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 $\alpha$ and $\beta$ are the roots of $ax^2+c=bx$, then the equation $(a+cy)^2=b^2y$ in $y$ has the roots
  1. $\alpha ^{-1}, \beta ^{-1}$
  2. $\alpha ^{2}, \beta ^{2}$
  3. $\alpha \beta^{-1}, \alpha ^{-1}\beta$
  4. $\alpha ^{-2}, \beta ^{-2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The roots of ax^2 - bx + c = 0 are alpha and beta. The equation (a + cy)^2 = b^2y can be rewritten by substituting y = 1/x^2. The roots of the transformed equation are the squares of the reciprocals of the original roots, which is alpha^-2 and beta^-2.

Multiple choice

If one root of the equation ${ x }^{ 2 }+px+q=0$ is the square of the other root, then
  1. ${ p }^{ 3 }+{ q }^{ 2 }-q(3p+1)=0$
  2. ${ p }^{ 3 }+{ q }^{ 2 }+q(1+3p)=0$
  3. ${ p }^{ 3 }+{ q }^{ 2 }+q(3p-1)=0$
  4. ${ p }^{ 3 }+{ q }^{ 2 }+q(1-3p)=0\quad $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let roots be a and a^2. Sum: a + a^2 = -p. Product: a^3 = q. Substituting a = q^(1/3) into the sum equation: q^(1/3) + q^(2/3) = -p. Cubing both sides: q + q^2 + 3*q^(1/3)q^(2/3)(q^(1/3)+q^(2/3)) = -p^3. q + q^2 + 3q(-p) = -p^3. p^3 + q^2 + q(1-3p) = 0.

Multiple choice

If one root of equation $ x^2 + ax + 12 = 0 $ is 4 while the equation $x^2 + ax + b = 0 $ has equal roots, then the value of b is :
  1. $\dfrac {4}{49}$
  2. $\dfrac {49}{4}$
  3. $\dfrac {7}{4}$
  4. $\dfrac {4}{7}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If 4 is a root of x^2 + ax + 12 = 0, then 16 + 4a + 12 = 0, so 4a = -28, a = -7. The second equation is x^2 - 7x + b = 0. For equal roots, the discriminant must be 0: (-7)^2 - 4(1)(b) = 0, so 49 - 4b = 0, b = 49/4.

Multiple choice

If $\left( 1-p \right) $ is a root of quadratic equation ${ x }^{ 2 }+px+\left( 1-p \right) =0$, then its roots are
  1. $0, 1$
  2. $-1, 1$
  3. $0, -1$
  4. $-1, 2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If (1-p) is a root, it satisfies the equation: (1-p)^2 + p(1-p) + (1-p) = 0. Factoring out (1-p), we get (1-p) * (1-p + p + 1) = 0, which implies (1-p) * 2 = 0, so p = 1. Substituting p = 1 into the original equation gives x^2 + x = 0, so x(x+1) = 0. The roots are 0 and -1.

Multiple choice

If $\alpha , \beta $ and $\gamma $ are the roots of the equation ${ x }^{ 3 }-8x+8=0$, then $\displaystyle\sum { { \alpha }^{ 2 } } $ and $\displaystyle\sum { \dfrac { 1 }{ \alpha \beta } } $ are respectively
  1. $0$ and $-16$
  2. $16$ and $8$
  3. $-16$ and $0$
  4. $16$ and $0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For x^3 - 8x + 8 = 0, the coefficients are a=1, b=0, c=-8, d=8. Sum of roots (alpha+beta+gamma) = -b/a = 0. Sum of roots taken two at a time (alpha*beta + beta*gamma + gamma*alpha) = c/a = -8. Product of roots (alpha*beta*gamma) = -d/a = -8. Sum of squares = (sum of roots)^2 - 2*(sum of roots taken two at a time) = 0^2 - 2*(-8) = 16. Sum of 1/(alpha*beta) = (alpha+beta+gamma)/(alpha*beta*gamma) = 0 / -8 = 0.

Multiple choice

If the equation ${ x }^{ 2 }+ix+a=0$, ${ x }^{ 2 }-2x+ia=0,a\neq 0$ have a common root, then
  1. $a$ is real
  2. $a=\dfrac{ 1 }{ 2 }+i$
  3. $a=\dfrac{ 1 }{ 2 }-i$

  4. The other root is also common
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let alpha be the common root. alpha^2 + i*alpha + a = 0 and alpha^2 - 2*alpha + i*a = 0. Subtracting: (i+2)alpha + a(1-i) = 0. alpha = -a(1-i)/(2+i). Substituting back into the first equation and solving for a yields a = 1/2 - i.

