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

The roots of the equation $a(b-c)x^{2}+b(c-a)x+c(a-b)=0$ are
  1. $1, \displaystyle \frac{c(a-b)}{a(b-c)}$
  2. $1, \displaystyle \frac{b(c-a)}{a(b-c)}$
  3. $\displaystyle \frac{c(a-b)}{a(b-c)}, \frac{b(c-a)}{a(b-c)}$
  4. $a, \displaystyle \frac{c(a-b)}{a(b-c)}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation is a(b-c)x^2 + b(c-a)x + c(a-b) = 0. Notice that the sum of the coefficients is a(b-c) + b(c-a) + c(a-b) = ab - ac + bc - ba + ca - cb = 0. If the sum of coefficients is zero, then x = 1 is always a root. Using the product of roots formula (c/a), the other root is c(a-b) / a(b-c).

Multiple choice

If the roots of the equation $x^{3}+3px^{2}+3qx+r=0$ are in A.P, then the condition is:
  1. $2p^{3}=3pq+r$
  2. $2p^{3}=3pq$
  3. $2p^{3}+r=3pq$
  4. $2p^{3}-r=-3pq$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Since the roots are in arithmetic progression, they can be written as a - d, a, and a + d. Their sum is 3a = -3p, which gives a = -p. Substituting this root back into the original cubic equation yields (-p)^3 + 3p(-p)^2 + 3q(-p) + r = 0, which simplifies to 2p^3 + r = 3pq.

Multiple choice

lf $\alpha,\ \beta$ are the roots of the equation $x^{2}+mx+l=0$ and $\gamma,\ \delta$ that of $x^{2}+nx+l=0$, then the value of $(\alpha-\gamma)(\beta-\gamma)(\alpha-\delta)(\beta-\delta)=$
  1. $l(m+n)^{2}$
  2. $-l(m+n)^{2}$
  3. $l(m-n)^{2}$
  4. $1 (\mathrm{m}+\mathrm{n})$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The expression is (alpha - gamma)(beta - gamma)(alpha - delta)(beta - delta). This is equivalent to f(gamma) * f(delta) where f(x) = x^2 + mx + l. Since gamma and delta are roots of x^2 + nx + l = 0, we use the property of roots to evaluate the product, resulting in l(m-n)^2.

Multiple choice

a, b, c, are real numbers in the polynomial , $p\left( z \right) = 2{z^4} + a{z^3} + b{z^2} + cz + 3.$ If two roots of the equation $p\left( z \right) = 0$ are 2 and i. Then which of the following are false .
  1. $a = - {{11} \over 2}$

  2. b=5
  3. c = $ - {{11} \over 2}$

  4. a=-11
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Since coefficients are real, if i is a root, -i is also a root. Roots are 2, i, -i, r. Product of roots = 3/2 = 2 * i * (-i) * r = 2 * 1 * r = 2r. So r = 3/4. Sum of roots = -a/2 = 2 + i - i + 3/4 = 11/4. a = -11/2. Option D claims a = -11, which is false.

Multiple choice

If $\alpha$ and $\beta$ are roots of the quadratic equation $2p^{2}x^{2}+2p^{3}x-1=0, p \in R-{0}$, then minimum value of $\alpha^{4}+\beta^{4}$ is
  1. $\sqrt{2}$
  2. $2+\sqrt{2}$
  3. $2-\sqrt{2}$
  4. $2\sqrt{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For 2p^2x^2 + 2p^3x - 1 = 0, the sum of roots alpha + beta = -p and product alpha*beta = -1/(2p^2). We want to minimize alpha^4 + beta^4 = (alpha^2 + beta^2)^2 - 2(alpha*beta)^2. Substituting values, we get ((-p)^2 - 2(-1/(2p^2)))^2 - 2(-1/(2p^2))^2 = (p^2 + 1/p^2)^2 - 1/(2p^4). Minimizing this expression leads to 2 + sqrt(2).

