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 $\alpha,\beta,\gamma,\delta $ are the roots of the equation $x^{4}-2x^{3}+2x^{2}+1=0$, then the equation whose roots are $2+\displaystyle \frac{1}{\alpha},2+\frac{1}{\beta},2+\frac{1}{\gamma},2+\frac{1}{\delta}$, is:
  1. $x^{4}-8x^{3}+12x^{2}-42x+29=0$
  2. $x^{4}+8x^3+ 6x^{2}-29=0$
  3. $x^{4}-14x+29=0$
  4. $x^{4}-8x^{3}+26x^{2}-42x+29=0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let y = 2 + 1/x, so 1/x = y - 2, or x = 1/(y - 2). Substitute this into the original equation x^4 - 2x^3 + 2x^2 + 1 = 0. Solving this transformation leads to the polynomial x^4 - 8x^3 + 26x^2 - 42x + 29 = 0.

Multiple choice

The number of integral values of $a$ for which the quadratic equation $(x+a)(x+1991)+1=0$ has integral roots are
  1. $3$
  2. $0$
  3. $1$
  4. $2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

(x+a)(x+1991) = -1. Let x+a = u and x+1991 = v. Then uv = -1. Since x is an integer, u and v are integers. The factors of -1 are (1, -1) and (-1, 1). Case 1: x+a = 1 and x+1991 = -1 => x = -1992, a = 1993. Case 2: x+a = -1 and x+1991 = 1 => x = -1990, a = 1989. Two values for a.

Multiple choice

If $ \alpha, \beta $ are the roots of $ ax^{2}+bx+c=0 $; $ \alpha+h,\beta+h $ are the roots of $ px^{2}+qx+r=0 $; and $ D_{1},D_{2} $ are the respective discriminants of these equations, then $ D_{1}:D_{2} $ is
  1. $ \displaystyle \frac{a^{2}}{p^{2}} $
  2. $ \displaystyle \frac{b^{2}}{q^{2}} $
  3. $ \displaystyle \frac{c^{2}}{r^{2}} $
  4. $1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If roots of ax^2+bx+c=0 are alpha, beta, then D1 = b^2 - 4ac. If roots of px^2+qx+r=0 are alpha+h, beta+h, the equation is p(x-h)^2 + q(x-h) + r = 0. Expanding this, the discriminant D2 = q^2 - 4pr. Given the relationship between roots, D1/D2 = a^2/p^2.

Multiple choice

Let $a,b,c,d$ be real number in G.P. If $u,v,w$ satisfy the system of equations $u+2v+3w=6$,$4u+5v+6w=12$,$6u+9v=4$. Then the roots of the equation $ \left( \dfrac { 1 }{ u } +\dfrac { 1 }{ v } +\dfrac { 1 }{ w } \right) { x }^{ 2 }+\left[ { \left( b-c \right) }^{ 2 }+{ \left( c-a \right) }^{ 2 }+{ \left( d-b \right) }^{ 2 } \right] x+u+v+w=0$ and $ { 20x }^{ 2 }+10{ \left( a-d \right) }^{ 2 }x-9=0$ are

  1. equal

  2. imaginary

  3. reciprocals

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

The system of equations for u, v, w is linear. Solving it gives specific values. The quadratic equations involve these values and properties of G.P. (a, b, c, d). The roots of the two equations being reciprocals is a standard property in such complex algebraic problems.

Multiple choice

Suppose $A, B, C$ are defined as $A=a^2b+ab^2-a^2c-ac^2, B=b^2c+bc^2-a^2b-ab^2$ and $C=a^2c+ac^2-b^2c-bc^2$, where $a > b > c > 0$ and the equation $Ax^2+Bx+C=0$ has equal roots, then $a, b, c$ are in
  1. $A.P.$
  2. $G.P.$
  3. $H.P.$
  4. $A.G.P.$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The equation Ax^2 + Bx + C = 0 has equal roots if B^2 = 4AC. Substituting the expressions for A, B, and C and simplifying shows that the condition holds when 1/a + 1/c = 2/b, which is the definition of a harmonic progression.

Multiple choice

The cubic equation which has three roots $1, 3-\sqrt{2}i, 3+\sqrt2i$ is
  1. $x^{3}+7x^{2}+17x-11=0$
  2. $x^{3}-7x^{2}+17x-11=0$
  3. $x^{3}-7x^{2}+17x+1=0$
  4. $x^{3}+7x^{2}-8=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The roots are 1, 3-sqrt(2)i, and 3+sqrt(2)i. The sum of roots = 1 + 3 - sqrt(2)i + 3 + sqrt(2)i = 7. The product of roots = 1 * (3^2 - (sqrt(2)i)^2) = 1 * (9 + 2) = 11. The sum of roots taken two at a time = 1(3-sqrt(2)i) + 1(3+sqrt(2)i) + (3-sqrt(2)i)(3+sqrt(2)i) = 6 + 11 = 17. The cubic equation is x^3 - (sum)x^2 + (sum of products)x - (product) = 0, which is x^3 - 7x^2 + 17x - 11 = 0.

Multiple choice

If two roots of the equation, $x^{3}-7x^{2}+4x+12=0$ are in the ratio $1:3 $, then the roots are
  1. $-6,-2,-1$
  2. $-1,2,6$
  3. $1,3,4 $
  4. $-2,3,6$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the roots be a, 3a, and b. By Vieta's formulas: a + 3a + b = 7 => 4a + b = 7. a * 3a * b = -12 => 3a^2 * b = -12 => a^2 * b = -4. Substituting b = 7 - 4a into the second: a^2(7 - 4a) = -4 => 4a^3 - 7a^2 - 4 = 0. Testing a = -1 gives 4(-1) - 7(1) - 4 = -15 (no). Testing a = 2 gives 4(8) - 7(4) - 4 = 32 - 28 - 4 = 0. So a=2, 3a=6, b=7-8=-1. The roots are -1, 2, 6.

Multiple choice

If $\displaystyle \Sigma\alpha=\frac{2}{3},\Sigma\alpha\beta=\frac{4}{3}$ and $\displaystyle \alpha\beta\gamma=\frac{8}{3}$, then the equation whose roots are $\alpha,\beta,\gamma$, is
  1. $x^{3}-2x^{2}+4x-8=0$
  2. $3x^{3}+2x^{2}-4x+8=0$
  3. $3x^{3}-2x^{2}+4x-8=0$
  4. $x^{3}+2x^{2}-4x+8=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The cubic equation with roots alpha, beta, gamma is x^3 - (sum)x^2 + (sum of products)x - (product) = 0. x^3 - (2/3)x^2 + (4/3)x - (8/3) = 0. Multiply by 3: 3x^3 - 2x^2 + 4x - 8 = 0.