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

Which of following equation has 4 as a root?
  1. $\displaystyle { x }^{ 2 }-4x+5=0$
  2. $\displaystyle { x }^{ 2 }-9x+20=0$
  3. $\displaystyle 2{ x }^{ 2 }-7x+6=0$
  4. $\displaystyle 3{ x }^{ 2 }-6x-2=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Substitute x = 4 into the equations. For B: 4^2 - 9(4) + 20 = 16 - 36 + 20 = 0. This confirms 4 is a root.

Multiple choice

If $a, b$ and $c$ are in arithmetic progression, then the roots of the equation $ax^{2} - 2bx + c = 0$ are
  1. $1$ and $\dfrac {c}{a}$
  2. $-\dfrac {1}{a}$ and $-c$
  3. $-1$ and $-\dfrac {c}{a}$
  4. $-2$ and $-\dfrac {c}{2a}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since a, b, c are in A.P., 2b = a + c. Substituting this into the equation ax^2 - (a+c)x + c = 0, we get ax^2 - ax - cx + c = 0, which factors to ax(x-1) - c(x-1) = 0. Thus, (ax-c)(x-1) = 0, yielding roots 1 and c/a.

Multiple choice

Given that $r$ and $s$ are constants , the solution of the quadratic equation, $r{x}^{2}=\dfrac{1}{s}x+3$ , is:
  1. $x=\dfrac { 1 }{ 2sr } \pm \dfrac { \sqrt { \dfrac { 1 }{ { s }^{ 2 } } +12r } }{ 2r } $
  2. $x=\dfrac { 1 }{ 2sr } \pm \dfrac { \sqrt { -\dfrac { 1 }{ { s }^{ 2 } } -12r } }{ 2sr } $
  3. $x=\dfrac { s }{ 2r } \pm \dfrac { \sqrt { \dfrac { 1 }{ { s }^{ 2 } } -12r } }{ 2r } $
  4. $x=\dfrac { s }{ 2r } \pm \dfrac { \sqrt { { s }^{ 2 }-12sr } }{ 2sr } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation is rx^2 - (1/s)x - 3 = 0. Using the quadratic formula x = [-b +/- sqrt(b^2 - 4ac)] / 2a, where a=r, b=-1/s, c=-3. x = [1/s +/- sqrt(1/s^2 - 4*r*(-3))] / 2r = [1/s +/- sqrt(1/s^2 + 12r)] / 2r. This simplifies to 1/(2sr) +/- sqrt(1/s^2 + 12r) / 2r.

Multiple choice

Find the root(s) of the equation $y+\sqrt{y+5}=7$

  1. 11

  2. 4

  3. 4 and 11
  4. $\pm 4$

  5. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

y + sqrt(y+5) = 7 => sqrt(y+5) = 7-y. Square both sides: y+5 = 49 - 14y + y^2 => y^2 - 15y + 44 = 0. Factors: (y-11)(y-4) = 0. y=11 (extraneous, 11+sqrt(16)=15!=7) or y=4 (4+sqrt(9)=7).

Multiple choice

If the roots of the equation $2x^2-3x + c = 0$ are real and irrational, a possible value of c is

  1. -2

  2. -1

  3. 0

  4. 1

  5. 2
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For roots to be real and irrational, the discriminant D = b^2 - 4ac must be a positive non-perfect square. Here, D = (-3)^2 - 4(2)(c) = 9 - 8c. For D > 0, 9 - 8c > 0, so c < 9/8. Checking options: if c = -1, D = 9 - 8(-1) = 17, which is positive and not a perfect square, making the roots irrational.

Multiple choice

The equation $4y^2 - 3y + C = 0$ has real roots. The value of C for which the product of the roots is a maximum is
  1. $\frac{9}{16}$
  2. $\frac{9}{4}$
  3. $\frac{4}{9}$
  4. $\frac{3}{4}$
  5. $-\frac{4}{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the equation 4y^2 - 3y + C = 0 to have real roots, the discriminant D = (-3)^2 - 4(4)(C) = 9 - 16C must be >= 0, so C <= 9/16. The product of the roots is C/4. Since this is an increasing function of C, the maximum value occurs at the largest possible value of C, which is 9/16.

Multiple choice

The equation $y+\sqrt{y-4}=6$ has

  1. two real roots

  2. only one real root

  3. no real roots

  4. one real root and one imaginary root

  5. only one imaginary root
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let sqrt(y-4) = x, so y = x^2 + 4. Substituting into the equation: x^2 + 4 + x = 6, which is x^2 + x - 2 = 0. Factoring gives (x+2)(x-1) = 0. Since x must be non-negative (as it is a square root), x = 1. Thus, sqrt(y-4) = 1, so y-4 = 1, y = 5. Only one real root exists.

Multiple choice

The equation $r+\dfrac{5}{r-1}=1+\dfrac{1}{r-1}$ has

  1. no root

  2. one integral root

  3. two equal roots

  4. two unequal, rational roots

  5. infinitely many roots
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation is r + 5/(r-1) = 1 + 1/(r-1). Subtract 1/(r-1) from both sides: r + 4/(r-1) = 1. Multiply by (r-1): r(r-1) + 4 = r-1. r^2 - r + 4 = r - 1. r^2 - 2r + 5 = 0. The discriminant is b^2 - 4ac = (-2)^2 - 4(1)(5) = 4 - 20 = -16. Since the discriminant is negative, there are no real roots.

Multiple choice

$ax^2+bx+c = 0$ is quadratic equations with real coefficients $a$ , $b$ and $c$.which of the following statement is/are not true ?
  1. If $a = b = c = 0$ then may be no roots are possible.
  2. If $a = b = c = 0$ then it must have infinite number of roots.
  3. if $a$ , $b$ and $c$ are rational then then irrational roots must be in conjugate pairs.
  4. if $a$ , $b$ and $c$ are irrational then irrational roots may be in conjugate pairs.
Reveal answer Fill a bubble to check yourself
A Correct answer