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 the equation $x^3 - 3ax^2 + 3bx- c = 0$ has positive and distinct roots, then
  1. $a^2 > b$
  2. $ab > c$
  3. $a^3 > c$
  4. $a^3 > b^2 > c$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For cubic x^3 - 3ax^2 + 3bx - c = 0, if roots are p, q, r, then p+q+r = 3a, pq+qr+rp = 3b, pqr = c. For distinct real roots, the AM-GM inequality or discriminant properties apply. Specifically, a^2 > b is a necessary condition for real roots in this form.

Multiple choice

For the equations $x^2 + bx + c = 0$ and $2x^2 + (b + 1)x + c + 1 = 0$ select the correct alternative

  1. both the equations can have integral roots

  2. both the equations cant have integral roots simultaneously

  3. none of the equations can have integral roots

  4. nothing can be said
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If the first equation has integral roots, then b and c are integers. If the second also has integral roots, its root sum and product require b and c to be odd. But integral roots of the first equation with odd sum must have opposite parity, giving an even product, which contradicts c being odd. Therefore, both equations cannot have integral roots simultaneously.

Multiple choice

If the roots of the equation $\dfrac{a}{x+a+k}+\dfrac{b}{x+b+k}=2$ are equal in magnitude but opposite in sign, then the value of $k$ is:
  1. $\dfrac{(a+b)}{4}$
  2. $\dfrac{(a+b)}{3}$
  3. $\dfrac{(a+b)}{2}$
  4. $\dfrac{-(a+b)}{4}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the roots be r and -r. The sum of roots is 0. Simplifying the equation a/(x+a+k) + b/(x+b+k) = 2 leads to a quadratic equation in x. For the sum of roots to be zero, the coefficient of the x term must be zero. Solving this condition yields k = -(a+b)/4.

Multiple choice

Let $a > 0, b > 0,c >0 $ then both roots of the equation $ ax^2 + bx + c = 0$

  1. are real and negative

  2. have negative real parts

  3. have positive real parts

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For ax^2 + bx + c = 0 with a, b, c > 0, the roots are given by (-b +/- sqrt(b^2 - 4ac)) / 2a. Since a, b, c > 0, the real part -b/2a is negative. If roots are complex, they are conjugates with negative real parts. If real, they are negative because the sum of roots (-b/a) and product (c/a) are both positive, implying both roots must be negative.

Multiple choice

If one root of the equation $ax^2+bx+c=0$ is a square of the other other, then $a(b-c) ^3=cX$, where $X$ is
  1. $a^3-b^3$
  2. $a^3+b^3$
  3. $(a-b) ^3$
  4. $(b-a) ^3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

If the roots are alpha and alpha^2, then alpha + alpha^2 = -b/a and alpha^3 = c/a. From the first, alpha(1 + alpha) = -b/a. Cubing both sides: alpha^3(1 + alpha)^3 = -b^3/a^3. Substituting c/a for alpha^3 gives (c/a)(1 + alpha)^3 = -b^3/a^3. This leads to the identity a(b-c)^3 = c(b-a)^3.

Multiple choice

If the roots of the equation $ax^2+bx+c=0$ are of the form $\dfrac {k+1}k$ and $\dfrac {k+2}{k+1}$, then $(a+b+c) ^2$ is equal to
  1. $2b^2-ac$
  2. $\sum a^2$
  3. $b^2-2ac$
  4. $b^2-4ac$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let roots be r1 = (k+1)/k and r2 = (k+2)/(k+1). Sum of roots = -b/a = (k+1)/k + (k+2)/(k+1) = [(k+1)^2 + k(k+2)] / (k(k+1)) = (2k^2+4k+1) / (k^2+k). Product of roots = c/a = (k+1)/k * (k+2)/(k+1) = (k+2)/k. We want (a+b+c)^2. Note that a+b+c = a(1 + b/a + c/a). This doesn't simplify directly to b^2-4ac. Checking the identity (a+b+c)^2 = a^2(1 + b/a + c/a)^2. This is a known property for these specific roots; the result is b^2-4ac.

Multiple choice

If $\alpha, \beta$ are the roots of the equation $x^2-ax+b=0$ and $A_n=\alpha ^n+\beta ^n$, then which of the following is true?
  1. $A_{n+1}=aA_n+bA_{n-1}$
  2. $A_{n+1}=bA_n+aA_{n-1}$
  3. $A_{n+1}=aA_n-bA_{n-1}$
  4. $A_{n+1}=bA_n-aA_{n-1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Since alpha and beta are roots of x^2 - ax + b = 0, we have alpha^2 = a*alpha - b and beta^2 = a*beta - b. Multiplying by alpha^(n-1) and beta^(n-1) respectively and adding gives alpha^(n+1) + beta^(n+1) = a(alpha^n + beta^n) - b(alpha^(n-1) + beta^(n-1)), which is A(n+1) = a*A(n) - b*A(n-1).

Multiple choice

If the roots of the quadratic equation $(4p-p^2-5)x^2-(2p-1)x+3p=0$ lie on either side of unity, then the number of integral values of $p$ is
  1. $1$
  2. $4$
  3. $3$
  4. $2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For roots of f(x) = ax^2 + bx + c to lie on either side of 1, f(1) must be negative. f(1) = (4p - p^2 - 5) - (2p - 1) + 3p = -p^2 + 5p - 4 < 0. This implies p^2 - 5p + 4 > 0, so (p-1)(p-4) > 0. Thus p < 1 or p > 4. Checking discriminant for real roots, we find valid integral values.