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 quadratic equation $x^2+(p-3)x-2p=3$ $(p\in R)$, then the minimum value of $(\alpha^2+\beta^2+\alpha\beta)$, is
  1. $2$
  2. $4$
  3. $8$
  4. $12$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Roots alpha, beta satisfy alpha+beta = -(p-3) and alpha*beta = -2p-3. We want to minimize alpha^2 + beta^2 + alpha*beta = (alpha+beta)^2 - alpha*beta. This is (p-3)^2 - (-2p-3) = p^2 - 6p + 9 + 2p + 3 = p^2 - 4p + 12. The minimum occurs at p = -(-4)/2 = 2. Value = 4 - 8 + 12 = 8.

Multiple choice

Find a quadratic equation whose product of roots $x_1$ and $x_2$ is equal to $4$ and satisfying the relation $\displaystyle \frac{x_1}{x_1 - 1} + \frac{x_2}{x_2 - 1} = 2$.
  1. $x^2 - 2x + 4 = 0$
  2. $x^2 - 3x + 4 = 0$
  3. $x^2 - 2x + 3 = 0$
  4. $x^2 - 3x + 2 = 0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given x1*x2 = 4. The equation is x1/(x1-1) + x2/(x2-1) = 2. Simplifying: (x1(x2-1) + x2(x1-1)) / ((x1-1)(x2-1)) = 2. This leads to 2x1x2 - (x1+x2) / (x1x2 - (x1+x2) + 1) = 2. Substituting x1x2=4: (8 - (x1+x2)) / (5 - (x1+x2)) = 2. 8 - (x1+x2) = 10 - 2(x1+x2). So x1+x2 = 2. The quadratic is x^2 - 2x + 4 = 0.

Multiple choice

Find a quadratic equation whose roots $x_1$ and $x_2$ satisfy the condition $x_1^2 + x_2^2 = 5, 3 (x_1^5+x_2^5) = 11 (x_1^3 + x_2^3)$. (Assume that $x_1, x_2$ are real)
  1. $x^2 \pm 3x + 2 = 0$
  2. $x^2 \pm 6x + 12 = 0$
  3. $x^2 \pm 9x + 12 = 0$

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

Given x1^2 + x2^2 = 5. Let s = x1+x2, p = x1*x2. s^2 - 2p = 5. Using the second condition 3(x1^5+x2^5) = 11(x1^3+x2^3), we can solve for s and p. The resulting quadratic is x^2 - sx + p = 0. Testing option A: x^2 - 3x + 2 = 0, roots 1, 2. 1^2+2^2 = 5. 3(1+32) = 99, 11(1+8) = 99. Matches.

Multiple choice

If $\alpha, \beta$ are the roots of the equation $ax^2 + bx + c = 0$, then the value of $\dfrac{(a \alpha^2 + c)}{(a \alpha + b)} + \dfrac{(a \beta^2 + c)}{(a \beta + b)}$ is
  1. $\displaystyle \frac{b (b^2 - 2ac)}{4a}$
  2. $\displaystyle \frac{b^2 - 4ac}{2a}$
  3. $\displaystyle \frac{b (b^2 - 2ac)}{a^2 c}$

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

Given ax^2+bx+c=0, roots alpha, beta. a*alpha^2+c = -b*alpha. The expression becomes (-b*alpha)/(a*alpha+b) + (-b*beta)/(a*beta+b). This simplifies to -b * (alpha/(a*alpha+b) + beta/(a*beta+b)). Using alpha+beta = -b/a and alpha*beta = c/a, the calculation leads to b(b^2-2ac)/(a^2*c).

Multiple choice

Let $k$ be a real number such that $k\neq 0$. If $\alpha$ and $\beta$ are non zero complex numbers satisfying $\alpha +\beta=-2k$ and $\alpha^2+\beta^2=4k^2-2k$, then a quadratic equation having $\dfrac{\alpha+\beta}{\alpha}$ and $\dfrac{\alpha+\beta}{\beta}$ as its roots is equal to
  1. $4x^2-4kx+k=0$
  2. $x^2-4kx+4k=0$
  3. $4kx^2-4x+k=0$
  4. $4kx^2-4kx+1=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Roots are (alpha+beta)/alpha and (alpha+beta)/beta. Sum = (alpha+beta)(1/alpha + 1/beta) = (alpha+beta)(alpha+beta)/(alpha*beta) = (-2k)^2 / (alpha*beta). From alpha^2+beta^2 = 4k^2-2k, (alpha+beta)^2 - 2alpha*beta = 4k^2-2k, so 4k^2 - 2alpha*beta = 4k^2-2k, alpha*beta = k. Sum = 4k^2/k = 4k. Product = (alpha+beta)^2 / (alpha*beta) = 4k^2/k = 4k. Equation: x^2 - (sum)x + product = x^2 - 4kx + 4k = 0.

Multiple choice

If a and b are the roots of a quadratic equation such that ${a^2}b + {b^2}a = 84,$ also a and b are natural numbers, then equation can be
  1. ${x^2} + 21x - 4 = 0$
  2. ${x^2} - 21x + 4 = 0$
  3. ${x^2} - 7x + 12 = 0$
  4. ${x^2} - 12x + 7 = 0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

a^2b + b^2a = ab(a+b) = 84. Since a, b are natural numbers, we test factors of 84. If a=3, b=4: 3*4*(3+4) = 12*7 = 84. The quadratic equation with roots 3 and 4 is (x-3)(x-4) = x^2 - 7x + 12 = 0.

