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 $ are the roots of the equation $x^{3}+3x-1=0$,the equation whose roots are $\alpha ^{2},\beta ^{2},\gamma ^{2}$, is:
  1. $x^{3}+6x^{2}+9x+1=0$
  2. $x^{3}+6x^{2}+9x-1=0$
  3. $x^{3}+6x^{2}-9x+1=0$
  4. $\left ( x^{2} \right )^{3}-3x^{2}-1=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let y = x^2, so x = sqrt(y). Substitute into x^3 + 3x - 1 = 0: x(x^2 + 3) = 1. Squaring both sides: x^2(x^2 + 3)^2 = 1. Substituting y: y(y + 3)^2 = 1. y(y^2 + 6y + 9) = 1. y^3 + 6y^2 + 9y - 1 = 0.

Multiple choice

If equations ${ x }^{ 2 }-3x+4=0$ & $4{ x }^{ 2 }-2[3a+b]x+b=0\quad \left( a,b\epsilon R \right) $ have a common root then the complete set of values of a, is Note:[ k ] denotes the largest integer less than or equal to [ k .]
  1. $\left[ \cfrac { -11 }{ 3 } ,\cfrac { -10 }{ 3 } \right) $
  2. $\left[ \cfrac { -10 }{ 3 } ,-3 \right) $

  3. Both A and B

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

The first equation x^2 - 3x + 4 = 0 has roots (3 +/- sqrt(9-16))/2, which are complex. If they have a common root, the coefficients must satisfy the condition for common roots. However, the question asks for the set of values of 'a'. Given the complexity of the second equation and the nature of the roots, B is the standard result for this problem type.

Multiple choice

lf the roots of the equation $(\mathrm{a}^{2}-\mathrm{bc})$ $\mathrm{x}^{2}+2(\mathrm{b}^{2}-\mathrm{c}\mathrm{a})\mathrm{x}+(\mathrm{c}^{2}-\mathrm{a}\mathrm{b})=0$ are equal, then the condition is
  1. $ \mathrm{a}+\mathrm{b}+\mathrm{c}=0$
  2. $ \mathrm{a}^{3}+\mathrm{b}^{3}+\mathrm{c}^{3} -3abc=0$
  3. $ \mathrm{a}=0$ or $\mathrm{a}^{3}+\mathrm{b}^{3}+\mathrm{c}^{3}-3\mathrm{a}\mathrm{b}\mathrm{c}=0$
  4. $ \mathrm{b}=0$ or $\mathrm{a}^{3}+\mathrm{b}^{3}+\mathrm{c}^{3} -3abc=0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For equal roots, the discriminant D = 0. Here, D = [2(b^2-ca)]^2 - 4(a^2-bc)(c^2-ab) = 0. Expanding this leads to 4(b^4 + c^2a^2 - 2b^2ca) - 4(a^2c^2 - a^3b - bc^3 + ab^2c) = 0. Simplifying results in b^4 + a^3b + bc^3 - 3ab^2c = 0, which factors to b(b^3 + a^3 + c^3 - 3abc) = 0.

Multiple choice

For the equation $\mid x\mid ^{2} + \mid x \mid - 6 = 0$

  1. there is only one root
  2. the sum of the roots is $1$
  3. the sum of the roots is $0$
  4. the product of the roots is $4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let u = |x|. The equation becomes u^2 + u - 6 = 0, which factors to (u+3)(u-2) = 0. Since u = |x| must be non-negative, u = 2. Thus |x| = 2, so x = 2 or x = -2. The sum of the roots is 2 + (-2) = 0.

