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 $\alpha$ and $\beta$ are the roots of the equation $ax^{2}\, +\, bx\, +\, c\, =\, 0$, then the value of $\displaystyle \frac{\alpha}{\beta}\, +\, \displaystyle \frac{\beta}{\alpha}$ is $\displaystyle \frac{b^{2}\, -\, 2ac}{ac}$. State true or false.

  1. True

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

The sum of roots is -b/a and the product of roots is c/a. The expression (alpha/beta) + (beta/alpha) simplifies to (alpha^2 + beta^2) / (alpha * beta), which is ((alpha + beta)^2 - 2 * alpha * beta) / (alpha * beta). Substituting the values gives ((-b/a)^2 - 2(c/a)) / (c/a) = (b^2/a^2 - 2c/a) / (c/a) = (b^2 - 2ac) / a^2 * (a/c) = (b^2 - 2ac) / ac.

Multiple choice

If the quadratic equation $ax^2 + bx + b = 0$ has the integer roots, where a and b are also intergers, then which of the following cannot be true
  1. $4a-b=0$
  2. $4a+b=0$
  3. $b=0$
  4. $b\neq 0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If roots are integers, the discriminant must be a perfect square. For ax^2 + bx + b = 0, D = b^2 - 4ab. If b = 0, x = 0 (integer). If b != 0, we test options. 4a+b=0 implies b = -4a, D = 16a^2 - 4a(-4a) = 32a^2, not a perfect square unless a=0.

Multiple choice

One root of the equation $ \displaystyle ax^{2}+bx+c=0 $ is square of the other if
  1. $ \displaystyle a^{2}c+b^{3}+ac^{2}=3abc $
  2. $ \displaystyle a^{3}+b^{3}+c^{3}=3abc $
  3. $ \displaystyle b^{2}c+c^{2}a+a^{2}b=abc $
  4. $ \displaystyle a^{2}+b^{2}+c^{2}-ab-bc-ca=0 $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If roots are alpha and alpha^2, then alpha + alpha^2 = -b/a and alpha * alpha^2 = c/a. From the second, alpha^3 = c/a. Substituting into the first: alpha^2 + alpha = -b/a. Cubing both sides: (alpha^2 + alpha)^3 = (-b/a)^3. This leads to the condition a^2c + b^3 + ac^2 = 3abc.