Multiple choice

If $a,b,c$ are in GP, then the equation $a{ x }^{ 2 }+2bx+c=0$ and $d{ x }^{ 2 }+2ex+f=0$ have a common root, if $\cfrac { d }{ a } ,\cfrac { e }{ b } ,\cfrac { f }{ c } $ are in

  1. AP

  2. GP

  3. HP

  4. A.G.P.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If a, b, c are in GP, then b^2 = ac. The equation ax^2 + 2bx + c = 0 becomes ax^2 + 2bx + b^2/a = 0, which is (sqrt(a)x + b/sqrt(a))^2 = 0, so x = -b/a. Since this is a common root, it must satisfy the second equation. The condition for the coefficients leads to the conclusion that d/a, e/b, f/c are in AP.

Multiple choice

If $\alpha ,\beta $ are the roots of equation $a{ x }^{ 2 }+bx+c=0$, then $\log _{ e }{ a } +2\log _{ e }{ x } -\cfrac { 1 }{ x } \left( \alpha +\beta \right) -\cfrac { 1 }{ 2{ x }^{ 2 } } \left( { \alpha }^{ 2 }+{ \beta }^{ 2 } \right) -\cfrac { 1 }{ 3{ x }^{ 2 } } \left( { \alpha }^{ 3 }+{ \beta }^{ 3 } \right) ....$ is equal to
  1. $\log _{ e }{ a } $
  2. $\log _{ e }{ \beta } $
  3. $\log _{ e }{ \left( a{ x }^{ 2 }+bx+c \right) } $
  4. $\log _{ e }{ \cfrac { \beta }{ \alpha } } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice

Let $\alpha ,\beta $ be the roots of $a{ x }^{ 2 }+bx+c=0;\gamma ,\delta $ be the roots of $ p{ x }^{ 2 }+qx+r=0$ and ${ D }{ 1 },{ D }{ 2 }$ are the respective discriminants of these equations. If the $\alpha ,\beta ,\gamma ,\delta $ are in AP, then ${ D }{ 1 }:{ D }{ 2 }$ is equal to
  1. $\cfrac { { a }^{ 2 } }{ { b }^{ 2 } } $
  2. $\cfrac { { a }^{ 2 } }{ { p }^{ 2 } } $
  3. $\cfrac { { b }^{ 2 } }{ { q }^{ 2 } } $
  4. $\cfrac { { c }^{ 2 } }{ { r }^{ 2 } } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If roots are in AP, let them be a-3d, a-d, a+d, a+3d. The discriminants D1 = b^2 - 4ac and D2 = q^2 - 4pr relate to the squared differences of roots. Using properties of AP and quadratic coefficients, the ratio simplifies to a^2/p^2.

Multiple choice

Let $a=\cos { \cfrac { 2\pi }{ 7 } } +i\sin { \cfrac { 2\pi }{ 7 } } ,\alpha =a+{ a}^{ 2 }+{ a}^{ 4 }$ and $\beta ={ a }^{ 3 }+{ a }^{ 5 }+{ a }^{ 6 } $. Then, the equation whose roots are $\alpha ,\beta $ is
  1. $\quad { x }^{ 2 }-x+2=0$
  2. ${ x }^{ 2 }+x-2=0$
  3. ${ x }^{ 2 }-x-2=0$
  4. ${ x }^{ 2 }+x+2=0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given a = exp(i*2pi/7). alpha and beta are sums of powers of a. Using properties of roots of unity, alpha + beta = -1 and alpha * beta = 2. The quadratic equation is x^2 - (sum)x + (product) = 0, which is x^2 + x + 2 = 0.

Multiple choice

Let $z_1, z_2$ be two roots of the equation $z^2 + az + b = 0$, $z $ being a complex number. Further assume that the origin, $z_1$ and $z_2$ form an equilateral triangle. Then,
  1. $a^2 = b$
  2. $a^2 = 2b$
  3. $a^2 = 3b$
  4. $a^2 = 4b$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For the origin, z1, and z2 to form an equilateral triangle, the condition is z1^2 + z2^2 - z1*z2 = 0. Since z1 and z2 are roots of z^2 + az + b = 0, we have z1 + z2 = -a and z1*z2 = b. Substituting (z1+z2)^2 - 3*z1*z2 = 0, we get (-a)^2 - 3*b = 0, so a^2 = 3b.

Multiple choice

Let $\alpha$ and $\beta$ be two real roots of the equation $(k+1)\tan^2x-\sqrt{2}\lambda\tan x=(1-k)$, where $k(\neq\,\,-1)$ and $\lambda$ are real numbers. If $tan^2(\alpha+\beta)=50$, then a value of $\lambda$ is:
  1. $10\sqrt{2}$
  2. $5\sqrt{2}$
  3. $10$
  4. $5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Equation: (k+1)tan^2x - sqrt(2)lambda tan x + (k-1) = 0. Let tan alpha, tan beta be roots. tan alpha + tan beta = sqrt(2)lambda / (k+1). tan alpha * tan beta = (k-1)/(k+1). tan(alpha+beta) = (tan alpha + tan beta) / (1 - tan alpha * tan beta). tan^2(alpha+beta) = 50. Solving leads to lambda = 10.