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 quadratic equation $\displaystyle a x^{2}+b x +c=0$ will always have imaginary root in which of the following cases?
  1. $a < -1 , \ \ 0 < c < 1, \ \ b > 0$
  2. $a < -1, \ \ 1 < c < 0, \ \ 0 < b <1$
  3. $a < -1 , \ \ c < 0, \ \ b > 1$
  4. $a < -1, \ \ c < - 1, \ \ 1 < b < 2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A quadratic has imaginary roots when its discriminant is negative, so b^2 - 4ac < 0. In option D, ac > 1 and b^2 < 4, making 4ac > b^2 and the discriminant negative.

Multiple choice

If the equation $\displaystyle x^{2}-2kx-2x+k^{2}=0 $ has equal roots the value of k must be

  1. zero
  2. either zero or $\displaystyle -\frac{1}{2} $
  3. $\displaystyle -\frac{1}{2} $
  4. either $\displaystyle \frac{1}{2} $ or $\displaystyle -\frac{1}{2} $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For x^2 - 2(k+1)x + k^2 = 0 to have equal roots, the discriminant must be 0. D = (2(k+1))^2 - 4(1)(k^2) = 4(k^2 + 2k + 1) - 4k^2 = 8k + 4. Setting 8k + 4 = 0 gives k = -1/2.

Multiple choice

lf ${a},b,c,{d}$ are real and no two of them are simultaneously zero, then the equation $(x^{2}+ax -3b)(x^{2}-cx +b)(x^{2}-dx +2b)=0$ has

  1. at least two real roots

  2. at least four real roots

  3. all roots real

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

If b >= 0, the first quadratic has discriminant a^2 + 12b, which is nonnegative. If b < 0, the second and third quadratics have positive discriminants, so in every case there are at least two real roots.

Multiple choice

If $tan\theta\;and\;sec\theta$ are the roots of the quadratic equation $ax^2+bx+c=0$ then
  1. $a^4=b^2(b^2-4ac)$
  2. $b^4=a^2(b^2+4ac)$
  3. $a^4=b^2(b^2+4ac)$
  4. $b^4=a^2(b^2-4ac)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Roots are tan(t) and sec(t). Sum = tan(t) + sec(t) = -b/a. Product = tan(t)sec(t) = c/a. Use identity sec^2(t) - tan^2(t) = 1. (sec(t) - tan(t))(sec(t) + tan(t)) = 1. So (sec(t) - tan(t)) = 1/(-b/a) = -a/b. Now, 2sec(t) = -b/a - a/b = -(a^2+b^2)/ab. 2tan(t) = -b/a + a/b = (a^2-b^2)/ab. Square both: 4sec^2(t) = (a^2+b^2)^2/a^2b^2, 4tan^2(t) = (a^2-b^2)^2/a^2b^2. Subtracting: 4(sec^2-tan^2) = 4 = [(a^2+b^2)^2 - (a^2-b^2)^2]/a^2b^2 = 4a^2b^2/a^2b^2 = 4. This leads to a^4 = b^2(b^2-4ac) via further manipulation of the product c/a.

Multiple choice

lf the roots of the equation $x^{2}+2ax+b=0$ are real and distinct and they differ by at most $2m$, then $b$ lies in the interval
  1. $(\mathrm{a}^{2}-\mathrm{m}^{2}, \mathrm{a}^{2})$
  2. $[\mathrm{a}^{2}-\mathrm{m}^{2}, \mathrm{a}^{2})$
  3. $( \mathrm{a}^{2} -\mathrm{m}^{2}, \infty)$
  4. $[\mathrm{a}^{2}-\mathrm{m}^{2},\infty ]$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Roots are x = -a +/- sqrt(a^2 - b). Difference = 2 * sqrt(a^2 - b). Given 2 * sqrt(a^2 - b) <= 2m, so sqrt(a^2 - b) <= m. Squaring gives a^2 - b <= m^2, so b >= a^2 - m^2. Also, for real distinct roots, a^2 - b > 0, so b < a^2.

Multiple choice

The set of values of $a$ for which one root of the equation $x^{2}+(2a+1)x+(a-1)=0$ is greater than $1$ and the other root is less than zero is
  1. $a < -\displaystyle \frac{1}{3}$
  2. $0 < a <1$
  3. $a > 1$
  4. $a \geq 1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let f(x) = x^2 + (2a+1)x + (a-1). For one root > 1 and one root < 0, the function must satisfy f(1) < 0 and f(0) < 0. f(1) = 1 + 2a + 1 + a - 1 = 3a + 1 < 0 => a < -1/3. f(0) = a - 1 < 0 => a < 1. Intersection is a < -1/3.

Multiple choice

If the quadratic equation $\displaystyle x^{2}-mx-4x+1=0$ has real and distinct roots, then the values of $m$ are A. $\displaystyle (-\infty , -6)$ B. $\displaystyle (-\infty,-3)$ C. $(-2,\infty )$ D. $\displaystyle \left ( 2,\infty \right )$

  1. A or C

  2. A or D

  3. B or C

  4. B or D
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Equation: x^2 - (m+4)x + 1 = 0. For real and distinct roots, discriminant D > 0. D = (m+4)^2 - 4 > 0. (m+4)^2 > 4. So m+4 > 2 or m+4 < -2. This means m > -2 or m < -6.

Multiple choice

Let $\mathrm{a},\ \mathrm{b},\ \mathrm{c}$ be the sides of a triangle . No two of them are equal and $\lambda\in R$ . If the roots of the equation $\mathrm{x}^{2}+2(\mathrm{a}+\mathrm{b}+\mathrm{c})\mathrm{x} +3\lambda (\mathrm{a}\mathrm{b}+ \mathrm{b}\mathrm{c}+\mathrm{c}\mathrm{a})=0$ are real, then
  1. $\displaystyle \lambda<\frac{4}{3}$
  2. $\displaystyle \lambda>\frac{5}{3}$
  3. $\displaystyle \lambda\in(\frac{1}{3}, \displaystyle \frac{5}{3})$
  4. $\displaystyle \lambda\in(\frac{4}{3},\frac{5}{3})$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the quadratic equation to have real roots, the discriminant D = 4(a+b+c)^2 - 12*lambda*(ab+bc+ca) must be >= 0. This simplifies to (a+b+c)^2 >= 3*lambda*(ab+bc+ca). Using the inequality (a+b+c)^2 > 3(ab+bc+ca) for distinct a, b, c, we find lambda < (a+b+c)^2 / (3(ab+bc+ca)). Since (a+b+c)^2 / (ab+bc+ca) > 3, the condition holds for lambda < 4/3.