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

Statement-I : If $a + b + c > 0$ and $a < 0 < b < c$, then the roots of the equation $a(x - b) (x- c) + b(x- c)(x- a) + c(x- a) (x- b) = 0$ are of both negative. Statement-II : If both roots are negative, then sum of roots $< 0$ and product of roots $> 0$.

  1. Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I

  2. Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for statement-I

  3. Statement-I is true, Statement-II is false

  4. Statement-I is false, Statement-II is true
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The equation is a cyclic expression. If x=a, x=b, or x=c, the expression simplifies to a(a-b)(a-c) + b(b-c)(b-a) + c(c-a)(c-b) = 0. This is a known identity for quadratic equations where the roots are a, b, and c, but since it is quadratic, the roots are actually a, b, and c. However, the statement claims roots are negative; given a < 0 < b < c, the roots are a (negative), b (positive), and c (positive). Thus, Statement-I is false. Statement-II is a standard property of quadratic equations.

Multiple choice

If $p, q$ are the roots of equation $x^2 + px + q = 0$, then the value of $p$ must be equal to
  1. $0,1$
  2. $1$
  3. $2$
  4. $1, -2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the equation x^2 + px + q = 0, the sum of roots is -p and the product of roots is q. Since the roots are p and q, we have p + q = -p (so 2p + q = 0) and pq = q. From pq = q, either q = 0 or p = 1. If q = 0, then 2p = 0, so p = 0. If p = 1, then 2 + q = 0, so q = -2. The values for p are 0 and 1.

Multiple choice

If $\alpha$, $\beta$ are roots of the equation $ax^2 - bx - c = 0$, then $ \alpha^2 - \alpha \beta + \beta^2$ is equal to
  1. $\displaystyle \frac{b^2 + 3ac}{a^2}$
  2. $\displaystyle \frac{b^2 - 3ac}{a^2}$
  3. $\displaystyle \frac{b^2 + 2ac}{a^2}$
  4. $\displaystyle \frac{b^2 - 2ac}{a^2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

alpha + beta = b/a, alpha * beta = -c/a. alpha^2 - alpha*beta + beta^2 = (alpha + beta)^2 - 3*alpha*beta = (b/a)^2 - 3(-c/a) = b^2/a^2 + 3c/a = (b^2 + 3ac) / a^2.

Multiple choice

If $\alpha, \beta$ are the roots of equation $4x^2 - \sqrt{13}x - 7 = 0$, then the value of $\alpha - \beta$ is
  1. $\displaystyle \frac{5 \sqrt{5} }{4}$
  2. $\displaystyle \frac{5}{6}$
  3. $\displaystyle \frac{\sqrt{5}}{6}$
  4. $2 \sqrt{5}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For 4x^2 - sqrt(13)x - 7 = 0, sum of roots alpha + beta = sqrt(13)/4, product alpha * beta = -7/4. (alpha - beta)^2 = (alpha + beta)^2 - 4*alpha*beta = (13/16) - 4(-7/4) = 13/16 + 7 = 125/16. alpha - beta = sqrt(125)/4 = 5*sqrt(5)/4.

Multiple choice

If $\alpha$ and $\beta$ are the roots of the equation $3x^2-6x+1=0$, form an equation whose roots are (i) $\dfrac{1}{\alpha }, \dfrac{1}{\beta }$
  1. $x^2-6x+3=0$
  2. $x^2+6x+3=0$
  3. $x^2-6x-3=0$

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

For the equation 3x^2 - 6x + 1 = 0, the sum of roots alpha + beta = 6/3 = 2 and product alpha*beta = 1/3. The new roots are 1/alpha and 1/beta. Their sum is (alpha + beta) / (alpha*beta) = 2 / (1/3) = 6. Their product is 1 / (alpha*beta) = 1 / (1/3) = 3. The new equation is x^2 - (sum)x + (product) = 0, which is x^2 - 6x + 3 = 0.

Multiple choice

If $\alpha$ and $\beta$ are the roots of the equation $x^2-3x-1=0$, then form a quadratic equation whose roots are$\dfrac{1}{\alpha^2} $ and $\dfrac{1}{\beta^2}$.
  1. $x^2-11x+1=0$
  2. $x^2+11x+1=0$
  3. $x^2-11x-1=0$

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

For x^2-3x-1=0, sum of roots alpha+beta = 3 and product alpha*beta = -1. The new roots are 1/alpha^2 and 1/beta^2. Their sum is (alpha^2+beta^2)/(alpha*beta)^2 = ((alpha+beta)^2 - 2alpha*beta)/(alpha*beta)^2 = (9+2)/1 = 11. Their product is 1/(alpha*beta)^2 = 1/1 = 1. The equation is x^2 - 11x + 1 = 0.

Multiple choice

If $\alpha$ and $\beta$ are the roots of the equation $3x^2-5x+2=0$, then find the values of $\dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha }$
  1. $\dfrac{13}{6}$
  2. $\dfrac{-13}{6}$
  3. $\dfrac{12}{6}$

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

For 3x^2 - 5x + 2 = 0, alpha + beta = 5/3 and alpha * beta = 2/3. alpha/beta + beta/alpha = (alpha^2 + beta^2) / (alpha * beta) = ((alpha+beta)^2 - 2*alpha*beta) / (alpha*beta). = ((25/9) - 4/3) / (2/3) = (13/9) / (2/3) = 13/6.

Multiple choice

If $\alpha$ and $\beta$ are the roots of the equation $2x^2-3x-5=0$, then form a quadratic equation whose roots are $\alpha^2 $ and $\beta^2$.
  1. $4x^2-29x+25=0$
  2. $4x^2+29+25=0$
  3. $4x^2-29-25=0$

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

For 2x^2 - 3x - 5 = 0, alpha + beta = 3/2 and alpha * beta = -5/2. The new roots are alpha^2 and beta^2. Their sum is (alpha + beta)^2 - 2(alpha * beta) = (3/2)^2 - 2(-5/2) = 9/4 + 5 = 29/4. Their product is (alpha * beta)^2 = 25/4. The equation is x^2 - (29/4)x + 25/4 = 0, or 4x^2 - 29x + 25 = 0.