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 and b are the roots of the quadratic equation $\displaystyle { 2x }^{ 2 }-6x+3=0$, find the value of $\displaystyle { a }^{ 3 }+{ b }^{ 3 }-3ab\left( { a }^{ 2 }+{ b }^{ 2 } \right) -3ab\left( a+b \right) $.

  1. -27

  2. -25

  3. 27

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For 2x^2 - 6x + 3 = 0, a+b = 3, ab = 1.5. The expression is a^3 + b^3 - 3ab(a^2 + b^2) - 3ab(a+b). Note that a^3 + b^3 = (a+b)(a^2 - ab + b^2) = (a+b)((a+b)^2 - 3ab). a^3 + b^3 = 3 * (9 - 4.5) = 13.5. a^2 + b^2 = (a+b)^2 - 2ab = 9 - 3 = 6. Expression = 13.5 - 3(1.5)(6) - 3(1.5)(3) = 13.5 - 27 - 13.5 = -27.

Multiple choice

If $\displaystyle \alpha ,\beta $ are the roots of the equation $\displaystyle { x }^{ 2 }-x-4=0$, find the value of $\displaystyle \frac { 1 }{ \alpha } +\frac { 1 }{ \beta } -\alpha \beta $.
  1. $\displaystyle \frac { 4 }{ 15 } $
  2. $\displaystyle \frac { 15 }{ 4 } $
  3. $\displaystyle \frac { 4 }{ 5 } $
  4. $\displaystyle \frac { 5 }{ 4 } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For x^2 - x - 4 = 0, sum of roots (alpha + beta) = 1, product (alpha * beta) = -4. The expression is (alpha + beta) / (alpha * beta) - (alpha * beta) = 1 / -4 - (-4) = -0.25 + 4 = 3.75 = 15/4.

Multiple choice

If $\displaystyle \alpha ,\beta $ are the roots of a quadratic equation $\displaystyle { x }^{ 2 }-3kx+{ k }^{ 2 }=0$, find the values of k. If $\displaystyle { \alpha }^{ 2 }+{ \beta }^{ 2 }=\frac { 7 }{ 4 } $.
  1. $\displaystyle \pm \frac { 1 }{ 3 } $
  2. $\displaystyle \pm \frac { 1 }{ 2 } $
  3. $\displaystyle \pm \frac { 1 }{ 4 } $
  4. $\displaystyle \pm \frac { 1 }{ 5 } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For x^2 - 3kx + k^2 = 0, alpha+beta = 3k and alpha*beta = k^2. alpha^2+beta^2 = (alpha+beta)^2 - 2*alpha*beta = (3k)^2 - 2(k^2) = 9k^2 - 2k^2 = 7k^2. Given 7k^2 = 7/4, so k^2 = 1/4, k = +/- 1/2.

Multiple choice

Sum of the cubes of the roots of the quadratic equation $\displaystyle { 3x }^{ 2 }-5x+2=0$
  1. $\displaystyle -\frac { 35 }{ 27 } $
  2. $\displaystyle \frac { 7 }{ 9 } $
  3. $\displaystyle \frac { 27 }{ 35 } $
  4. $\displaystyle \frac { 35 }{ 27 } $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For 3x^2 - 5x + 2 = 0, the sum of roots (a+b) = 5/3 and product (ab) = 2/3. The sum of cubes is (a+b)^3 - 3ab(a+b) = (5/3)^3 - 3(2/3)(5/3) = 125/27 - 10/3 = 125/27 - 90/27 = 35/27.

Multiple choice

Arrange of the following quadratic equations in the ascending order of their number of real roots: A: $x^{2}-5x+6=0$ B: $x^{2}-7|x|=0$ C: $x^{2}-4x+5=0$ D: $x^{2}-5|x|+6=0$
  1. $C, A, B, D$
  2. $D, C, B, A$
  3. $A, D, B, C$
  4. $D, A, B, C$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A: x^2 - 5x + 6 = 0 has 2 roots (x=2, 3). B: x^2 - 7|x| = 0 has 3 roots (x=0, 7, -7). C: x^2 - 4x + 5 = 0 has 0 real roots (discriminant < 0). D: x^2 - 5|x| + 6 = 0 has 4 roots (x=2, -2, 3, -3). Ascending order: C(0), A(2), B(3), D(4).

Multiple choice

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

For the first equation, the discriminant D1 = (-2c)^2 - 4(ab) = 4(c^2 - ab) > 0, so c^2 > ab. For the second equation, the discriminant D2 = [-2(a+b)]^2 - 4(a^2 + b^2 + 2c^2) = 4(a^2 + 2ab + b^2 - a^2 - b^2 - 2c^2) = 4(2ab - 2c^2) = 8(ab - c^2). Since c^2 > ab, ab - c^2 < 0, meaning D2 < 0. Therefore, the roots are imaginary.