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 $x^4+x^2+1=0$ then the equation whose roots are $\alpha^2 , \beta^2, \gamma^2, \delta^2$ is
  1. $(x^2-x+1)^2=0$
  2. $(x^2+x+1)^2=0$
  3. $x^4-x^2+1=0$
  4. $x^2-x+1=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Roots of x^4 + x^2 + 1 = 0 are roots of (x^2 + 1)^2 - x^2 = 0, which is (x^2 - x + 1)(x^2 + x + 1) = 0. Roots are omega, omega^2, -omega, -omega^2. Squares are omega^2, omega^4=omega, omega^2, omega. The roots are omega, omega, omega^2, omega^2. The equation is (x - omega)^2 (x - omega^2)^2 = (x^2 + x + 1)^2 = 0.

Multiple choice

If $a < 0$, the positive root of the equation $x^2-2a|x-a| -3a^2=0$ is
  1. $a(-1-\sqrt 6)$
  2. $a(1-\sqrt 2)$
  3. $a(-1+\sqrt 6)$
  4. $a(1+\sqrt 2)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

x^2 - 2a|x-a| - 3a^2 = 0. If x >= a, x^2 - 2ax + 2a^2 - 3a^2 = 0, x^2 - 2ax - a^2 = 0. Roots: x = (2a +/- sqrt(4a^2 + 4a^2))/2 = a +/- a*sqrt(2). Since a < 0, a(1+sqrt(2)) is negative and a(1-sqrt(2)) is positive. Checking x < a: x^2 + 2ax - 2a^2 - 3a^2 = 0, x^2 + 2ax - 5a^2 = 0. Roots: x = (-2a +/- sqrt(4a^2 + 20a^2))/2 = -a +/- a*sqrt(6). Both are negative. The positive root is a(1-sqrt(2)).

Multiple choice

Let $ \alpha,\beta,\gamma $ be the roots of the equation ${x}^{3}+{3ax}^{2}+3bx+c=0$. If $ \alpha,\beta,\gamma $are in $H.P$ then $\beta$ is equal to-
  1. $\dfrac{-c}{b}$
  2. $\dfrac{c}{b}$
  3. $-a$
  4. $a$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If roots are in HP, their reciprocals are in AP. Let roots be 1/p, 1/q, 1/r. The equation for reciprocals is cx^3 + 3bx^2 + 3ax + 1 = 0. For roots in AP, the middle term is related to the coefficients. For a cubic x^3 + Ax^2 + Bx + C = 0, if roots are in AP, the middle root is -A/3. Here, the middle root of the reciprocal equation is -(3b/c)/3 = -b/c. Thus, the middle root of the original equation is -c/b.

Multiple choice

Consider the equation $x ^ { 3 } - 3 x ^ { 2 } + 6 x - 1 = 0 ,$ let $\alpha , \beta , \gamma$ are the roots of the equation then The centroid of the triangle, the coordinate of whose vertices are $\left( \alpha ^ { 2 } , \frac { 1 } { a ^ { 2 } } \right) \cdot \left( \beta ^ { 2 } , \frac { 1 } { \beta ^ { 2 } } \right)$ and $\left( \gamma ^ { 2 } , \frac { 1 } { \gamma ^ { 2 } } \right)$ is
  1. $( 0,3 )$
  2. $( - 1,10 )$
  3. $( 2,5 )$
  4. $( - 3,8 )$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice

Roots of the equation $f(x)=x^6-12x^5+bx^4+cx^3+dx^2+ex+64=0$ are positive. Which of the following has the greatest absolute value?
  1. $b$
  2. $c$
  3. $d$
  4. $e$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

By Vieta's formulas for x^6 - 12x^5 + bx^4 + cx^3 + dx^2 + ex + 64 = 0, the product of the roots is 64. Since there are 6 positive roots, their geometric mean is (64)^(1/6) = 2. If all roots are 2, the polynomial is (x-2)^6 = x^6 - 12x^5 + 60x^4 - 160x^3 + 240x^2 - 192x + 64. Comparing coefficients: b=60, c=-160, d=240, e=-192. The absolute values are 60, 160, 240, 192. The greatest is 240 (d).

Multiple choice

If $a$ and $c$ are the lengths of segments of any focal chord of the parabola $y^2 = 2bx (b > 0)$, then the roots of the equation $ax^2 + bx + c = 0$ are

  1. real and distinct

  2. real and equal

  3. imaginary

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

For a focal chord of y^2=2bx, the segments a and c satisfy the property that the semi-latus rectum is the harmonic mean of the segments, so 2/b = 1/a + 1/c = (a+c)/ac, meaning b = 2ac/(a+c). The discriminant of ax^2+bx+c is D = b^2 - 4ac = (2ac/(a+c))^2 - 4ac = 4a^2c^2/(a+c)^2 - 4ac = 4ac(ac/(a+c)^2 - 1). Since (a+c)^2 > 4ac, ac/(a+c)^2 < 1/4, so D < 0, meaning roots are imaginary.