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

Solve the equation using formula. $2x^{2}\, +\, \displaystyle \frac{x\, -\, 1}{5}\, =\, 0$
  1. $x\, =\, \displaystyle \frac{-1\, \pm\, \sqrt{10}}{4}$
  2. $x\, =\, \displaystyle \frac{-1\, \pm\, \sqrt{41}}{20}$
  3. $x\, =\, \displaystyle \frac{1\, \pm\, \sqrt{41}}{20}$
  4. $x\, =\, \displaystyle \frac{1\, \pm\, \sqrt{10}}{4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Multiply by 5: 10x^2 + x - 1 = 0. Using the quadratic formula x = (-b +/- sqrt(b^2 - 4ac)) / 2a, we get x = (-1 +/- sqrt(1 - 4(10)(-1))) / 20 = (-1 +/- sqrt(41)) / 20.

Multiple choice

Form the quadratic equation whose roots are: $3$ and $10$
  1. $x^{2}\, -\, 30x\, +\, 13\, =\, 0$
  2. $x^{2}\, +\, 13x\, +\, 30\, =\, 0$
  3. $x^{2}\, -\, 13x\, +\, 30\, =\, 0$
  4. $x^{2}\, +\, 30x\, +\, 13\, =\, 0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Roots are 3 and 10. Sum = 13. Product = 30. Equation is x^2 - (sum)x + product = 0. x^2 - 13x + 30 = 0.

Multiple choice

Form the quadratic equation whose roots are: $-3$ and $\displaystyle \frac{5}{2}$
  1. $2x^{2}\, +\, x\, -\, 15\, =\, 0$
  2. $2x^{2}\, -\, x\, -\, 15\, =\, 0$
  3. $x^{2}\, +\, x\, -\, 45\, =\, 0$
  4. $x^{2}\, -\, x\, -\, 45\, =\, 0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A quadratic equation with roots r1 and r2 is given by x^2 - (r1 + r2)x + (r1 * r2) = 0. Here, r1 = -3 and r2 = 5/2. The sum is -3 + 2.5 = -0.5, and the product is -3 * 2.5 = -7.5. The equation is x^2 + 0.5x - 7.5 = 0. Multiplying by 2 gives 2x^2 + x - 15 = 0.

Multiple choice

Solve the equation using formula. $6m^{2}\, -\, 4m\, =\, 3$
  1. $m\, =\, \displaystyle \frac{-2\, \pm\, \sqrt{22}}{6}$
  2. $m\, =\, \displaystyle \frac{-4\, \pm\, \sqrt{22}}{12}$
  3. $m\, =\, \displaystyle \frac{2\, \pm\, \sqrt{22}}{6}$
  4. $m\, =\, \displaystyle \frac{4\, \pm\, \sqrt{22}}{12}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

6m^2 - 4m - 3 = 0. Using quadratic formula: m = [-b +/- sqrt(b^2 - 4ac)] / 2a = [4 +/- sqrt(16 - 4*6*(-3))] / 12 = [4 +/- sqrt(16 + 72)] / 12 = [4 +/- sqrt(88)] / 12 = [4 +/- 2*sqrt(22)] / 12 = (2 +/- sqrt(22)) / 6.

Multiple choice

Form the quadratic equation whose roots are: $-5$ and $9$
  1. $x^{2}\, +\, 4x\, -\, 45\, =\, 0$
  2. $x^{2}\, -\, 14x\, -\, 45\, =\, 0$
  3. $x^{2}\, +\, 14x\, -\, 45\, =\, 0$
  4. $x^{2}\, -\, 4x\, -\, 45\, =\, 0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A quadratic equation with roots r1 and r2 is given by x^2 - (r1 + r2)x + (r1 * r2) = 0. Here, the sum of roots is -5 + 9 = 4 and the product is -5 * 9 = -45, resulting in x^2 - 4x - 45 = 0.

Multiple choice

Form the quadratic equation with rational coefficients whose one of the root is: $3\, -\, 2\sqrt3$
  1. $x^{2}\, -\, 3x\, -\, 6\,  =\, 0$
  2. $x^{2}\, +\, 6x\, -\, 3\,  =\, 0$
  3. $x^{2}\, +\, 3x\, -\, 6\,  =\, 0$
  4. $x^{2}\, -\, 6x\, -\, 3\,  =\, 0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

If one root is 3 - 2*sqrt(3), the other must be 3 + 2*sqrt(3) for rational coefficients. Sum of roots = 6, product = 9 - 12 = -3. Equation: x^2 - (sum)x + product = 0, which is x^2 - 6x - 3 = 0.

