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

$\alpha,\beta $ are the roots of the equation $x^{2} -7x +2 =0, r $ is the arithmetic mean of $\alpha,\beta $ The locus of the point $P$ which is equidistant from the points $(\alpha,\beta)$ and $(\alpha + r,\beta + r)$ is $2x + 2y-21 =0.$

  1. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

  2. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

  3. Assertion is correct but Reason is incorrect

  4. Both Assertion and Reason are incorrect
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice

The equation whose roots are $2,3-5$ is ?
  1. $x^3-19x+30=0$
  2. $x^3-19x^2+16x-2=0$
  3. $x^3-4x^2-8x+8=0$
  4. $x^3-2x^2+x-2=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The roots are 2, 3, -5. The equation is (x-2)(x-3)(x+5) = (x^2-5x+6)(x+5) = x^3 + 5x^2 - 5x^2 - 25x + 6x + 30 = x^3 - 19x + 30 = 0.

Multiple choice

Find a if equations $x^3+ax+1=0$ and $x^4+ax^2+1=0$ have a common root

  1. 0

  2. 1

  3. -2

  4. 2
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If x is a common root, x^3 + ax + 1 = 0 and x^4 + ax^2 + 1 = 0. Multiplying the first by x gives x^4 + ax^2 + x = 0. Subtracting this from the second equation gives 1 - x = 0, so x = 1. Substituting x = 1 into the first equation: 1 + a + 1 = 0, so a = -2.

Multiple choice

Solve the cubic equation : $x^3+x^2-16x=16$
  1. $x=-1, \pm 4$
  2. $x=2, \pm 3$
  3. $x=1,\pm 2$
  4. $x=2, \pm 1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

x^3 + x^2 - 16x - 16 = 0. x^2(x+1) - 16(x+1) = 0. (x^2-16)(x+1) = 0. (x-4)(x+4)(x+1) = 0. Roots are -1, 4, -4.

Multiple choice

If $2$ and $3$ are the roots of the equation $2 x ^ { 3 } + m x ^ { 2 } - 13 x + n = 0 ,$ the third root of this equation is
  1. $-5$
  2. $30$
  3. $- \dfrac { 5 } { 2 }$
  4. $- 15$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the roots be 2, 3, and r. By Vieta's formulas, the sum of roots 2 + 3 + r = -m/2, and the product of roots 2 * 3 * r = -n/2. Also, the sum of roots taken two at a time is 2*3 + 2*r + 3*r = -13/2. So, 6 + 5r = -6.5, which means 5r = -12.5, so r = -2.5.

Multiple choice

Find the sum of the squares of the roots of the cubic equation $x^3+3x^2+3x=3$.

  1. 1

  2. 2

  3. 3

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

The equation is x^3 + 3x^2 + 3x - 3 = 0. This is (x+1)^3 - 4 = 0. Let y = x+1, then y^3 = 4. The roots are y1, y2, y3 where y = cube_root(4) * omega^k. x = y - 1. Sum of squares of roots = sum(yi - 1)^2 = sum(yi^2 - 2yi + 1) = sum(yi^2) - 2*sum(yi) + 3. For y^3 - 4 = 0, sum(yi) = 0 and sum(yi*yj) = 0. sum(yi^2) = (sum yi)^2 - 2*sum(yi*yj) = 0. Thus, sum(xi^2) = 0 - 0 + 3 = 3.

Multiple choice

If $(x+ 1) (x + 2) (x + 3) (x + 4) =24$, then which of the following is true?

  1. Zero is one of the roots of the equation

  2. The given equation has no roots
  3. The given equation has exactly $24$ roots

  4. The given equation is an identity
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Rearrange the expression as ((x+1)(x+4)) * ((x+2)(x+3)) = 24, which is (x^2 + 5x + 4)(x^2 + 5x + 6) = 24. Let u = x^2 + 5x, then (u+4)(u+6) = 24, so u^2 + 10u + 24 = 24, leading to u(u+10) = 0. If u = 0, x^2 + 5x = 0, so x(x+5) = 0, meaning 0 is a root.

Multiple choice

The cubic equation whose roots are each 2 more than the roots of the equation ${x^3} + 4{x^2} - 3x + 12 = 0$ is
  1. ${x^3} + 2{x^2} - 31x - 2 = 0$
  2. ${x^3} + 6{x^2} - x + 14 = 0$
  3. ${x^3} - 2{x^2} - 7x + 26 = 0$
  4. ${x^3} + 2{x^2} - 5x + 10 = 0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If the roots of the new equation are x = y + 2, then y = x - 2. Substituting (x - 2) for x in the original equation (x^3 + 4x^2 - 3x + 12 = 0) gives (x - 2)^3 + 4(x - 2)^2 - 3(x - 2) + 12 = 0. Expanding this yields x^3 - 6x^2 + 12x - 8 + 4(x^2 - 4x + 4) - 3x + 6 + 12 = x^3 - 2x^2 - 7x + 26 = 0.

Multiple choice

Let a, b, c be three roots of the equation $x^3+x^2-333x-1002=0$, then $(\displaystyle\sum (a^3)-2\displaystyle\sum a)$ equals to?
  1. $2008$
  2. $2000$
  3. $2006$
  4. $2002$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For x^3 + x^2 - 333x - 1002 = 0, the sum of roots a+b+c = -1. Using Newton's sums or the fact that the roots satisfy the equation, a^3 = -a^2 + 333a + 1002. Summing for a, b, c gives sum(a^3) = -sum(a^2) + 333*sum(a) + 3006. Since sum(a^2) = (sum a)^2 - 2*sum(ab) = (-1)^2 - 2(-333) = 667, we find sum(a^3) = -667 + 333(-1) + 3006 = 2006. Then 2006 - 2(-1) = 2008.