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 the equation $ ax^2 + bx + c = 0, a > 0 $ has two distinct real roots $ \alpha $ and $ \beta $ such that $ \alpha < - 5 $ and $ \beta > 5, $ then :
  1. $ c > 0 $
  2. $ c = 0 $
  3. $ c = \dfrac {a+b}{2} $
  4. $ c < 0 $
  5. $ c = a +b $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For a quadratic f(x) = ax^2 + bx + c with a > 0, if the roots are on opposite sides of 0 (specifically alpha < -5 and beta > 5), then f(0) must be negative. Since f(0) = c, c must be less than 0.

Multiple choice

If the equations $ x^2 + ax + bc = 0 $ and $ x^2 + bx + ca = 0$ have a common root and if $a, b $ and $c$ are non-zero distinct real numbers, then their other roots satisfy the equation :
  1. $ x^2 + x + abc = 0 $
  2. $ x^2 - (a+b) x + ab = 0 $
  3. $ x^2 + (a+b) x + ab = 0 $
  4. $ x^2 + x+ ab = 0 $
  5. $ x^2 + abx+ abc = 0 $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If x^2 + ax + bc = 0 and x^2 + bx + ca = 0 have a common root 'k', then k^2 + ak + bc = k^2 + bk + ca, which implies k(a-b) = c(a-b). Since a != b, k = c. Substituting k=c into the first equation: c^2 + ac + bc = 0, so c(c + a + b) = 0. Since c != 0, c = -(a+b). The roots of the first equation are c and b, and the second are c and a. The other roots are a and b, which are roots of x^2 - (a+b)x + ab = 0.

Multiple choice

Let $b$ be a non- zero real number. Suppose the quadratic equation $2x^2 + bx + \dfrac{1}{b} = 0$ has two distinct real roots. Then
  1. $b + \dfrac{1}{b} > \dfrac{5}{2}$
  2. $b + \dfrac{1}{b} < \dfrac{5}{2}$
  3. $b^2 - 3b > -2$
  4. $b^2 + \dfrac{1}{b^2} < 4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For the quadratic equation 2x^2 + bx + 1/b = 0 to have two distinct real roots, the discriminant D = b^2 - 4(2)(1/b) must be greater than 0. This simplifies to b^2 - 8/b > 0, which means (b^3 - 8)/b > 0. Testing values or analyzing the inequality shows that b^2 - 3b > -2 is a valid consequence of the root conditions.

Multiple choice

Let $r$ be a root of the equation $x^2+2x+6=0$. The value of $(r+2)(r+3)(r+4)(r+5)$ is equal to.
  1. $51$
  2. $-51$
  3. $-126$
  4. $126$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

r^2 + 2r + 6 = 0. We want (r+2)(r+3)(r+4)(r+5). Group as [(r+2)(r+5)] * [(r+3)(r+4)] = (r^2 + 7r + 10)(r^2 + 7r + 12). Since r^2 = -2r - 6, substitute: (-2r - 6 + 7r + 10)(-2r - 6 + 7r + 12) = (5r + 4)(5r + 6) = 25r^2 + 30r + 20r + 24 = 25r^2 + 50r + 24. Substitute r^2 again: 25(-2r - 6) + 50r + 24 = -50r - 150 + 50r + 24 = -126.

Multiple choice

If $p,q$ be non-zero real numbers and $f\left(x\right)\neq 0$ in $\left[0,2\right]$ and ${\int}{0}^{1}f\left(x\right).\left(x^{2}+px+q\right)dx={\int}{0}^{2}f\left(x\right).\left(x^{2}+px+q\right)dx=0$ then equation $x^{2}+px+q=0$ has

  1. Two imaginary roots
  2. No root in $\left(0,2\right)$
  3. One root in $\left(0,1\right)$ and other in $\left(1,2\right)$
  4. One root in $\left(-\infty,0\right)$ and other in $\left( 2,\infty\right)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice

Consider the two quadratic polynomials: $\displaystyle C_{a}:y=\frac{x^{2}}{4}-ax+a^{2}+a-2$ and $\displaystyle C:y=2-\frac{x^{2}}{4}$ If the origin lies between the zeroes of the polynomial $ C_{a},$ then the number of integral value(s) of $'a'$ is
  1. $1$
  2. $2$
  3. $3$
  4. more than $3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For the origin to lie between the roots of Ca, the product of the roots must be negative, or f(0) < 0. f(0) = a^2 + a - 2 < 0. (a+2)(a-1) < 0. This holds for -2 < a < 1. The integers are -1 and 0. There are 2 such values.

Multiple choice

A quadratic equation whose roots are $ \displaystyle \text{cosec}^{2} \theta $ and $\displaystyle \sec ^{2}\theta $ can be
  1. $\displaystyle x^{2}-5x+2= 0$
  2. $\displaystyle x^{2}-2\sin 2\theta x+6\sin \theta= 0$
  3. $\displaystyle x^{2}-4\text{cosec} 2\theta x+4\text{cosec} 2\theta= 0$

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

Roots are csc^2(theta) and sec^2(theta). Sum = 1/sin^2 + 1/cos^2 = (sin^2 + cos^2) / (sin^2*cos^2) = 1 / (sin^2*cos^2) = 4 / sin^2(2*theta) = 4*csc^2(2*theta). Product = 1 / (sin^2*cos^2) = 4*csc^2(2*theta). Equation: x^2 - (sum)x + product = 0 => x^2 - 4*csc^2(2*theta)x + 4*csc^2(2*theta) = 0. Option C is x^2 - 4*csc(2*theta)x + 4*csc(2*theta) = 0, which is missing the square on csc.