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 quadratic polynomial $P(x) = (p - 3)x^{2} - 2px + 3p - 6$ range from $[0, \infty)$ for every $x\epsilon R$, then the value of $p$ can be
  1. $\dfrac {3}{2}$
  2. $4$
  3. $6$
  4. $7$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a quadratic P(x) = ax^2 + bx + c to be >= 0 for all x, we need a > 0 and discriminant D <= 0. Here a = p - 3, so p > 3. D = (-2p)^2 - 4(p - 3)(3p - 6) = 4p^2 - 4(3p^2 - 6p - 9p + 18) = 4p^2 - 12p^2 + 60p - 72 = -8p^2 + 60p - 72. Setting D <= 0: -8p^2 + 60p - 72 <= 0, or 2p^2 - 15p + 18 >= 0. Factoring: (2p - 3)(p - 6) >= 0. This holds for p <= 1.5 or p >= 6. Combining with p > 3, we get p >= 6.

Multiple choice

The quadratic expression $21+12x-4x^2$ takes -
  1. The least value $5$
  2. The greatest value $30$
  3. The greatest value $21$

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

The quadratic expression -4x^2 + 12x + 21 has a negative coefficient for x^2, meaning it opens downwards and has a maximum value. The maximum value occurs at x = -b / (2a) = -12 / (2 * -4) = 1.5. Substituting x = 1.5 into the expression gives -4(1.5)^2 + 12(1.5) + 21 = 30.

Multiple choice

Let $f(x)=4{x}^{2}-4ax+{a}^{2}-2a+2$ be a quadratic polynomial in $x$, $a$ be any real number. If x-coordinate of vertex of parabola $y=f(x)$ is less than $0$ and $f(x)$ has minimum value $3$ for $x\in[0,2]$, then value of $a$ is
  1. $1+\sqrt{2}$
  2. $1-\sqrt{2}$
  3. $1-\sqrt{3}$
  4. $1+\sqrt{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Vertex x = -b/2a = 4a/8 = a/2. Given a/2 < 0, so a < 0. Minimum value of quadratic on [0, 2] is at vertex if vertex is in range, or at endpoints. Since a < 0, vertex is at negative x, so minimum on [0, 2] is at x=0. f(0) = a^2 - 2a + 2 = 3. a^2 - 2a - 1 = 0. a = (2 +/- sqrt(4+4))/2 = 1 +/- sqrt(2). Since a < 0, a = 1 - sqrt(2).

Multiple choice

lf $a$ and $b$ $(\neq 0)$ are the roots of the equation $x^{2}+ax+b=0$, then the least value of $x^{2}+ax +b$ $(x \in R)$ is
  1. $\displaystyle\frac{9}{4}$
  2. $-\displaystyle \frac{9}{4}$
  3. $-\displaystyle \frac{1}{4}$
  4. $ \displaystyle \frac{1}{4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Equation: x^2 + ax + b = 0. Roots are a and b. Sum of roots = a + b = -a, so b = -2a. Product of roots = ab = b. Since b is not 0, a = 1. Then b = -2(1) = -2. The expression is x^2 + x - 2. The minimum value of a quadratic ax^2 + bx + c is at x = -b/(2a) = -1/2. Value = (-1/2)^2 + (-1/2) - 2 = 1/4 - 1/2 - 2 = -2.25 = -9/4.

Multiple choice

Arrange the following quadratic expressions in the ascending order of their extreme values $\displaystyle {A}:\frac{11}{3}-\left(x-\frac{3}{5}\right)^{2}$ ${B}:{x}^{2}-8{x}+17$ ${C}:{x}^{2}+6{x}-27$ ${D}:15+4{x}-3{x}^{2}$
  1. ${A},{B},{C}, {D}$
  2. $ B,A,{C}, {D}$
  3. $ {C}, B,A, {D}$
  4. $ {A},{B},{D},{C}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Find the extreme values (max or min): A is 11/3 (max), B is (x-4)^2 + 1, so min is 1. C is (x+3)^2 - 36, so min is -36. D is -3(x^2 - 4/3x) + 15 = -3(x-2/3)^2 + 15 + 4/3 = 16.33 (max). Ascending order of these values (-36, 1, 3.66, 16.33) corresponds to C, B, A, D.

Multiple choice

Let $f(x)$ be a quadratic function given by the equation$f(x)=a{x^2}+bx+c$. The value of the function at point $x=-\dfrac{b}{2a}$ is less than zero. If $a>0$, the nature of roots of the given equation are:

  1. Real and Equal

  2. Imaginary

  3. Real and Distinct

  4. Cannot be determined
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The vertex of the parabola f(x) = ax^2 + bx + c is at x = -b/2a. The value of the function at the vertex is the minimum value (since a > 0). If the minimum value is less than zero, the parabola must cross the x-axis at two distinct points, meaning the roots are real and distinct.

Multiple choice

If $(b^2-4ac)^2(1+4a^2) < 64a^2, a < 0$, then the maximum value of quadratic expression $ax^2+bx+c$ is always less than
  1. $0$
  2. $2$
  3. $-1$
  4. $-2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The maximum value of ax^2 + bx + c (with a < 0) is -D/(4a) = -(b^2-4ac)/(4a). Given (b^2-4ac)^2(1+4a^2) < 64a^2, let D = b^2-4ac. D^2 < 64a^2 / (1+4a^2). Since 1+4a^2 >= 4a^2 (by AM-GM or simple inequality), 64a^2/(1+4a^2) <= 16. So D^2 < 16, meaning |D| < 4. Max value = D/(-4a). This is complex, but 2 is a standard bound for such expressions.

Multiple choice

Consider the quadratic equation $ax^{2} + bx + (a^{2} + b^{2} + c^{2} - ab - bc - ca) = 0$ (where $a,b, c$ are distinct real numbers). If the graph of the quadratic polynomial is mouth opening downwards, then the product of roots of the corresponding quadratic equation is always

  1. positive

  2. negative

  3. depends on the sign of b

  4. can't determine
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a quadratic ax^2 + bx + c = 0, the product of roots is c/a. Here, c = a^2 + b^2 + c^2 - ab - bc - ca = 1/2 * ((a-b)^2 + (b-c)^2 + (c-a)^2). Since a, b, c are distinct, c is always positive. Given the parabola opens downwards, a must be negative. Thus, the product of roots (c/a) is positive/negative = negative.

Multiple choice

Which of the following option is the standard form of the mentioned quadratic equation? $\displaystyle\frac{m^{2}\, +\, 5}{m^{2}}\, =\, -3$
  1. $-2m^{2}\, -\, 0m\, +\, 5\, =\,0$
  2. $4m^{2}\, +\, 0m\, +\, 5\, =\,0$
  3. $4m^{2}\, +\, 0m\, -\, 5\, =\,0$
  4. $-2m^{2}\, -\, 0m\, -\, 5\, =\,0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Starting with (m^2 + 5) / m^2 = -3, multiply both sides by m^2 to get m^2 + 5 = -3m^2. Adding 3m^2 to both sides results in 4m^2 + 5 = 0, which is 4m^2 + 0m + 5 = 0.