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 the quadratic equations $x^2+abx+c=0$ and $x^2+acx+b=0$ have a common root, the equation containing their other roots is/are:
  1. $x^2+a(b+c)x-a^2 bc=0$
  2. $x^2-a (b+c)x+a^2 bc=0$
  3. $a(b+c) x^2-(b+c)x+abc=0$
  4. $a(b+c) x^2+(b+c)x-abc=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let alpha be the common root. alpha^2+ab*alpha+c=0 and alpha^2+ac*alpha+b=0. Subtracting: a(b-c)*alpha + (c-b) = 0. a(b-c)*alpha = b-c. If b!=c, alpha = 1/a. Substituting alpha=1/a into the first equation: 1/a^2 + b + c = 0, so 1+a^2(b+c)=0. The product of roots for eq1 is c, so the other root is c/alpha = ac. For eq2, product is b, other root is b/alpha = ab. The equation with roots ac and ab is x^2 - (ac+ab)x + a^2bc = 0, or x^2 - a(b+c)x + a^2bc = 0.

Multiple choice

If the equations $x^2+4x+5=0$ and $ax^2+bx+c=0$ have a common root (where $a,b,c \in N$), then the least value of $a+b+c$ is equal to:
  1. $10$
  2. $15$
  3. $20$
  4. $25$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

x^2+4x+5=0 has roots -2 +/- i. If ax^2+bx+c=0 has a common root, it must have both roots (since coefficients are real). So ax^2+bx+c = k(x^2+4x+5). For a,b,c to be in N, k must be a positive integer. Min value is k=1, giving a=1, b=4, c=5. Sum = 1+4+5 = 10.

Multiple choice

For $a,\ b,\ c\ \in\ R$ and $a\ \neq 0$, the equation $ax^{2}+bx+c=0$ and $x^{2}+2x+3=0$ have atlest one common root if $\dfrac{a}{\lambda}=\dfrac{b}{\mu}=\dfrac{c}{\Psi }$ are positive integers. The least value of $(\lambda+\mu+\Psi )$ is
  1. $6$
  2. $5$
  3. $7$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For common roots, the ratio of coefficients must be equal. x^2+2x+3=0 has roots -1 +/- i*sqrt(2). For ax^2+bx+c=0 to share a root, a/1 = b/2 = c/3 = k. Since a,b,c are positive integers, the smallest k=1 gives a=1, b=2, c=3. Sum = 1+2+3 = 6.

Multiple choice

If $a, b, c$ are in AP and if quadratic equations $(b - c)x^2 + (c - a)x + (a - b) = 0, 2(c + a)x = 0$ have a common root, then
  1. $a^2, b^2, c^2$ are in AP
  2. $a^2, c^2, b^2$ are in AP
  3. $a^2, c^2, b^2$ are in GP

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

If a, b, c are in AP, then 2b = a + c. The common root condition leads to the relation that the coefficients satisfy a specific property, which simplifies to showing that a^2, b^2, c^2 are in AP when specific conditions are met.

Multiple choice

The equation ${ ax }^{ 2 }+bx+c=0,{ bx }^{ 2 }+cx+a=0$ have a common root then $\dfrac { { a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 } }{ abc } $

  1. 1

  2. 3

  3. 2

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

If the quadratic equations ax^2 + bx + c = 0 and bx^2 + cx + a = 0 share a common root, then either a + b + c = 0 or a = b = c. In both mathematical scenarios, the algebraic identity a^3 + b^3 + c^3 is equal to 3abc. Dividing both sides by abc yields a value of 3.

Multiple choice

If the equation $\displaystyle k \left ( 6x^{2} + 3 \right ) + rx + \left ( 2x^{2} - 1 \right ) = 0$ and $\displaystyle 6k\left ( 2x^{2} + 1 \right ) + px + 4x^{2} - 2 = 0$ have symmetrical roots, then $\displaystyle 2r - p$ equals
  1. $\displaystyle 1$
  2. $\displaystyle 2$
  3. $\displaystyle 3$
  4. $\displaystyle 0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Symmetrical roots imply the equations are proportional. Comparing coefficients of the two equations: k(6x^2 + 3) + rx + (2x^2 - 1) = 0 and 6k(2x^2 + 1) + px + 4x^2 - 2 = 0. Simplifying and equating ratios leads to 2r - p = 0.

Multiple choice

I. lf ${f}({x})$ be a quadratic expression which is positive for all real ${x}$ and ${g}({x})={f}({x})+{f'}({x})+{f''}({x})$ for any real ${x}$, then ${g}({x})>0$. II. lf the equations ${p}{x}^{2}-7{x}+3{p}=0$ and $2{x}^{2}+{q}{x}+6=0$ have the same roots, then ${pq}=-14$.Which of the above statement(s) is/are true?

  1. Only I

  2. Only II

  3. Both I and II

  4. Neither I nor II
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

I: If f(x) is a quadratic with positive values, its discriminant is negative. g(x) is also a quadratic with the same leading coefficient, and its discriminant can be shown to be negative, so g(x) > 0. II: If roots are the same, coefficients are proportional. 2/p = q/-7 = 6/3p. Solving gives p=1, q=-14, so pq=-14.

Multiple choice

If the equation $x^2+ax+bc=0$ and $x^2+bx+ca=0$ have one common root, then their remaining roots are-
  1. $1,b$
  2. $b,a$
  3. $b,c$
  4. $c,a$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let alpha be the common root. alpha^2 + a*alpha + bc = 0 and alpha^2 + b*alpha + ca = 0. Subtracting: (a-b)*alpha + c(b-a) = 0 => (a-b)*alpha = c(a-b). If a != b, alpha = c. Substituting c into the first: c^2 + ac + bc = 0 => c(c+a+b) = 0. The roots of the first are c and b. The roots of the second are c and a.