Multiple choice

If $\alpha$ and $\beta$ are the roots of $x^2+px+q=0$ and $\alpha^4, \beta^4$ are the roots of $x^2-rx+5=0$, then the equation $x^2-4qx+2q^2-r=0$ has always

  1. one positive and one negative root

  2. two positive roots

  3. two negative roots

  4. cannot say anything
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given the roots of x^2 + px + q = 0 are alpha and beta, alpha^4 and beta^4 are roots of x^2 - rx + 5 = 0. Using symmetric properties, the product of roots alpha^4 * beta^4 = q^4 = 5. The equation x^2 - 4qx + 2q^2 - r = 0 has a discriminant D = 16q^2 - 4(2q^2 - r) = 8q^2 + 4r. Since q^4 = 5, q^2 = sqrt(5). The constant term is 2q^2 - r. Analysis shows the product of roots is negative, implying one positive and one negative root.

Multiple choice

All the values of $\mathrm { m }$ for which both roots of the equation $\mathrm { x } ^ { 2 } - 2 \mathrm { m } \mathrm { x } + \mathrm { m } ^ { 2 } - 1 = 0$ are greater than $- 2$ but less than $4 ,$ lie in the interval-
  1. $- 1 < \mathrm { m } < 3$
  2. $1 < m < 4$
  3. $- 2 < \mathrm { m } < 0$
  4. $m > 3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The roots of x^2 - 2mx + m^2 - 1 = 0 are (m-1) and (m+1). Given -2 < m-1 < 4 and -2 < m+1 < 4, we solve the inequalities: -1 < m < 5 and -3 < m < 3. The intersection is -1 < m < 3.

Multiple choice

Let $\displaystyle \alpha_1$ and $\displaystyle \alpha_2$ be the roots of the equation $\displaystyle x^2 - 4x + P_1 = 0$, and $\displaystyle \alpha_3$ and $\displaystyle \alpha_4$ be the roots of the equation $\displaystyle x^2 - 36x + P_2 = 0$. If $\displaystyle \alpha_1 < \alpha_2 < \alpha_3 < \alpha_4$ and $\displaystyle \alpha_1, \alpha_2, \alpha_3, \alpha_4$ are in G.P., then the product $\displaystyle P_1P_2$ equals

  1. 81

  2. 243

  3. 729

  4. 27
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice

If $\alpha,\beta$ are the roots of $\displaystyle x^{2}-px+q=0$, then the product of the roots of the quadratic equation whose roots are $\displaystyle \alpha^{2}-\beta^{2}$ and $\alpha^{3}-\beta^{3}$ is
  1. $p(p^{2}-q)^{2}$
  2. $p(p^{2}-q)(p^{2}-4q)$
  3. $p(p^{2}-4q)(p^{2}+q)$

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

Roots are alpha, beta. alpha+beta = p, alpha*beta = q. New roots: alpha^2-beta^2 = (alpha-beta)(alpha+beta) and alpha^3-beta^3 = (alpha-beta)(alpha^2+alpha*beta+beta^2). Product = (alpha-beta)^2 * (alpha+beta) * (alpha^2+alpha*beta+beta^2). (alpha-beta)^2 = (alpha+beta)^2 - 4alpha*beta = p^2-4q. alpha^2+alpha*beta+beta^2 = (alpha+beta)^2 - alpha*beta = p^2-q. Product = (p^2-4q) * p * (p^2-q).

Multiple choice

The product of the real roots of the equation ${ \left| 2x+3 \right| }^{ 2 }-3\left| 2x+3 \right| +2=0$ is
  1. $\displaystyle\frac{5}{4}$
  2. $\displaystyle\frac{5}{2}$
  3. $5$
  4. $2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let u = |2x+3|. u^2 - 3u + 2 = 0 => (u-1)(u-2) = 0. So |2x+3| = 1 or |2x+3| = 2. Case 1: 2x+3 = 1 => x = -1; 2x+3 = -1 => x = -2. Case 2: 2x+3 = 2 => x = -0.5; 2x+3 = -2 => x = -2.5. Roots are -1, -2, -0.5, -2.5. Product = (-1)(-2)(-0.5)*(-2.5) = 2 * 1.25 = 2.5 = 5/2.

Multiple choice

Let $\displaystyle \alpha ,\beta $ be the roots of the equation $\displaystyle x^{2}-px+r=0$ and $\displaystyle {\alpha }/{2},2\beta$ be the roots of the equation $\displaystyle x^{2}-qx+r=0.$ Then the value of $r$ is
  1. $\displaystyle {2}/{9}\left ( p-q \right )\left ( 2q-p \right )$
  2. $\displaystyle {2}/{9}\left ( q-p \right )\left ( 2p-q \right )$
  3. $\displaystyle {2}/{9}\left ( q-2p \right )\left ( 2q-p \right )$
  4. $\displaystyle {2}/{9}\left ( 2p-q \right )\left ( 2q-p \right )$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For x^2 - px + r = 0, alpha + beta = p, alpha * beta = r. For x^2 - qx + r = 0, alpha/2 + 2beta = q, (alpha/2) * 2beta = r. So alpha * beta = r. From alpha/2 + 2beta = q, substitute beta = r/alpha: alpha/2 + 2r/alpha = q. alpha^2 - 2q*alpha + 4r = 0. Using roots of first eq, alpha^2 - p*alpha + r = 0. Subtracting: (p-2q)alpha = 3r. alpha = 3r/(p-2q). Substitute back to get r.