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

$3 - 2i$ and $3 + 2i$ are roots of the quadratic equation:
  1. $x^{2} + 6x + 13 = 0$
  2. $x^{2} - 6x + 13 = 0$
  3. $x^{2} - 6x - 13 = 0$
  4. $x^{2} + 6x + 7 = 0$
  5. $x^{2} + 6x - 4 = 0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If roots are 3 +/- 2i, the sum is 6 and the product is 3^2 + 2^2 = 13. The equation is x^2 - (sum)x + (product) = 0, which is x^2 - 6x + 13 = 0.

Multiple choice

Find the quadratic equation in $x$, whose roots are $(m + n)$ and $(m - n)$
  1. $x^{2} + 2mx + (m^{2} + n^{2}) = 0$
  2. $x^{2} - 2mx + (m^{2} - n^{2}) = 0$
  3. $x^{2} - mx + (m^{2} - n^{2}) = 0$
  4. $x^{2} - mnx + (m^{2} - n^{2}) = 0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A quadratic equation with roots r1 and r2 is given by x^2 - (r1+r2)x + (r1*r2) = 0. Here, r1+r2 = (m+n) + (m-n) = 2m, and r1*r2 = (m+n)(m-n) = m^2 - n^2. Substituting these gives x^2 - 2mx + (m^2 - n^2) = 0.

Multiple choice

If $\alpha$ and $\beta$ are the roots of $ax^2 + bx + c = 0$ find the equation having roots $\dfrac{1}{\alpha}$ and $\dfrac{1}{\beta}$.
  1. $bx^2 + cx + a = 0$
  2. $cx^2 + ax + b = 0$
  3. $cx^2 + bx + a = 0$
  4. $cx^2 - bx + a = 0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If the roots are alpha and beta, the equation is a(x-alpha)(x-beta) = ax^2 - a(alpha+beta)x + a(alpha*beta) = 0. Comparing to ax^2 + bx + c = 0, we have alpha+beta = -b/a and alpha*beta = c/a. For roots 1/alpha and 1/beta, the new sum is (alpha+beta)/(alpha*beta) = (-b/a)/(c/a) = -b/c, and the product is 1/(alpha*beta) = a/c. The equation is x^2 - (sum)x + (product) = 0, which is x^2 + (b/c)x + (a/c) = 0, or cx^2 + bx + a = 0.

Multiple choice

Find the value of $\lambda$ for which the quadratic equation $3{ x }^{ 2 }+2({ \lambda }^{ 2 }+1)x+({ \lambda }^{ 2 }-3\lambda +2)=0$ has roots of opposite signs
  1. $1< \lambda < 2$
  2. $1< \lambda < 7$
  3. $1< \lambda < 8$

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

For a quadratic equation ax^2 + bx + c = 0 to have roots of opposite signs, the product of roots c/a must be less than 0. So (lambda^2 - 3*lambda + 2) / 3 < 0. (lambda - 1)(lambda - 2) < 0. This holds when 1 < lambda < 2.

Multiple choice

If the roots of the equation ${ x }^{ 2 }+2(3a+5)x+2(9{ a }^{ 2 }+25)=0$ are real, then find $a$.
  1. $\cfrac{5}{3}$
  2. $\cfrac{7}{3}$
  3. $\cfrac{2}{3}$

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

For real roots, discriminant D >= 0. D = (2(3a+5))^2 - 4(1)(2(9a^2+25)) >= 0. 4(9a^2 + 30a + 25) - 8(9a^2 + 25) >= 0. 36a^2 + 120a + 100 - 72a^2 - 200 >= 0. -36a^2 + 120a - 100 >= 0. Dividing by -4: 9a^2 - 30a + 25 <= 0. (3a-5)^2 <= 0. This only holds if 3a-5 = 0, so a = 5/3.

Multiple choice

The equation $2(a^{2} + b^{2}) x^{2} + 2(a + b)x + 1 = 0$ has which type of roots, when $a\neq b$.

  1. Two distinct real roots

  2. No real roots

  3. Two equal roots

  4. Two real roots
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Discriminant D = b^2 - 4ac. Here, b = 2(a+b), a = 2(a^2+b^2), c = 1. D = [2(a+b)]^2 - 4 * 2(a^2+b^2) * 1 = 4(a^2 + 2ab + b^2) - 8(a^2 + b^2) = 4a^2 + 8ab + 4b^2 - 8a^2 - 8b^2 = -4a^2 + 8ab - 4b^2 = -4(a^2 - 2ab + b^2) = -4(a-b)^2. Since a != b, (a-b)^2 > 0, so D < 0. Thus, no real roots.