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

Ruhi and Tara were asked to solve a quadratic equation $ax^{2} + bx + c = 0, a\neq 0$$. Ruhi made some mistake in writing b and found the roots as $3$ and $-\dfrac {1}{2}$. Tara, too, made mistake in writing c and found the roots $-1$ and $-\dfrac {1}{4}$. Find the correct roots of the given equation
  1. $-2, \dfrac {3}{4}$
  2. $3, -1$
  3. $-\dfrac {1}{2}, -1$
  4. $3, -\dfrac {1}{4}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Ruhi's roots 3 and -1/2 imply the sum is 5/2, but she erred in b, so the product c/a = 3 * (-1/2) = -3/2 is correct. Tara's roots -1 and -1/4 imply the product 1/4 is wrong, but the sum -b/a = -1 + (-1/4) = -5/4 is correct. Thus, -b/a = -5/4 and c/a = -3/2. The equation is x^2 + (5/4)x - 3/2 = 0, or 4x^2 + 5x - 6 = 0. Factoring gives (4x-3)(x+2) = 0, so roots are 3/4 and -2.

Multiple choice

If the equation $9x^2-4kx + 4 = 0$ has two equal roots, k is equal to
  1. $\pm 1$
  2. $2$
  3. $\pm 3$
  4. $\pm 4$
  5. $5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a quadratic equation ax^2 + bx + c = 0 to have equal roots, the discriminant D = b^2 - 4ac must be 0. Here, a=9, b=-4k, c=4. So, (-4k)^2 - 4(9)(4) = 0, which simplifies to 16k^2 - 144 = 0. Thus, k^2 = 9, meaning k = +/- 3.

Multiple choice

If $a,b$ are the roots of a equation of a quadratic equation $x^2-3x+5=0$ then the equation whose roots are $(a^2-3a+7)$ and $(b^2-3b+7)$ is-
  1. $x^2+4x+1=0$
  2. $x^2-4x-1=0$
  3. $x^2+4x+4=0$
  4. $x^2-4x+4=0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given a, b are roots of x^2 - 3x + 5 = 0, so a^2 - 3a + 5 = 0 and b^2 - 3b + 5 = 0. Thus, a^2 - 3a = -5 and b^2 - 3b = -5. The new roots are (-5 + 7) = 2 and (-5 + 7) = 2. The equation with roots 2, 2 is (x-2)^2 = x^2 - 4x + 4 = 0.

Multiple choice

If $A,B$ are roots of the quadratic equation $x^2-5x+6=0$ then the equation whose roots are $A+3$ and $B+3$ is-
  1. $x^2-11x+30=0$
  2. $(x-3)^2-5(x-3)+6=0$
  3. Both $(A)$ and $(B)$

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

Roots of x^2 - 5x + 6 = 0 are 2 and 3. If roots are A+3 and B+3, they are 5 and 6. The equation is (x-5)(x-6) = x^2 - 11x + 30 = 0. Also, replacing x with x-3 in the original equation gives (x-3)^2 - 5(x-3) + 6 = 0, which is also correct.

Multiple choice

The equations ${ x }^{ 3 }+5{ x }^{ 2 }+px+q=0$ and ${ x }^{ 3 }+7{ x }^{ 2 }+px+r=0$ have two roots in common. If the third root of each equation is represented by ${x}{1}$ and ${x}{2}$ respectively, then the ordered pair $({x}{1},{x}{2})$ is:
  1. $(-5,-7)$
  2. $(1,-1)$
  3. $(-1,1)$
  4. $(5,7)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Subtracting the two equations: (x^3+5x^2+px+q) - (x^3+7x^2+px+r) = 0 => -2x^2 + q - r = 0. This doesn't help directly. Let roots be a, b, x1 and a, b, x2. Sum of roots: a+b+x1 = -5 and a+b+x2 = -7. Thus x1 - x2 = 2. Also product of roots: abx1 = -q and abx2 = -r. The common roots satisfy x^2(x+5+x1) = 0? No, use Vieta's. The common roots a, b satisfy a+b = -5-x1 and a+b = -7-x2. Also ab + a(x1) + b(x1) = p and ab + a(x2) + b(x2) = p. So ab + (a+b)x1 = ab + (a+b)x2 => (a+b)(x1-x2) = 0. Since x1 != x2, a+b = 0. Then x1 = -5 and x2 = -7.

