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 roots of equation $2 x ^ { 4 } - 3 x ^ { 3 } + 2 x ^ { 2 } - 7 x - 1 = 0$ are $\alpha , \beta , \gamma \text { and } \delta$ then value of $\displaystyle \sum \dfrac { \alpha + 1 } { \alpha }$ is equal to ?

  1. -3

  2. 3
  3. $\frac { 11 } { 2 }$

  4. -11
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the equation 2x^4 - 3x^3 + 2x^2 - 7x - 1 = 0, the sum of (alpha + 1)/alpha is sum(1 + 1/alpha) = 4 + sum(1/alpha). The sum of reciprocals of roots 1/alpha + 1/beta + 1/gamma + 1/delta is - (coefficient of x) / (constant term) = -(-7) / (-1) = -7. Thus, 4 + (-7) = -3.

Multiple choice

The number of values of $a$ for which equation $x^3+ax+1=0$ and $x^4+ax^2+1=0$ have a common root, is:
  1. $0$
  2. $1$
  3. $2$
  4. $\infty$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If x is a common root, x^3 + ax + 1 = 0 and x^4 + ax^2 + 1 = 0. Multiply the first by x: x^4 + ax^2 + x = 0. Subtracting the second equation: (x^4 + ax^2 + x) - (x^4 + ax^2 + 1) = 0 => x - 1 = 0, so x = 1. Substituting x = 1 into the first equation: 1 + a + 1 = 0, so a = -2. There is only one value for a.

Multiple choice

Let $p(x)=0$ be a polynomial equation of the least possible degree, with rational coefficients, having $\sqrt [3]{7}+\sqrt [3]{49}$ as one of its roots, then the product of all the roots of $p(x)=0$, is:
  1. $56$
  2. $63$
  3. $7$
  4. $49$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let x = 7^(1/3) + 49^(1/3). Cubing both sides: x^3 = 7 + 49 + 3 * 7^(1/3) * 49^(1/3) * (7^(1/3) + 49^(1/3)) = 56 + 3 * 7 * x. So x^3 - 21x - 56 = 0. The product of the roots of a cubic ax^3 + bx^2 + cx + d = 0 is -d/a. Here, -(-56)/1 = 56.

Multiple choice

If the equation $\displaystyle 2x^{3}+3x^{2}-12x+d=0$ has 3 real and distinct roots, then $d$ belongs to
  1. $\displaystyle (-\infty ,-7)\cup (20,\infty )$
  2. $(-20, 7 )$
  3. $\displaystyle (-\infty ,-20)\cup (7,\infty )$
  4. $(20, 7)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let f(x) = 2x^3 + 3x^2 - 12x + d. For 3 distinct real roots, the local maximum and minimum must have opposite signs. f'(x) = 6x^2 + 6x - 12 = 6(x^2 + x - 2) = 6(x+2)(x-1). Critical points at x = -2 and x = 1. f(-2) = 2(-8) + 3(4) - 12(-2) + d = -16 + 12 + 24 + d = 20 + d. f(1) = 2 + 3 - 12 + d = d - 7. For opposite signs, (20+d)(d-7) < 0, so -20 < d < 7.

Multiple choice

If $\alpha =\log _{ 3 }{ 4 }$ is one of the roots of equation ${ x }^{ 3 }+a{ x }^{ 2 }+bx+1=0$, $a,b\in R$, then equation which always have a root $\beta =\log _{ 2 }{ 48 }$, is
  1. ${ x }^{ 3 }+2\left( b-6 \right) { x }^{ 2 }+4\left( a-4b+12 \right) x+8\left( 4b-2a-7 \right) =0$
  2. ${ x }^{ 3 }+b{ x }^{ 2 }+ax+1=0$
  3. $\left( 65+16a+4b \right) { x }^{ 2 }+2\left( 48+8a+b \right) { x }^{ 2 }+4\left( 12+a \right) x+8=0$
  4. $\left( 65+16a+4b \right) { x }^{ 2 }+\left( 48+8a+b \right) { x }^{ 2 }+\left( 12+a \right) x+8=0 $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let t = log2(3). Then alpha = 2/t and beta = 4 + t, so alpha(beta - 4) = 2. Substituting alpha = 2/(beta - 4) into the given cubic produces the equation in option A.

Multiple choice

If $2m^{2}n +m(2n^{2}+1)+n =135$ and $m+n=15$, then the quadratic equation whose roots are m and n can be in the form of k (where k is any constant)
  1. $y^{2}-15y+135=0$
  2. $y^{2}-15y+4=0$
  3. $y^{2}-9y+15=0$
  4. $y^{2}-15y-4=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given m+n=15 and 2m^2n + m(2n^2+1) + n = 135. Expand: 2mn(m+n) + m + n = 135. Substitute m+n=15: 2mn(15) + 15 = 135. 30mn = 120, so mn = 4. The quadratic equation with roots m and n is y^2 - (m+n)y + mn = 0, which is y^2 - 15y + 4 = 0.

Multiple choice

Let $b,a, c,p,q $ be real numbers in ascending order. Suppose $\alpha ,\beta $ are the roots of the equation $x^2+2px+q=0$ and $\alpha ,\cfrac{1}{\beta }$ are the roots of the equation $ax^2+2bx+c=0$, where $\beta ^2\notin {-1,0,1}$ $(p^2 -q) (b^2 - ac) \geq 0$

  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. Assertion is incorrect but Reason is correct
Reveal answer Fill a bubble to check yourself
A Correct answer