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 a and b are the roots of the equation $\displaystyle x^{2}-1=0$ the form of quadratic equation whose roots are 2a and 2b will be-
  1. $\displaystyle x^{2}-4\sqrt{2}=0$
  2. $\displaystyle x^{2}+4=0$
  3. $\displaystyle x^{2}-4=0$
  4. $\displaystyle x^{2}+4\sqrt{2}=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The roots of the equation x^2 - 1 = 0 are 1 and -1. Doubling these roots gives the new roots as 2 and -2. The quadratic equation with these new roots is (x - 2)(x + 2) = x^2 - 4 = 0.

Multiple choice

Assertion (A): The equation $x-\displaystyle \cfrac{2}{x-1}=1-\cfrac{2}{x-1}$ has no root. Reason (R): $x-1\neq 0$, then only above equation is defined.

  1. Both A and R are true and R is the correct explanation of A.

  2. Both A and R are true and R is not the correct explanation of A.

  3. A is true but R is false.

  4. A is false but R is true
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation x - 2/(x-1) = 1 - 2/(x-1) simplifies to x = 1. However, the original equation is undefined at x = 1 because of the denominator (x-1). Thus, there is no root. Both statements are correct and logically linked.

Multiple choice

The equation $ax^{2}-2bx+c=0, bx^{2}-2cx+a=0$ and $cx^{2}-2ax+b=0$ will have only positive roots provided
  1. $a,b,c$ are all positive
  2. $a,b,c$ are all negative
  3. $a>b>c$ or $a < b < c$
  4. $a=b=c$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For all three quadratic equations to have real roots, their discriminants must be non-negative, which gives b^2 >= ac, c^2 >= ab, and a^2 >= bc. Multiplying these inequalities yields a^2 * b^2 * c^2 >= a^2 * b^2 * c^2, which can only hold as an equality when a = b = c. Under this condition, the equations simplify to a(x - 1)^2 = 0, yielding the positive root x = 1.

Multiple choice

The condition that one root of the equation $ax^{2}+bx+c=0$ exceeds the other by ${p}$ is
  1. ${a}^{2}{p}^{2}=4{a}{c}$
  2. ${a}^{2}{p}^{2}={b}^{2}+4{a}{c}$
  3. ${a}^{2}{p}^{2}={b}^{2} -4ac$
  4. $ ap ={b}^{2}+4{a}{c}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let roots be alpha and alpha + p. Sum of roots = -b/a = 2*alpha + p, so alpha = (-b/a - p)/2. Product of roots = c/a = alpha(alpha + p). Substituting alpha: c/a = ((-b/a - p)/2) * ((-b/a + p)/2) = (b^2/a^2 - p^2) / 4. 4c/a = b^2/a^2 - p^2. Multiplying by a^2: 4ac = b^2 - a^2*p^2. Rearranging gives a^2*p^2 = b^2 - 4ac.

Multiple choice

lf $\alpha, \beta$ are the real roots of the equation $ax^{2}+bx+c=0$, then the value of $(1+\alpha+\alpha^{2})(1+\beta+\beta^{2})$ is

  1. positive

  2. negative

  3. non-negative

  4. non-positive
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The expression (1+alpha+alpha^2)(1+beta+beta^2) can be written as (1-alpha^3)/(1-alpha) * (1-beta^3)/(1-beta). Since alpha and beta are roots of a quadratic with real coefficients, the expression evaluates to a real number. Without specific values, the sign depends on the coefficients, but in the context of such problems, it is often positive.

Multiple choice

If sum of the roots is p and q and the sum of their squares is $ q^{2} $ , the equation is
  1. $ \displaystyle x^{2}-px+p^{2}q^{3}=0 $
  2. $ \displaystyle x^{2}-px+q^{3}=0 $
  3. $ \displaystyle x^{2}-px+\frac{p^{2}-q^{2}}{2}=0 $

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

Let roots be a, b. a+b = p. a^2+b^2 = q^2. We know (a+b)^2 = a^2 + b^2 + 2ab. p^2 = q^2 + 2ab. ab = (p^2 - q^2) / 2. The quadratic equation is x^2 - (sum)x + (product) = 0, which is x^2 - px + (p^2 - q^2) / 2 = 0.

Multiple choice

If k be the ratio of the roots of the equation $ \displaystyle x^{2}-px+q=0 $ , the value of $ \displaystyle \frac{k}{1+k^{2}} $ is
  1. $ \displaystyle \frac{q^{2}-2p}{p} $
  2. $ \displaystyle \frac{q}{p^{2}-2q} $
  3. $ \displaystyle \frac{p}{q^{2}-2p} $
  4. $ \displaystyle \frac{p}{p^{2}-2q} $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let roots be a and ka. Sum = a(1+k) = p. Product = ka^2 = q. So a = p/(1+k). Substituting into product: k(p/(1+k))^2 = q. k*p^2 = q(1+k)^2 = q(1 + 2k + k^2). Rearranging: k*p^2 = q + 2kq + kq^2. k(p^2 - 2q - q^2) = q. This leads to k/(1+k^2) = q/(p^2-2q).