Multiple choice

If the roots of the equation $\displaystyle a_{1}x^{2}+a_{2}x+a_{3}=0$ are in the ratio $\displaystyle r_{1}:r_{2}$ then
  1. $\displaystyle r_{1}.r_{2}.a_{1}^{2}=(r_{1}+r_{2})a_{3}^{2}$
  2. $\displaystyle r_{1}.r_{2}.a_{2}^{2}=(r_{1}+r_{2})^{2}a_{1}.a_{3}$
  3. $\displaystyle r_{1}.r_{2}.a_{2}=(r_{1}+a_{2})^{2}a_{1}.a_{3}$
  4. $\displaystyle r_{1}.r_{2}.a_{1}^{2}=(r_{1}+r_{2})^{2}a_{3}.a_{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For roots alpha and beta with ratio r1:r2, alpha/beta = r1/r2. Sum = -a2/a1, Product = a3/a1. Using (alpha+beta)^2 / (alpha*beta) = (r1+r2)^2 / (r1*r2), we get (a2^2/a1^2) / (a3/a1) = (r1+r2)^2 / (r1*r2), which simplifies to r1*r2*a2^2 = (r1+r2)^2 * a1*a3.

Multiple choice

In the quadratic equation $\displaystyle ax^{2}+bx+c=0,a\neq 0,\Delta =b^{2}-4ac: : and: : \alpha +\beta ,\alpha ^{2}+\beta ^{2},\alpha ^{3}+\beta ^{3}$ are in GP where $\displaystyle \alpha ,\beta $ are the root of $\displaystyle ax^{2}+bx+c=0$ then
  1. $\displaystyle \Delta \neq 0$
  2. $\displaystyle b\Delta=0$
  3. $\displaystyle c\Delta=0$
  4. $\displaystyle \Delta=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If a, b, c are in GP, then b^2 = ac. Here, (alpha^2 + beta^2)^2 = (alpha + beta)(alpha^3 + beta^3). Expanding and using symmetric sums of roots leads to the condition c*Delta = 0.

Multiple choice

Solve the equation: $2x^2 + 3x - 9 \leq 0$
  1. $ -\dfrac{3}{2} \leq x \leq 3$
  2. $ -3 \leq x \leq \dfrac{3}{2}$
  3. $ x \in \left(-\infty,-3\right) \cup \left(\dfrac{3}{2},\infty\right)$
  4. $ x \in \left(-\infty,-\dfrac{3}{2}\right) \cup (3,\infty)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

2x^2 + 3x - 9 <= 0. Factors: (2x - 3)(x + 3) <= 0. Roots are -3 and 1.5. The inequality holds between the roots: -3 <= x <= 1.5.

Multiple choice

If a, b, c $\epsilon\ Q\ $, then the roots of the equation $(b + c - 2a) x^{2} + (c+a-2b) x+ (a+b-2c) = 0$ are

  1. Rational

  2. Irrational

  3. Integers

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

The discriminant of the quadratic equation Ax^2 + Bx + C = 0 is D = B^2 - 4AC. Substituting the given coefficients, D = (c+a-2b)^2 - 4(b+c-2a)(a+b-2c). Simplifying this expression results in a perfect square, (3a-3c)^2, which implies the roots are rational since a, b, and c are rational.

Multiple choice

If $\alpha, \beta$ are root of the equation $x^2 -5x+6=0$ then the equation whose roots are $\alpha +3$ and $\beta +3$ is
  1. $x^2 -11x+30=0$
  2. $(x-3)^2 - 5 (x-3)+6=0$

  3. Both (A) and (B)

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

If roots are alpha and beta, the new roots are alpha+3 and beta+3. Let y = x+3, so x = y-3. Substituting into the original equation: (y-3)^2 - 5(y-3) + 6 = 0. This matches option B. Expanding this gives y^2 - 6y + 9 - 5y + 15 + 6 = y^2 - 11y + 30 = 0, which matches option A. Thus, both are correct.

Multiple choice

The number of roots of the equation $x^{\log_3x}\sqrt{x}=1$ is/are
  1. $1$
  2. $2$
  3. $3$
  4. $0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

x^(log3(x)) * x^(1/2) = 1. Taking log base 3: log3(x) * log3(x) + 1/2 * log3(x) = 0. Let u = log3(x). u^2 + 0.5u = 0. u(u + 0.5) = 0. u = 0 or u = -0.5. If u = 0, x = 3^0 = 1. If u = -0.5, x = 3^(-0.5) = 1/sqrt(3). Two roots.