Multiple choice

Two candidates attempt to solve a quadratic equation $\displaystyle x^{2} + px + q = 0$. One starts with a wrong value of $\displaystyle p$ and gets $2$, $6$ as its roots and other starts with wrong value of $\displaystyle q$ and obtains roots $2$, $-9$. The correct roots are
  1. $-3, \: -4$
  2. $3, \: 4$
  3. $-3, \: 4$
  4. $-4, \: 3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

First candidate gets roots 2, 6, so x^2 - 8x + 12 = 0. q is correct, so q = 12. Second candidate gets roots 2, -9, so x^2 + 7x - 18 = 0. p is correct, so p = 7. Correct equation is x^2 + 7x + 12 = 0. Roots are (x+3)(x+4) = 0, so -3, -4.

Multiple choice

$Let\quad a,b,c\in R\quad such\quad that\quad no\quad two\quad of\quad them\quad are\quad equal\quad and\ satisfy\left[ \begin{matrix} 2a & b & c \ b & c & 2a \ c & 2a & b \end{matrix} \right] =0,\quad then\quad equation\quad 24a{ x }^{ 2 }+4bx+c=0\quad has$
  1. $At\quad least\quad one\quad root\quad in\quad [0,1]$
  2. $At\quad least\quad one\quad root\quad in\quad [\cfrac{-1}{2},\cfrac{1}{2}]$
  3. $At\quad least\quad one\quad root\quad in\quad [-1,0]$
  4. $At\quad least\quad two\quad root\quad in\quad [0,2]$
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice

For the equation $\begin{vmatrix} 1 & x & { x }^{ 2 } \ { x }^{ 2 } & 1 & x \ x & { x }^{ 2 } & 1 \end{vmatrix}=0$

  1. There are exactly two distinct roots

  2. There is one pair of equation real roots

  3. There are three pairs of equal roots

  4. Modulus of each root is 2
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The determinant of the matrix is -(x^3-1)^2 = 0, which implies x^3 = 1. The roots are 1, omega, and omega^2, where omega is a complex cube root of unity. Each root appears twice, leading to three pairs of equal roots.

Multiple choice

Solve the system of the equations: $ax+by+cz=d,\quad { a }^{ 2 }x+{ b }^{ 2 }y+{ c }^{ 2 }z={ d }^{ 2 },\quad { a }^{ 3 }x+{ b }^{ 3 }y+{ c }^{ 3 }z={ d }^{ 3 }$.
  1. $x=\cfrac { d(d+b)(c-d) }{ a(a-b)(c-a) } ,\quad y=\cfrac { d(a-d)(d-c) }{ b(a-b)(b-c) } ,\quad z=\cfrac { d(b-d)(d-a) }{ d(b-c)(c-a) } $
  2. $x=\cfrac { d(d-b)(c-d) }{ a(a-b)(c-a) } ,\quad y=\cfrac { d(a+d)(d-c) }{ b(a-b)(b-c) } ,\quad z=\cfrac { d(b-d)(d-a) }{ d(b-c)(c-a) } $
  3. $x=\cfrac { d(d-b)(c-d) }{ a(a-b)(c-a) } ,\quad y=\cfrac { d(a-d)(d-c) }{ b(a-b)(b-c) } ,\quad z=\cfrac { d(b-d)(d-a) }{ d(b-c)(c-a) } $

  4. None of these.
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice

If $a,b,c\epsilon R,a\neq 0$ and $\left(b-1\right)^{2}<4ac$, then the number of real roots of the system of equation $\left(in\ three\ unknowns\ x_{1}, x_{2}, x_{3}\right)ax_{1}^{2}+bx_{1}+c=x_{2},ax_{2}^{2}+bx_{2}+c=x_{3},ax_{3}^{2}+bx_{3}+c=x_{1}$ is
  1. $0$
  2. $1$
  3. $2$
  4. $3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Rewrite each equation as x2 - x1 = ax1^2 + (b - 1)x1 + c, and similarly for the other two variables. The quadratic on the right has discriminant (b - 1)^2 - 4ac < 0, so it always has the same strict sign. This would force x2, x3, and x1 to change in one direction cyclically, which is impossible, so there are no real solutions.