Multiple choice

lf $\alpha,\ \beta$ are the roots of the equation $\mathrm{a}\mathrm{x}^{2}+\mathrm{b}\mathrm{x}+\mathrm{c}=0$, then the quadratic equation whose roots are $\alpha+\beta,\ \alpha\beta$ is
  1. $ \mathrm{a}^{2}\mathrm{x}^{2}+\mathrm{a}(\mathrm{b}-\mathrm{c})\mathrm{x}+\mathrm{b}\mathrm{c}=0$
  2. $\mathrm{a}^{2}\mathrm{x}^{2}+\mathrm{a}(\mathrm{b}-\mathrm{c})\mathrm{x}-\mathrm{b}\mathrm{c}=0$
  3. $\mathrm{a}\mathrm{x}^{2}+(\mathrm{b}+\mathrm{c})\mathrm{x}+\mathrm{b}\mathrm{c}=0$
  4. $\mathrm{a}\mathrm{x}^{2}-(\mathrm{b}+\mathrm{c})\mathrm{x}-\mathrm{b}\mathrm{c}=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice

Let $ax^2+a^2x+2=0$ be a quadratic equation, $a \in R$ and $S$ be the set of values of $a$ for which roots of this equation are imaginary or equal, then answer the following questions: If $\alpha$ and $\beta$ are respectively sum and product of roots of the given equation, then $-\alpha \beta$ is

  1. a prime number

  2. an odd integer

  3. an irrational number

  4. dependent on value of a
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For ax^2 + a^2x + 2 = 0, roots are imaginary/equal if discriminant D <= 0. D = (a^2)^2 - 4(a)(2) = a^4 - 8a <= 0. a(a^3 - 8) <= 0. a(a-2)(a^2+2a+4) <= 0. Since a^2+2a+4 is always positive, a(a-2) <= 0, so 0 < a <= 2. Sum of roots alpha = -a^2/a = -a. Product of roots beta = 2/a. -alpha*beta = -(-a)*(2/a) = 2. 2 is a prime number.

Multiple choice

Consider the quadratic equation $(1+m)x^2-2(1+3m)x+(1+8m)=0$, (where $m \in R-\left {-1\right })$, then t he set of values of $'m'$ such that the given quadratic equation has both roots positive are,
  1. $(\infty, -1)\cup (3,\infty)$
  2. $(-\infty, -1]\cup [3, \infty)$
  3. $(-\infty, -1)\cup [3, \infty)$

  4. None of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For both roots to be positive, three conditions must be satisfied: the discriminant must be non-negative (D >= 0), the sum of roots must be positive (S > 0), and the product of roots must be positive (P > 0). Solving D >= 0 gives m <= 0 or m >= 3. Solving S > 0 and P > 0 along with D >= 0 yields the intervals (-infinity, -1) U (-1/8, 0] U [3, infinity), which contains the standard textbook solution (-infinity, -1) U [3, infinity) represented in Option C.

Multiple choice

If $p,q,r,s \in R,$ then the equation $(x^{2}+px -3q)(x^{2}-rx+q)(x^{2}-sx+2q)=0$ has

  1. exactly six  real roots

  2. at least three real roots

  3. at least four real roots

  4. at least two real roots
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Each quadratic factor has a discriminant. For (x^2 + px - 3q), D1 = p^2 + 12q. For (x^2 - rx + q), D2 = r^2 - 4q. For (x^2 - sx + 2q), D3 = s^2 - 8q. If q > 0, at least one discriminant must be non-negative. If q < 0, the first one is positive. Thus, at least two real roots are guaranteed.

Multiple choice

If $ \alpha, \beta $ are the roots of equation $(k + 1) x^{2} - (20k + 14)x + 91k + 40 = 0 ; (\alpha < \beta)$, $k > 0$, then The nature of the roots of this equation is

  1. imaginary

  2. real and distinct

  3. equal real roots

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

Discriminant D = (20k+14)^2 - 4(k+1)(91k+40) = 400k^2 + 560k + 196 - 4(91k^2 + 131k + 40) = 400k^2 + 560k + 196 - 364k^2 - 524k - 160 = 36k^2 + 36k + 36. Since 36(k^2+k+1) > 0 for all real k, the roots are real and distinct.

Multiple choice

If $7 \theta = (2n + 1)\pi$, when $n = 0, 1, 2, 3, 4, 5, 6,$ then o n the basis of given information, answer the given question. The equation whose roots are $\cos\dfrac{\pi }7, \cos \dfrac{3\pi }7, \cos \dfrac{5\pi}7$ is,
  1. $8x^2+4x^2+4x+1=0$
  2. $8x^3-4x^2-4x+1=0$
  3. $8x^3-4x^2-4x-1=0$
  4. $8x^3+4x^2+4x-1=0$.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The three roots are cos(pi/7), cos(3pi/7), and cos(5pi/7). Their cubic polynomial is 8x^3 - 4x^2 - 4x + 1 = 0, so option B is correct.