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

Number of real roots of the equation ${ 3 }^{ {\log }_{ 3 }\left( { x }^{ 2 }-4x+3 \right) }=(x-3)$ is:
  1. $0$
  2. $1$
  3. $2$
  4. $3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation 3^(log3(x^2 - 4x + 3)) = x - 3 simplifies to x^2 - 4x + 3 = x - 3, which is x^2 - 5x + 6 = 0. Factoring gives (x - 2)(x - 3) = 0, so x = 2 or x = 3. However, the logarithm argument x^2 - 4x + 3 must be positive; for x = 2, 4 - 8 + 3 = -1 (invalid), and for x = 3, 9 - 12 + 3 = 0 (invalid). Thus, there are no real roots.

Multiple choice

lf both the roots of the equation ${x}^{2}+{x}+{a}=0$ exceed ${a}$, then
  1. $2<{a}<3$
  2. ${a}>3$
  3. $-3<{a}<3$
  4. $ a<-2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For roots of x^2 + x + a = 0 to exceed a, the discriminant must be >= 0 (1 - 4a >= 0 => a <= 1/4), the vertex -b/2a = -1/2 must be > a (a < -1/2), and f(a) > 0 (a^2 + a + a > 0 => a^2 + 2a > 0 => a(a+2) > 0). Combining a < -1/2 and a < -2 or a > 0, we get a < -2.

Multiple choice

The number of real roots of the equation $\displaystyle \left ( x+3 \right )^{4}+\left ( x+5 \right )^{4}=16$, is:
  1. $0$
  2. $2$
  3. $4$

  4. none
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let y = x + 4. Then the equation becomes (y-1)^4 + (y+1)^4 = 16. Expanding gives 2y^4 + 12y^2 + 2 = 16, or y^4 + 6y^2 - 7 = 0. Letting u = y^2, u^2 + 6u - 7 = 0, so (u+7)(u-1) = 0. Since u = y^2 >= 0, u = 1, so y^2 = 1, y = +/- 1. Thus x+4 = 1 or x+4 = -1, giving x = -3 or x = -5.

Multiple choice

If $\alpha_{1},\alpha_{2},\ \alpha_{3},\cdots \ \alpha_{n}$ are the roots of the equation $(x-\beta_{1})(x-\beta_{2}) \cdots \ldots(x-\beta_{n})=A$ and if the equation having $\beta_{1},\ \beta_{2},\ \beta_{3},\ \beta_{n}$ as the roots is $(x-\alpha_{1})(x-\alpha_{2})\ldots\ldots\ldots(x-\alpha_{n})=k$, then $k =$
  1. $A$
  2. $-A$
  3. $\displaystyle \frac{A}{\alpha_{1}\alpha_{2}\ldots\alpha_{n}}$
  4. $\displaystyle -\frac{A}{\alpha_{1}\alpha_{2}\ldots\alpha_{n}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given the polynomial identity (x-beta1)(x-beta2)...(x-betan) = (x-alpha1)(x-alpha2)...(x-alphan) + A, if we swap the roles of alpha and beta, we get (x-alpha1)...(x-alphan) = (x-beta1)...(x-betan) - A. Thus, k = -A.

Multiple choice

lf $pr=2(q+s)$, then among the equations $\mathrm{x}^{2}+ px +\mathrm{q}=0$ and $\mathrm{x}^{2}+ rx +\mathrm{s}=0$

  1. both have real roots

  2. both have imaginary roots

  3. at least one has real roots

  4. at least one has imaginary roots
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Discriminants are D1 = p^2 - 4q and D2 = r^2 - 4s. Sum of discriminants = p^2 + r^2 - 4(q+s). Since pr = 2(q+s), sum = p^2 + r^2 - 2pr = (p-r)^2. Since the sum of discriminants is >= 0, at least one discriminant must be >= 0.

Multiple choice

The equations $x^{2}+3x+5=0$ and $ax^{2}+bx+c=0$ have a common root. lf $a,\ b,\ c\in N$, then the least possible value of $a+b+c$ is equal to
  1. $3$
  2. $6$
  3. $9$
  4. $12$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If x^2 + 3x + 5 = 0 and ax^2 + bx + c = 0 have a common root, the roots of the first are (-3 +/- sqrt(9-20))/2 = (-3 +/- i*sqrt(11))/2. Since the coefficients a, b, c are integers, the other root must be the conjugate. The quadratic is k(x^2 + 3x + 5) = 0. For a, b, c in N, the smallest k=1 gives a=1, b=3, c=5. Sum = 1+3+5 = 9.

Multiple choice

If $\alpha, \beta$ are the root of quadratic equation $ax^2+bx+c=0$, then $\displaystyle \left ( a\alpha +b \right )^{-3}+\left ( a\beta +b \right )^{-3}=$
  1. $\displaystyle \frac{b^{2}-2abc}{a^{3}c^{3}}$
  2. $\displaystyle \frac{b^{3}-2abc}{a^{2}c^{2}}$
  3. $\displaystyle \frac{b^{3}-3abc}{a^{3}c^{3}}$

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

Roots are alpha, beta. a*alpha+b = -c/alpha (from quadratic properties). The expression becomes (-alpha/c)^3 + (-beta/c)^3 = -(alpha^3 + beta^3)/c^3. Using alpha^3 + beta^3 = (alpha+beta)^3 - 3*alpha*beta*(alpha+beta) and Vieta's formulas, this simplifies to the expression in option C.

Multiple choice

The equation $kx^2+x+k=0$ and $kx^2+kx+1=0$ have exactly one root in common for
  1. $\displaystyle k=-\frac { 1 }{ 2 } ,1$
  2. `$k=1$
  3. $\displaystyle k=-\frac { 1 }{ 2 } $
  4. $\displaystyle k=\frac { 1 }{ 2 } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Subtracting the two equations: (k-k)x^2 + (1-k)x + (k-1) = 0 => (1-k)x = 1-k. If k=1, 0=0 (infinitely many solutions). If k != 1, x=1. Substituting x=1 into kx^2+x+k=0 gives 2k+1=0, so k = -1/2.

Multiple choice

All the roots of the equation $11z^{10}+10iz^9+10iz-11=0$ lie
  1. Inside $|z|=1$
  2. On $|z|=1$
  3. Outside $|z|=1$

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

The equation is 11z^10 - 11 + 10iz^9 + 10iz = 0, which is 11(z^10 - 1) + 10iz(z^8 + 1) = 0. By Rouche's theorem or checking modulus, the roots lie on the unit circle |z|=1.

Multiple choice

Roots of the equation are $(z+1)^5 = (z - 1)^5$ are
  1. $\pm i\tan { \left( \displaystyle \frac { \pi }{ 5 } \right) ,\pm{ i\tan { \left( \displaystyle \frac { 2 }{ 5 } \right) } } } $
  2. $\pm i{ \cot { \left(\displaystyle \frac { \pi }{ 5 } \right) ,\pm {i \cot { \left(\displaystyle \frac { 2\pi }{ 5 } \right) } } } }$
  3. $\pm i{ \cot { \left(\displaystyle \frac { \pi }{ 5 } \right) ,\pm {i \tan { \left( \displaystyle \frac { 2\pi }{ 5 } \right) } } } }$

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

Set (z + 1)/(z - 1) equal to the non-real fifth roots of unity. Using the corresponding angle values gives roots ±i cot(pi/5) and ±i cot(2pi/5).