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 ratio of roots of the equation $x^2+px+q=0$ is equal to the ratio of roots of the equation $x^2+bx+c=0$, then $p^2c-b^2q=$
  1. $0$
  2. $2$
  3. $\dfrac {-3}{2}$
  4. $-2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the roots of x^2+px+q=0 be r1, r2 and x^2+bx+c=0 be r3, r4. Given r1/r2 = r3/r4. Using properties of roots, (r1+r2)^2 / (r1*r2) = p^2/q and (r3+r4)^2 / (r3*r4) = b^2/c. Since (r1+r2)^2 / (r1*r2) = (r1/r2 + r2/r1 + 2) and (r3/r4 + r4/r3 + 2) are equal, p^2/q = b^2/c, which implies p^2c = b^2q, so p^2c - b^2q = 0.

Multiple choice

For what values of $m\in R$, both roots of the equation $x^2-6mx+9m^2-2m+2=0$ exceed $3$?
  1. $[-7, -2]$
  2. $\left(\dfrac {11}{9}, \infty\right)$
  3. $[1, \infty]$
  4. $[0, \infty]$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For both roots to exceed 3, the vertex must be greater than 3, the discriminant must be non-negative, and the function value at x=3 must be positive. Solving -b/(2a) > 3 gives 3m > 3, so m > 1. The discriminant 36m^2 - 4(9m^2 - 2m + 2) >= 0 simplifies to 8m - 8 >= 0, or m >= 1. Finally, f(3) = 9 - 18m + 9m^2 - 2m + 2 = 9m^2 - 20m + 11 > 0, which factors to (9m - 11)(m - 1) > 0, implying m > 11/9 or m < 1.

Multiple choice

If $ \alpha , \beta $ are the roots of the equation $ x ^ { 2 } - 15 x + 1 = 0 $, then the value is $ \left( \frac { 1 } { \alpha } - 15 \right) ^ { - 2 } + \left( \frac { 1 } { \beta } - 15 \right) ^ { - 2 } $ is

  1. 225

  2. 900

  3. 223

  4. 0
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given x^2 - 15x + 1 = 0, roots alpha and beta satisfy alpha + beta = 15 and alpha*beta = 1. The expression is (1/alpha - 15)^-2 + (1/beta - 15)^-2. Since 1/alpha - 15 = (1 - 15alpha)/alpha = -alpha^2/alpha = -alpha, the expression simplifies to (-alpha)^-2 + (-beta)^-2 = 1/alpha^2 + 1/beta^2 = (alpha^2 + beta^2)/(alpha*beta)^2. With alpha^2 + beta^2 = (alpha+beta)^2 - 2alpha*beta = 225 - 2 = 223, the result is 223/1 = 223.

Multiple choice

Consider the quadratic equation $(1+m)x^2-2(1+3m)x+(1+8m)=0$, (where $m \in R-\left {-1\right })$, t hen the number of real values of $m$ such that the given quadratic equation has roots in the ratio $2: 3$ are,
  1. $0$
  2. $2$
  3. $3$

  4. Infinitely many
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let roots be 2k and 3k. Sum = 5k = 2(1+3m)/(1+m). Product = 6k^2 = (1+8m)/(1+m). From sum, k = 2(1+3m)/(5(1+m)). Substitute into product: 6 * [4(1+3m)^2 / 25(1+m)^2] = (1+8m)/(1+m). This simplifies to a quadratic in m, yielding two values.

Multiple choice

IF $3{p}^{2}=5p+2$ and $3{q}^{2}=5q+2$ where $p\ne q$, then the equation whose roots are $3p-2q$ and $3q-2p$ is
  1. $3{x}^{2}-5x-100=0$
  2. $5{X}^{2}+3x+100=0$
  3. $3{x}^{2}-5x+100=0$
  4. $5{x}^{2}-3x-100=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given 3p^2 - 5p - 2 = 0 and 3q^2 - 5q - 2 = 0, p and q are roots of 3x^2 - 5x - 2 = 0. Thus p+q = 5/3 and pq = -2/3. We need an equation with roots alpha = 3p-2q and beta = 3q-2p. Sum = alpha + beta = p+q = 5/3. Product = (3p-2q)(3q-2p) = 9pq - 6p^2 - 6q^2 + 4pq = 13pq - 6(p+q)^2 + 12pq = 25pq - 6(p+q)^2 = 25(-2/3) - 6(25/9) = -50/3 - 50/3 = -100/3. Equation: x^2 - (5/3)x - 100/3 = 0, or 3x^2 - 5x - 100 = 0.

Multiple choice

$\alpha ,\beta $ are roots of the equation $2{x^2} - 5x - 6 = 0$ then
  1. the equation whose roots are $\alpha ,\beta $ and 2 is $2{x^3} - 9{x^2} + 4x + 12 = 0$
  2. The value of $\left( {\frac{\alpha }{\beta } + \frac{\beta }{\alpha } + 1} \right)$ is $\frac{{ - 37}}{{12}}$
  3. the equation whose roots are $\alpha ,\beta $ and -1 is $2{x^3} - 3{x^2} - 11x - 6 = 0$
  4. The value of ${(\alpha - \beta )^2}$ is $\frac{{53}}{4}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Roots of 2x^2 - 5x - 6 = 0 are alpha and beta. Sum = 5/2, Product = -3. If roots are alpha, beta, and -1, the cubic equation is (x + 1)(2x^2 - 5x - 6) = 2x^3 - 5x^2 - 6x + 2x^2 - 5x - 6 = 2x^3 - 3x^2 - 11x - 6 = 0.

Multiple choice

If $\alpha $, $\beta $ are roots of the equation $6{x^2} + 11x + 3 = 0$, then
  1. ${\text{both}}\;{\cos ^{ - 1}}\alpha \;{\text{and}}\;{\cos ^{ - 1}}\beta \;{\text{are}}\;{\text{real}}$
  2. ${\text{both}}\;\cos {\text{e}}{{\text{c}}^{ - 1}}\alpha \;{\text{and}}\;{\cos ^{ - 1}}\beta \;{\text{are}}\;{\text{real}}$
  3. ${\text{both}}\;\cos {\text{e}}{{\text{c}}^{ - 1}}\alpha \;{\text{and}}\;{\cot ^{ - 1}}\beta \;{\text{are}}\;{\text{real}}$
  4. Both $B\ and\ C$
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice

If p and q are positive then the roots of the equation $x^2-px-q=0$ are-

  1. imaginary

  2. real & both positive

  3. real & both negative

  4. real & of opposite sign
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For x^2 - px - q = 0, the discriminant D = (-p)^2 - 4(1)(-q) = p^2 + 4q. Since p and q are positive, D > 0, so roots are real. The product of roots is -q. Since q > 0, the product is negative, meaning the roots have opposite signs.

Multiple choice

Equation $x^2-17x+66=0$ has real roots.

  1. True

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

The discriminant of x^2 - 17x + 66 = 0 is D = b^2 - 4ac = (-17)^2 - 4(1)(66) = 289 - 264 = 25. Since D > 0, the roots are real.