Multiple choice

Form the quadratic equation with rational coefficients whose one of the root is: $2\, +\, 3\sqrt5$
  1. $x^{2}\, -\, 4x\, -\, 41\,  =\, 0$
  2. $x^{2}\, +\, 4x\, -\, 41\,  =\, 0$
  3. $x^{2}\, -\, 4x\, -\, 45\,  =\, 0$
  4. $x^{2}\, +\, 4x\, -\, 45\,  =\, 0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If one root is 2 + 3*sqrt(5), the other must be 2 - 3*sqrt(5) for rational coefficients. Sum of roots = 4. Product of roots = 2^2 - (3*sqrt(5))^2 = 4 - 45 = -41. The equation is x^2 - (sum)x + (product) = 0, which is x^2 - 4x - 41 = 0.

Multiple choice

If $\alpha$ and $\beta$ are roots of $ax^2+bx+c=0$, then equation whose roots are $\displaystyle \alpha +\frac{1}{\beta }, \beta +\frac{1}{\alpha }$, is
  1. $\displaystyle acx^{2}-b\left ( a-c \right )x+\left ( a+c \right )^{2}= 0.$
  2. $\displaystyle acx^{2}+b\left ( a+b \right )x-\left ( bc \right )^{2}= 0.$
  3. $\displaystyle acx^{2}+b\left ( a+c \right )x+\left ( a+c \right )^{2}= 0.$

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

Given alpha+beta = -b/a and alpha*beta = c/a. The new roots are S = alpha+1/beta + beta+1/alpha = (alpha+beta) + (alpha+beta)/(alpha*beta) = -b/a + (-b/a)/(c/a) = -b/a - b/c = -b(a+c)/(ac). The product P = (alpha+1/beta)(beta+1/alpha) = alpha*beta + 1 + 1 + 1/(alpha*beta) = c/a + 2 + a/c = (c^2+2ac+a^2)/(ac) = (a+c)^2/(ac). The equation is x^2 - Sx + P = 0, which leads to acx^2 + b(a+c)x + (a+c)^2 = 0.

Multiple choice

If $\alpha , \beta , \gamma $ are the real roots of the equation $x^{3}-3px^{2}+3qx-1=0$, then the centroid of the triangle with vertices $\displaystyle \left ( \alpha , \frac{1}{\alpha } \right )$, $\left ( \beta , \dfrac{1}{\beta } \right )$ and $\displaystyle \left ( \gamma , \frac{1}{\gamma } \right )$ is at the point
  1. $(p, q)$
  2. $\left(\dfrac p3, \dfrac q3\right)$
  3. $(p+q, p-q)$
  4. $(3p, 3q)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The centroid of a triangle with vertices (x1, y1), (x2, y2), and (x3, y3) is ((x1+x2+x3)/3, (y1+y2+y3)/3). For the given cubic equation, the sum of roots alpha + beta + gamma = 3p and the sum of reciprocals (alpha*beta + beta*gamma + gamma*alpha) / (alpha*beta*gamma) = 3q/1 = 3q. Dividing the sums by 3 gives the centroid coordinates (p, q).

Multiple choice

If $\displaystyle \alpha \neq \beta$ but $\alpha ^{2}= 5\alpha -3, \beta ^{2}= 5\beta -3$, then the equation whose roots are $\displaystyle \frac{\alpha}{\beta} $ and $\dfrac{\beta}{\alpha} $ is
  1. $\displaystyle 3x^{2}-19x-3= 0.$
  2. $\displaystyle 3x^{2}+19x+3= 0.$
  3. $\displaystyle 3x^{2}-19x+3= 0.$

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

Given alpha^2 - 5alpha + 3 = 0 and beta^2 - 5beta + 3 = 0, alpha and beta are roots of x^2 - 5x + 3 = 0. Sum of roots = 5, product = 3. We need the equation with roots alpha/beta and beta/alpha. Sum = (alpha^2 + beta^2) / (alpha*beta) = ((alpha+beta)^2 - 2alpha*beta) / (alpha*beta) = (25 - 6) / 3 = 19/3. Product = 1. Equation: x^2 - (19/3)x + 1 = 0, or 3x^2 - 19x + 3 = 0.