt 3$
  3. All $p$ and $r$
  4. No $p$ and $r$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For px^2 + qx + r = 0 to have real roots, the discriminant D = q^2 - 4pr >= 0. Since p, q, r are in A.P., q = (p+r)/2. Substituting: ((p+r)/2)^2 - 4pr >= 0 -> (p+r)^2 - 16pr >= 0 -> p^2 + 2pr + r^2 - 16pr >= 0 -> p^2 - 14pr + r^2 >= 0. Dividing by r^2: (p/r)^2 - 14(p/r) + 1 >= 0. Solving this inequality leads to |p/r - 7| >= 4*sqrt(3).

Multiple choice

Quadratic equation ___ has equal roots.
  1. $x^{2}-10x+25=0$
  2. $x^{2}-10x-25=0$
  3. $x^{2}+10x-25=0$
  4. $x^{2}-25=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A quadratic equation has equal roots if the discriminant b^2 - 4ac = 0. For x^2 - 10x + 25 = 0, (-10)^2 - 4(1)(25) = 100 - 100 = 0.

Multiple choice

Roots of the equation ${x^3} - (a + b + c){x^2} + (ab + bc + ca)x - abc = 0$, if $x^2+2x+7=0$ and $ax^2+bx+c=0$ have a common root, where a,b,c$ \in $R, can be
  1. $4 , 8 , 28$
  2. $1 , 2 , 7$
  3. $1 , 4 , 36$

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The cubic equation factors to (x-a)(x-b)(x-c) = 0, so the roots are a, b, and c. If x^2 + 2x + 7 = 0 and ax^2 + bx + c = 0 share a root, and a, b, c are coefficients, the roots of the quadratic must be the roots of the cubic. By comparing coefficients, the roots are 1, 2, 7.

Multiple choice

If the roots of the equation $x^2-2cx+ab=0$ be real and unequal, the roots of the equation ${x^2} - 2(a + b)x + {a^2} + {b^2} + 2{c^2} = 0$ are

  1. real and distinct

  2. real and equal

  3. real

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

Roots of x^2 - 2cx + ab = 0 are real/unequal, so discriminant (2c)^2 - 4ab > 0 => c^2 > ab. For the second equation, discriminant D = [2(a+b)]^2 - 4(a^2 + b^2 + 2c^2) = 4(a^2 + 2ab + b^2) - 4a^2 - 4b^2 - 8c^2 = 8ab - 8c^2 = 8(ab - c^2). Since c^2 > ab, ab - c^2 < 0, so D < 0. Roots are imaginary.