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

Given $\alpha $ and $\beta $ are the roots of the equation ${x^2} - 4x + k = 0(k \ne 0)$. If $\alpha \beta ,\ \alpha {\beta ^2} + {\alpha ^2}\beta ,\ {\alpha ^3} + {\beta ^3}$ are in geometric progression then the value of $'k'$ equals
  1. $4$
  2. $\dfrac {16}{7}$
  3. $\dfrac {3}{7}$
  4. $12$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Roots alpha, beta satisfy alpha+beta=4, alpha*beta=k. The terms are k, alpha*beta(alpha+beta)=4k, alpha^3+beta^3=(alpha+beta)((alpha+beta)^2-3alpha*beta)=4(16-3k). For GP, (4k)^2 = k * 4(16-3k). 16k^2 = 64k - 12k^2. 28k^2 = 64k. Since k is not 0, k = 64/28 = 16/7.

Multiple choice

If $\alpha $ and $\beta $ are the zeros of the quadratic polynomial $f(x) =x^2- x -2$, find the value of $\frac{1}{\alpha}-\frac{1}{\beta}.$
  1. $\dfrac{-3}{2}$
  2. $\dfrac{1}{2} $
  3. $\dfrac{3}{2}$
  4. $1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For x^2 - x - 2, roots are (x-2)(x+1)=0, so 2 and -1. 1/alpha - 1/beta = (beta - alpha) / (alpha * beta). alpha + beta = 1, alpha * beta = -2. (beta - alpha)^2 = (alpha + beta)^2 - 4 * alpha * beta = 1 - 4(-2) = 9. beta - alpha = 3 or -3. Result = 3 / -2 = -1.5 or 1.5. Option C is 3/2.

Multiple choice

The condition for the equations $ax^{2} + bx + c = 0$ and $a'x^{2} + b'x + c' = 0$ to have reciprocal roots is $\dfrac{a}{c'}=\dfrac{b}{b'}=\dfrac{c}{a'}$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For roots to be reciprocals, the product of roots must be 1. For ax^2+bx+c=0, product is c/a. For a'x^2+b'x+c'=0, product is c'/a'. If roots are reciprocals, c/a = a'/c', which rearranges to a/c' = a'/c. The given condition is a standard result for reciprocal roots.

Multiple choice

If $b^2-4ac\geq 0$, then the root of quadratic equation $ax^2+bx+c=0$ is
  1. $\frac {b}{2a}\pm \frac {\sqrt {b^2-4a}}{2a}$
  2. $-\frac {b}{2a}\pm \frac {\sqrt {b^2+4ac}}{2a}$
  3. $\frac {b}{2a}\pm \frac {\sqrt {b^2+4ac}}{2a}$
  4. $-\frac {b}{2a}\pm \frac {\sqrt {b^2-4ac}}{2a}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The quadratic formula for ax^2 + bx + c = 0 is x = (-b +/- sqrt(b^2 - 4ac)) / 2a.

Multiple choice

The roots of the equation $bx^{2} + (b - c)x + (b -c -a) = 0$ are real if those of $ax^{2} + 2bx + b = 0$ are imaginary.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the second equation ax^2 + 2bx + b = 0 to have imaginary roots, the discriminant (2b)^2 - 4ab < 0, which means 4b^2 < 4ab, or b < a (assuming b>0). For the first equation bx^2 + (b-c)x + (b-c-a) = 0 to have real roots, the discriminant (b-c)^2 - 4b(b-c-a) >= 0 must hold. This is a logical statement about coefficients, and it is true.

Multiple choice

Which of the following is not a quadratic equation?
  1. $2(x-1)^2=4x^2-2x+1$
  2. $(x^2+1)^2=x^2+3x+9$
  3. $(x^2+2x)^2=x^4+3+4x^3$
  4. $x^2+9=3x^2-5x$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A quadratic equation must have a degree of 2. In option B, expanding (x^2+1)^2 results in x^4 + 2x^2 + 1, which leads to a degree of 4, making it not a quadratic equation.

Multiple choice

Given $\alpha$ and $\beta$ are the roots of the quadratic equation ${x}^{2}-4x+k=0(k\neq 0)$. If $ {\alpha \beta}^{2}+{\alpha}^{2}\beta$.
  1. $4k$
  2. $\dfrac {16}{7}$
  3. $\dfrac {3}{7}$
  4. $12$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The given expression is incomplete in the prompt text, but typically asks for alpha*beta^2 + alpha^2*beta. Factoring this gives alpha*beta(alpha + beta). From the equation x^2 - 4x + k = 0, the sum of roots is 4 and the product is k. Substituting these values yields k * 4 = 4k.