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

$a, b, c\in R, a\neq 0$ and the quadratic equation $ax^{2}+bx+c=0$ has no real roots, then which one of the following is not true?
  1. $a+b+c>0$
  2. $a(a+b+c)>0$
  3. $ac(a+b+c)>0$
  4. $c(a+b+c)>0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If ax^2 + bx + c = 0 has no real roots, the parabola never touches the x-axis. Thus, the function f(x) = ax^2 + bx + c is always positive (if a > 0) or always negative (if a < 0). Option A states a+b+c > 0, which is f(1). This is not necessarily true because f(x) could be entirely negative if a < 0.

Multiple choice

The equation $(a+2){x}^{2}+(a-3)x=2a-1,a\neq -2$ has rational roots for
  1. all rational values of $a$ except $a=-2$
  2. all real values of $a$ except $a=-2$
  3. rational values of $a> \displaystyle \frac{1}{2}$

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

For the roots of a quadratic Ax^2 + Bx + C = 0 to be rational, the discriminant D = B^2 - 4AC must be a perfect square. Here, D = (a - 3)^2 - 4(a + 2)(-(2a - 1)) = (a - 3)^2 + 4(a + 2)(2a - 1) = a^2 - 6a + 9 + 4(2a^2 + 3a - 2) = 9a^2 + 6a + 1 = (3a + 1)^2. Since the discriminant is a perfect square for any rational a, the roots are rational.

Multiple choice

The equation $\displaystyle x^{2}-6x+8+\lambda (x^{2}-4x+3)=0,\lambda \in R$, has
  1. real and unequal roots for all $\lambda $
  2. real roots for $\lambda < 0$ only
  3. real roots for $\lambda > 0$ only
  4. real and unequl roots for $\lambda =0$ only
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The discriminant of the quadratic equation (1+lambda)x^2 - (6+4lambda)x + (8+3lambda) = 0 is D = (6+4lambda)^2 - 4(1+lambda)(8+3lambda). Simplifying this yields D = 16lambda^2 + 48lambda + 36 - 4(3lambda^2 + 11lambda + 8) = 4lambda^2 + 4lambda + 4. Since the discriminant 4(lambda^2 + lambda + 1) is always positive for all real lambda, the roots are always real and unequal.

Multiple choice

If the roots of the equation $\displaystyle 4x^{2}+5k =(5k+1)x$ differ by unity, then the negative value of $k$ is
  1. $-3$
  2. $\displaystyle -\frac{1}{5}$
  3. $\displaystyle -\frac{3}{5}$
  4. $None\:of\:these$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Equation: 4x^2 - (5k+1)x + 5k = 0. Roots differ by 1, so |alpha - beta| = 1. (alpha - beta)^2 = 1, which is (alpha+beta)^2 - 4*alpha*beta = 1. ((5k+1)/4)^2 - 4*(5k/4) = 1. Solving (5k+1)^2 / 16 - 5k = 1 leads to 25k^2 + 10k + 1 - 80k = 16, 25k^2 - 70k - 15 = 0. Dividing by 5 gives 5k^2 - 14k - 3 = 0. Factoring: (5k+1)(k-3) = 0. Negative value is -1/5.

Multiple choice

If $X$ denotes the set of real numbers $p$ for which the equation $\displaystyle x^{2}=p(x+p)$ has its roots greater than $p$, then $X$ is equal to
  1. $\displaystyle \left ( -2,-\frac{1}{2} \right )$
  2. $\displaystyle \left ( -\frac{1}{2},\frac{1}{4} \right )$

  3. Null set 
  4. $(-\infty,0)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

x^2 - px - p^2 = 0. Roots = [p +/- sqrt(p^2 + 4p^2)] / 2 = p(1 +/- sqrt(5)) / 2. We need roots > p. p(1 + sqrt(5)) / 2 > p => (1 + sqrt(5))/2 > 1 (True for p > 0) or (1 + sqrt(5))/2 < 1 (False for p < 0). p(1 - sqrt(5)) / 2 > p => (1 - sqrt(5))/2 > 1 (False). Checking conditions shows no real p satisfies the condition.

Multiple choice

If $a, b, c \epsilon R, a>0 and c \neq 0$let $\alpha and \beta $ be the real and distinct roots of the equation $ ax^2+bx+c=\left | c \right | $ and p,q be the real and distinct roots of the equation $ax^2+bx+c=0$. Then
  1. p and q lie between $ \alpha $ and $\beta$
  2. p and q do not lie between $ \alpha $ and $\beta$
  3. only p lie between $ \alpha $ and $\beta$
  4. only q lie between $ \alpha $ and $\beta$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The roots of ax^2+bx+c=0 are p, q. The roots of ax^2+bx+c=|c| are alpha, beta. Since |c| > 0, the parabola ax^2+bx+c is shifted downwards by |c| to reach zero, meaning the roots of the first equation are 'inside' the roots of the second.

Multiple choice

The real number $k$ for which the equation $2{ x }^{ { 2 } }+3{ x }+{ k }=0$ has two distinct real roots in $[ 0,1 ].$
  1. lies between $1$ and $2 .$
  2. lies between $2$ and $3 .$
  3. lies between $-1$ and $0 .$

  4. does not exist
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let f(x) = 2x^2 + 3x + k. For roots in [0, 1], the discriminant D = 9 - 8k >= 0, so k <= 9/8. Also, the vertex -b/2a = -3/4 is not in [0, 1]. Since the parabola opens upward and the vertex is to the left of the interval, the function is increasing on [0, 1]. For roots in [0, 1], we need f(0) >= 0 and f(1) <= 0. f(0) = k >= 0. f(1) = 2 + 3 + k = 5 + k <= 0, so k <= -5. These conditions (k >= 0 and k <= -5) are contradictory.

Multiple choice

For what real values of a do the roots of the equation $x ^ { 2 } - 2 x - \left( a ^ { 2 } - 1 \right) = 0$ Lie between the roots of the equation $x ^ { 2 }- 2 x ( a + 1 ) + a ( a - 1 ) = 0$
  1. $( - 1 / 4 < a < 1 )$
  2. $( 1 / 4 < a < 1 ) \text { and } ( a < - 1 / 3 )$
  3. $( - 1 / 2 < a < 1 ) \text { and } ( a < - 1 / 3 )$
  4. $( - 1 / 2 < a < 1 )$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the roots of one quadratic to lie between the roots of another, the function values at the roots of the second must have specific signs. Solving the conditions for the given equations leads to the interval -1/4 < a < 1.